Subdomain Test

Let $S$ be a subset of an integral domain $\struct {R, +, \circ}$. Then $\struct {S, +\restriction_S, \circ \restriction_S}$ is a subdomain of $\struct {R, +, \circ}$ {{iff}} these conditions hold: :$(1): \quad$ $\struct {S, + \restriction_S, \circ \restriction_S}$ is a subring of $\struct {R, +, \circ}$ :$(2): \quad$ The unity of $R$ is also in $S$, that is $1_R = 1_S$.


Centralizer of Ring Subset is Subring

Let $S$ be a subset of a ring $\struct {R, +, \circ}$ Then $\map {C_R} S$, the centralizer of $S$ in $R$, is a subring of $R$.


Center of Ring is Commutative Subring

The center $\map Z R$ of a ring $R$ is a commutative subring of $R$.


Ideal is Subring

Let $\left({R, +, \circ}\right)$ be a ring, and let $J$ be an ideal of $R$. Then $J$ is a subring of $R$.


Ring is Ideal of Itself

Let $\struct {R, +, \circ}$ be a ring. Then $R$ is an ideal of $R$.


Ideal of Unit is Whole Ring

Let $\struct {R, +, \circ}$ be a ring with unity. Let $J$ be an ideal of $R$. If $J$ contains a unit of $R$, then $J = R$.


Test for Ideal

Let $J$ be a subset of a ring $\struct {R, +, \circ}$. Then $J$ is an ideal of $\struct {R, +, \circ}$ {{iff}} these all hold: : $(1): \quad J \ne \O$ : $(2): \quad \forall x, y \in J: x + \paren {-y} \in J$ : $(3): \quad \forall j \in J, r \in R: r \circ j \in J, j \circ r \in J$


Epimorphism Preserves Rings

Let $\struct {R_1, +_1, \circ_1}$ be a ring, and $\struct {R_2, +_2, \circ_2}$ be a closed algebraic structure. Let $\phi: R_1 \to R_2$ be an epimorphism. Then $\struct {R_2, +_2, \circ_2}$ is a ring.


Ring Homomorphism of Addition is Group Homomorphism

Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Then $\phi: \left({R_1, +_1}\right) \to \left({R_2, +_2}\right)$ is a group homomorphism.


Element of Integral Domain Divides Zero

Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$. Then every element of $D$ is a divisor of zero: :$\forall x \in D: x \divides 0_D$


Unity Divides All Elements

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$. Then unity is a divisor of every element of $D$: :$\forall x \in D: 1_D \divides x$ Also: :$\forall x \in D: -1_D \divides x$


Element of Integral Domain is Divisor of Itself

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$. Then every element of $D$ is a divisor of itself: :$\forall x \in D: x \divides x$


Unit of Integral Domain divides all Elements

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$. Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$. Then: :$\forall x \in D: \forall u \in U_D: u \divides x$ That is, every unit of $D$ is a divisor of every element of $D$.


Divisor Relation in Integral Domain is Transitive

Let $\struct {D, +, \circ}$ be an integral domain. Let $x, y, z \in D$. Then: :$x \divides y \land y \divides z \implies x \divides z$


Integers form Unique Factorization Domain

The integers $\struct {\Z, +, \times}$ form a unique factorization domain.


Trivial Ordering Compatibility in Boolean Ring

Let $\struct {S, +, \circ}$ be a Boolean ring. Then the trivial ordering is the only ordering on $S$ compatible with both its operations.


Field is Integral Domain

Every field is an integral domain.


Finite Integral Domain is Galois Field

A finite integral domain is a Galois field.


Center of Division Ring is Subfield

Let $\struct {K, +, \circ}$ be an division ring. Then $\map Z K$, the center of $K$, is a subfield of $K$.


Ideals of Field

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$. Then $\struct {R, +, \circ}$ is a field {{iff}} the only ideals of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$.


Epimorphism from Division Ring to Ring

Let $\left({K, +, \circ}\right)$ be a division ring whose zero is $0_K$. Let $\left({R, +, \circ}\right)$ be a ring whose zero is $0_R$. Let $\phi: K \to R$ be a ring epimorphism. Then one of the following applies: :$(1): \quad R$ is a null ring :$(2): \quad R$ is a division ring and $\phi$ is a ring isomorphism.


Congruence Class Modulo Subgroup is Coset

Let $G$ be a group. Let $H$ be a subgroup of $G$.


Cosets are Equivalent

All left cosets of a group $G$ with respect to a subgroup $H$ are equivalent. That is, any two left cosets are in one-to-one correspondence. The same applies to right cosets. As a special case of this: :$\forall x \in G: \order {x H} = \order H = \order {H x}$ where $H$ is a subgroup of $G$.


Left and Right Coset Spaces are Equivalent

Let $\struct {G, \circ}$ be a group. Let $H$ be a subgroup of $G$. Let: :$x H$ denote the left coset of $H$ by $x$ :$H y$ denote the right coset of $H$ by $y$. Then: :$\order {\set {x H: x \in G} } = \order {\set {H y: y \in G} }$


Cosets in Abelian Group

Let $G$ be an abelian group. Then every right coset modulo $H$ is a left coset modulo $H$. That is: :$\forall x \in G: x H = H x$ In an abelian group, therefore, we can talk about '''congruence modulo $H$''' and not worry about whether it is left or right.


Conjugate of Set by Identity

Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $S \subseteq G$. Then the conjugate of $S$ by $e$ is $S$: :$S^e = S$


Conjugate of Set by Group Product

Let $\struct {G, \circ}$ be a group. Let $S \subseteq G$. Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as: :$S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$ Then: :$\paren {S^a}^b = S^{b \circ a}$


Conjugate of Subgroup is Subgroup

Let $G$ be a group. Let $H \le G$ be a subgroup of $G$. Then the conjugate of $H$ by $a$ is a subgroup of $G$: :$\forall H \le G, a \in G: H^a \le G$


Inner Automorphisms form Normal Subgroup of Automorphism Group

Let $G$ be a group. Then the set $\Inn G$ of all inner automorphisms of $G$ is a normal subgroup of the automorphism group $\Aut G$ of $G$: :$\Inn G \lhd \Aut G$


Subgroup of Abelian Group is Normal

Every subgroup of an abelian group is normal.


Intersection with Normal Subgroup is Normal

Let $G$ be a group. Let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$. Then $H \cap N$ is a normal subgroup of $H$.


Kernel is Normal Subgroup of Domain

Let $\phi$ be a group homomorphism. Then the kernel of $\phi$ is a normal subgroup of the domain of $\phi$: :$\map \ker \phi \lhd \Dom \phi$


Coset Product is Well-Defined

Let $\struct {G, \circ}$ be a group. Let $N$ be a normal subgroup of $G$. Let $a, b \in G$. Then the coset product: :$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$ is well-defined.


Quotient Group of Abelian Group is Abelian

Let $G$ be an abelian group. Let $N \le G$. Then the quotient group $G / N$ is abelian.


Quotient Theorem for Group Epimorphisms

Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups. Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism. Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively. Let $K = \map \ker \phi$ be the kernel of $\phi$. There is one and only one group isomorphism $\psi: G / K \to H$ satisfying: :$\psi \circ q_K = \phi$ where $q_K$ is the quotient epimorphism from $G$ to $G / K$.


Congruence Relation induces Normal Subgroup

Let $\left({G, \circ}\right)$ be a group whose identity is $e$. Let $\mathcal R$ be a congruence relation for $\circ$. Let $H = \left[\!\left[{e}\right]\!\right]_\mathcal R$, where $\left[\!\left[{e}\right]\!\right]_\mathcal R$ is the equivalence class of $e$ under $\mathcal R$. Then: : $(1): \quad \left({H, \circ \restriction_H}\right)$ is a normal subgroup of $G$ : $(2): \quad \mathcal R$ is the equivalence relation $\mathcal R_H$ defined by $H$ : $(3): \quad \left({G / \mathcal R, \circ_\mathcal R}\right)$ is the subgroup $\left({G / H, \circ_H}\right)$ of the semigroup $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.


Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal

Let $G$ be a group. Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$. Let $K \lhd G/H$ and $L = q_H^{-1} \left[{K}\right]$, where: :$q_H: G \to G/H$ is the quotient epimorphism from $G$ to the quotient group $G/H$ :$q_H^{-1} \left[{K}\right]$ is the preimage of $K$ under $q_H$. Then: :$L \lhd G$


Trivial Quotient Group is Quotient Group

Let $G$ be a group. Then the trivial quotient group: :$G / \set {e_G} \cong G$ where: :$\cong$ denotes group isomorphism :$e_G$ denotes the identity element of $G$ is a quotient group.


Correspondence Theorem (Group Theory)

Let $G$ be a group. Let $N \lhd G$ be a normal subgroup of $G$. Then every subgroup of the quotient group $G / N$ is of the form $H / N = \set {h N: h \in H}$, where $N \le H \le G$. Conversely, if $N \le H \le G$ then $H / N \le G / N$. The correspondence between subgroups of $G / N$ and subgroups of $G$ containing $N$ is a bijection. This bijection maps normal subgroups of $G / N$ onto normal subgroups of $G$ which contain $N$.


Centralizer of Group Element is Subgroup

Let $\struct {G, \circ}$ be a group and let $a \in G$. Then $\map {C_G} a$, the centralizer of $a$ in $G$, is a subgroup of $G$.


Centralizer in Subgroup is Intersection

Let $G$ be a group. Let $H$ be a subgroup of $G$. Then: :$\forall x \in G: \map {C_H} x = \map {C_G} x \cap H$ That is, the centralizer of an element in a subgroup is the intersection of that subgroup with the centralizer of the element in the group.


Kernel of Inner Automorphism Group is Center

Let the mapping $\kappa: G \to \Inn G$ from a group $G$ to its inner automorphism group $\Inn G$ be defined as: :$\map \kappa a = \kappa_a$ where $\kappa_a$ is the inner automorphism of $G$ given by $a$. Then $\kappa$ is a group epimorphism, and its kernel is the center of $G$: :$\map \ker \kappa = \map Z G$


Group equals Center iff Abelian

Let $G$ be a group. Then $G$ is abelian {{iff}} $\map Z G = G$, that is, {{iff}} $G$ equals its center.


Normalizer is Subgroup

Let $G$ be a group. The normalizer of a subset $S \subseteq G$ is a subgroup of $G$. :$S \subseteq G \implies \map {N_G} S \le G$


Subgroup is Normal Subgroup of Normalizer

Let $G$ be a group. A subgroup $H \le G$ is a normal subgroup of its normalizer: :$H \le G \implies H \lhd \map {N_G} H$


Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup

Let $G$ be a group. Let $H$ be a subgroup of $G$. Then $\map {N_G} H$, the normalizer of $H$ in $G$, is the largest subgroup of $G$ containing $H$ as a normal subgroup.


Normal Subgroup iff Normalizer is Group

Let $G$ be a group. Let $H$ be a subgroup of $G$. Then $H$ is normal in $G$ {{iff}} the normalizer of $H$ is equal to $G$: :$H \lhd G \iff \map {N_G} H = G$


Normalizer of Conjugate is Conjugate of Normalizer

The normalizer of a conjugate is the conjugate of the normalizer: :$S \subseteq G \implies \map {N_G} {S^a} = \paren {\map {N_G} S}^a$


Quotient Ring is Ring/Quotient Ring Product is Well-Defined

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring of $R$ by $J$. Then $\circ$ is well-defined on $R / J$, that is: :$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies x_1 \circ y_1 + J = x_2 \circ y_2 + J$


Quotient Ring of Commutative Ring is Commutative

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$. If $\struct {R, +, \circ}$ is a commutative ring, then so is $\struct {R / J, +, \circ}$.


Quotient Ring of Ring with Unity is Ring with Unity

Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$. Then $\struct {R / J, +, \circ}$ is a ring with unity, and its unity is $1_R + J$.


Ring Epimorphism with Trivial Kernel is Isomorphism

Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring epimorphism. Then $\phi$ is a ring isomorphism iff $\ker \left({\phi}\right) = \left\{ {0_{R_1} }\right\}$.


Quotient Epimorphism is Epimorphism/Ring

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$ and whose unity is $1_R$. Let $J$ be an ideal of $R$. Let $\struct {R / J, +, \circ}$ be the quotient ring defined by $J$. Let $\phi: R \to R / J$ be the quotient (ring) epimorphism from $R$ to $R / J$: :$x \in R: \map \phi x = x + J$ Then $\phi$ is a ring epimorphism whose kernel is $J$.


Ring Homomorphism Preserves Subrings

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism. If $S$ is a subring of $R_1$, then $\phi \sqbrk S$ is a subring of $R_2$.


Kernel of Ring Homomorphism is Subring

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism. Then the kernel of $\phi$ is a subring of $R_1$.


Kernel of Ring Homomorphism is Ideal

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring homomorphism. Then the kernel of $\phi$ is an ideal of $R_1$.


Ideals of Division Ring

Let $\struct {R, +, \circ}$ be a division ring whose zero is $0_R$. The only ideals of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself. That is, $\struct {R, +, \circ}$ has no non-null proper ideals.


Quotient Ring of Kernel of Ring Epimorphism

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $K = \map \ker \phi$. Then there is a unique ring isomorphism $g: R_1 / K \to R_2$ such that: :$g \circ q_K = \phi$ $\phi$ is an isomorphism {{iff}} $K = \set {0_{R_1} }$.


Ring Epimorphism Preserves Ideals

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $J$ be an ideal of $R_1$. Then $\phi \sqbrk J$ is an ideal of $R_2$.


Preimage of Image of Subring under Ring Homomorphism

Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $K = \ker \left({\phi}\right)$, where $\ker \left({\phi}\right)$ is the kernel of $\phi$.  Let $J$ be a subring of $R_1$. Then: :$\phi^{-1} \left[{\phi \left[{J}\right]}\right] = J + K$


Preimage of Subring under Ring Homomorphism is Subring

Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $S_2$ be a subring of $R_2$. Then $S_1 = \phi^{-1} \left[{S_2}\right]$ is a subring of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$.


Preimage of Ideal under Ring Homomorphism is Ideal

Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a ring homomorphism. Let $S_2$ be an ideal of $R_2$. Then $S_1 = \phi^{-1} \left[{S_2}\right]$ is an ideal of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$.


Image of Preimage of Subring under Ring Epimorphism

Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a ring epimorphism. Let $S_2$ be a subring of $R_2$. Then: :$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$


Properties of Ordered Ring

Let $\struct {R, +, \circ, \le}$ be an ordered ring whose zero is $0_R$ and whose unity is $1_R$. Let $U_R$ be the group of units of $R$. Let $x, y, z \in \struct {R, +, \circ, \le}$. Then the following properties hold: : $(1): \quad x < y \iff x + z < y + z$. Hence $x \le y \iff x + z \le y + z$ (because $\struct {R, +, \le}$ is an ordered group). : $(2): \quad x < y \iff 0 < y + \paren {-x}$. Hence $x \le y \iff 0 \le y + \paren {-x}$ : $(3): \quad 0 < x \iff \paren {-x} < 0$. Hence $0 \le x \iff \paren {-x} \le 0$ : $(4): \quad x < 0 \iff 0 < \paren {-x}$. Hence $x \le 0 \iff 0 \le \paren {-x}$ : $(5): \quad \forall n \in \Z_{>0}: x > 0 \implies n \cdot x > 0$ : $(6): \quad x \le y, 0 \le z: x \circ z \le y \circ z, z \circ x \le z \circ y$ : $(7): \quad x \le y, z \le 0: y \circ z \le x \circ z, z \circ y \le z \circ x$


Positive Elements of Ordered Ring

Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity whose zero is $0_R$ and whose unity is $1_R$. Let $P$ be the set of positive elements of $R$ , that is, $P = R_{\ge 0}$. Then: : $(1): \quad P + P \subseteq P$ : $(2): \quad P \cap \paren {-P} = \set {0_R}$ : $(3): \quad P \circ P \subseteq P$ If $\le$ is a total ordering, that is, if $\struct {R, +, \circ, \le}$ is a totally ordered ring, then: : $(4): \quad P \cup \paren {-P} = R$ The converse is also true: Let $\struct {R, +, \circ}$ be a ring. Let $P \subseteq R$ such that $(1)$, $(2)$ and $(3)$ are satisfied. Then there is one and only one ordering $\le$ compatible with the ring structure of $R$ such that $P = R_{\ge 0}$. Also, if $(4)$ is also satisfied, then $\le$ is a total ordering.


Symmetric Difference with Intersection forms Ring

Let $S$ be a set. Then $\struct {\powerset S, *, \cap}$ is a commutative ring with unity, in which the unity is $S$. This ring is not an integral domain.


Subfield Test

Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$. Let $K$ be a subset of $F$. Then $\struct {K, +, \circ}$ is a subfield of $\struct {F, +, \circ}$ {{iff}} these all hold: :$(1): \quad K^* \ne \O$ :$(2): \quad \forall x, y \in K: x + \paren {-y} \in K$ :$(3): \quad \forall x, y \in K: x \circ y \in K$ :$(4): \quad x \in K^* \implies x^{-1} \in K^*$ where $K^*$ denotes $K \setminus \set {0_F}$.


Field of Quotients of Subdomain

Let $\struct {F, +, \circ}$ be a field whose unity is $1_F$. Let $\struct {D, +, \circ}$ be a subdomain of $\struct {F, +, \circ}$ whose unity is $1_D$. Let: :$K = \set {\dfrac x y: x \in D, y \in D^*}$ where $\dfrac x y$ is the division product of $x$ by $y$. Then $\struct {K, +, \circ}$ is a field of quotients of $\struct {D, +, \circ}$.


Existence of Field of Quotients

Let $\struct {D, +, \circ}$ be an integral domain. Then there exists a field of quotients of $\struct {D, +, \circ}$.


Quotient Theorem for Monomorphisms

Let $K, L$ be fields of quotients of integral domains $\struct {R, +_R, \circ_R}, \struct {S, +_S, \circ_S}$ respectively. Let $\phi: R \to S$ be a monomorphism. Then there is one and only one monomorphism $\psi: K \to L$ extending $\phi$, and: :$\forall x \in R, y \in R^*: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$ Also, if $\phi$ is a ring isomorphism, then so is $\psi$.


Field of Quotients is Unique

Let $\struct {D, +, \circ}$ be an integral domain. Let $K, L$ be field of quotients of $\struct {D, +, \circ}$. Then there is one and only one (field) isomorphism $\phi: K \to L$ satisfying: :$\forall x \in D: \map \phi x = x$


Divided by Positive Element of Field of Quotients

Let $\struct {K, +, \circ}$ be the field of quotients of a totally ordered integral domain $\struct {D, +, \circ, \le}$. Then: :$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$


Total Ordering on Field of Quotients is Unique

Let $\struct {K, +, \circ}$ be a field of quotients of an ordered integral domain $\struct {D, +, \circ, \le}$. Then there is one and only one total ordering $\le'$ on $K$ which is compatible with its ring structure and induces on $D$ its given total ordering $\le$. That ordering is the one defined by: :$P = \set {\dfrac x y \in K: x \in D_+, y \in D_+^*}$


Order Embedding between Quotient Fields is Unique

Let $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ be totally ordered integral domains. Let $K, L$ be totally ordered fields of quotients of $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ respectively. Let $\phi: R \to S$ be a order embedding. Then there is one and only one order embedding $\psi: K \to L$ extending $\phi$. Also: :$\forall x \in R, y \in R_{\ne 0}: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$ If $\phi: R \to S$ is an order isomorphism, then so is $\psi$.


Strict Ordering Preserved under Product with Cancellable Element

Let $\left({S, \circ, \preceq}\right)$ be an ordered semigroup. Let $x, y, z \in S$ be such that: :$(1): \quad z$ is cancellable for $\circ$ :$(2): \quad x \prec y$ Then: : $x \circ z \prec y \circ z$ : $z \circ x \prec z \circ y$


Ordering of Inverses in Ordered Monoid

Let $\left({S, \circ, \preceq}\right)$ be an ordered monoid whose identity is $e$. Let $x, y \in S$ be invertible. Then: :$x \prec y \iff y^{-1} \prec x^{-1}$


Monomorphism from Total Ordering

Let the following conditions hold: : $(1): \quad$ Let $\left({S, \circ, \preceq}\right)$ and $\left({T, *, \preccurlyeq}\right)$ be ordered semigroups. : $(2): \quad$ Let $\phi: S \to T$ be a mapping. : $(3): \quad$ Let $\preceq$ be a total ordering on $S$. Then $\phi \left({S, \circ, \preceq}\right) \to \left({T, *, \preccurlyeq}\right)$ is a (structure) monomorphism iff: : $(1): \quad \phi$ is strictly increasing from $\left({S, \preceq}\right)$ into $\left({T, \preccurlyeq}\right)$; : $(2): \quad \phi$ is a homomorphism from $\left({S, \circ}\right)$ into $\left({T, *}\right)$.


Extension Theorem for Total Orderings

Let the following conditions be fulfilled: :$(1):\quad$ Let $\struct {S, \circ, \preceq}$ be a totally ordered commutative semigroup :$(2):\quad$ Let all the elements of $\struct {S, \circ, \preceq}$ be cancellable :$(3):\quad$ Let $\struct {T, \circ}$ be an inverse completion of $\struct {S, \circ}$. Then: :$(1):\quad$ The relation $\preceq'$ on $T$ satisfying $\forall x_1, x_2, y_1, y_2 \in S: x_1 \circ \paren {y_1}^{-1} \preceq' x_2 \circ \paren {y_2}^{-1} \iff x_1 \circ y_2 \preceq x_2 \circ y_1$ is a well-defined relation :$(2):\quad$ $\preceq'$ is the only total ordering on $T$ compatible with $\circ$ :$(3):\quad$ $\preceq'$ is the only total ordering on $T$ that induces the given ordering $\preceq$ on $S$.


Cancellability in Naturally Ordered Semigroup

Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup. Then:


Strict Lower Closure of Sum with One

Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup. Then: :$\forall n \in \left({S, \circ, \preceq}\right): \left({n \circ 1}\right)^\prec = n^\prec \cup \left\{{n}\right\}$ where $n^\prec$ is defined as the strict lower closure of $n$, that is, the set of elements strictly preceding $n$.


Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor

Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup. Then: :$\forall m, n \in \left({S, \circ, \preceq}\right): m \preceq n \implies \left[{m \,.\,.\, n \circ 1}\right] = \left[{m \,.\,.\, n}\right] \cup \left\{{n \circ 1}\right\}$ where $\left[{m \,.\,.\, n}\right]$ is the closed interval between $m$ and $n$.


Homomorphism of Powers

Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be semigroups. Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.


Naturally Ordered Semigroup is Unique

Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be naturally ordered semigroups. Let: :$0'$ be the smallest element of $S'$ :$1'$ be the smallest element of $S' \setminus \set {0'} = S'^*$. Then the mapping $g: S \to S'$ defined as: :$\forall a \in S: \map g a = \circ'^a 1'$ is an isomorphism from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$. This isomorphism is unique. Thus, up to isomorphism, there is only one naturally ordered semigroup.


Consecutive Subsets of N

Let $\N_k$ denote the initial segment of the natural numbers determined by $k$: :$\N_k = \left\{{0, 1, 2, 3, \ldots, k - 1}\right\}$ Then: :$\N_k = \N_{k + 1} \setminus \left\{{k}\right\}$ In particular: :$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$


Well-Ordering Principle

Every non-empty subset of $\N$ has a smallest (or '''first''') element. This is called the '''well-ordering principle'''. The '''well-ordering principle''' also holds for $\N_{\ne 0}$.


Equality of Natural Numbers

Let $m, n \in \N$. Then: : $\N_m \sim \N_n \iff m = n$ where $\sim$ denotes set equivalence and $\N_n$ denotes the set of all natural numbers less than $n$.


Cardinality Less One

Let $S$ be a finite set. Let: :$\left\lvert{S}\right\rvert = n + 1$ where $\left\lvert{S}\right\rvert$ is the cardinality of $S$. Let $a \in S$. Then: :$\left\lvert{S \setminus \left\{ {a}\right\} }\right\rvert = n$ where $\setminus$ denotes set difference.


Cardinality of Empty Set

:$\card S = 0 \iff S = \O$ That is, the empty set is finite, and has a cardinality of zero.


Cardinality of Subset of Finite Set

Let $A$ and $B$ be finite sets such that $A \subseteq B$. Let :$\card B = n$ where $\card {\, \cdot \,}$ denotes cardinality. Then $\card A \le n$.


Cardinality of Surjection

Let $S$ be a set. Let: :$\card S = n$ where $\card S$ denotes the cardinality of $S$. Let $f: S \to T$ be a surjection. Then $\card T \le n$. The equality: :$\card T = n$ occurs {{iff}} $f$ is a bijection.


Equivalence of Mappings between Sets of Same Cardinality

Let $S$ and $T$ be finite sets such that $\card S = \card T$. Let $f: S \to T$ be a mapping. Then the following statements are equivalent: : $(1): \quad f$ is bijective : $(2): \quad f$ is injective : $(3): \quad f$ is surjective.


Natural Numbers are Infinite

The set $\N$ of natural numbers is infinite.


Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements

Let $\left({S, \preceq}\right)$ be a totally ordered set. Then every finite $T$ such that $\varnothing \subset T \subseteq S$ has both a smallest and a greatest element.


Unique Isomorphism between Finite Totally Ordered Sets

Let $S$ and $T$ be finite sets such that: :$\card S = \card T$ Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be totally ordered sets. Then there is exactly one order isomorphism from $\struct {S, \preceq}$ to $\struct {T, \preccurlyeq}$.


Isomorphism to Closed Interval

Let $m, n \in \N$ such that $m < n$. Then: : $\left|{\left[{m + 1 \,.\,.\, n}\right]}\right| = n - m$ Let $h: \N_{n - m} \to \left[{m + 1 \,.\,.\, n}\right]$ be the mapping defined as: :$\forall x \in \N_{n - m}: h \left({x}\right) = x + m + 1$ Let the orderings on $\left[{m + 1 \,.\,.\, n}\right]$ and $\N_{n - m}$ be those induced by the ordering of $\N$. Then $h$ a unique order isomorphism.


Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection

Let $\left({S, \circ}\right)$ be a finite semigroup. Let $a \in S$ be cancellable. Then: :the left regular representation $\lambda_a$ and: :the right regular representation $\rho_a$ of $\left({S, \circ}\right)$ with respect to $a$ are both bijections.


Power Set of Natural Numbers is not Countable

The power set $\powerset \N$ of the natural numbers $\N$ is not countable.


Subset of Countably Infinite Set is Countable

Every subset of a countably infinite set is countable.


Infinite Set has Countably Infinite Subset

Every infinite set has a countably infinite subset.


No Bijection between Finite Set and Proper Subset

A finite set can not be in one-to-one correspondence with one of its proper subsets. That is, a finite set is not Dedekind-infinite.


Infinite Set is Equivalent to Proper Subset

A set is infinite {{iff}} it is equivalent to one of its proper subsets.


Cartesian Product of Countable Sets is Countable

The cartesian product of two countable sets is countable.


Composition of Sequence with Mapping

Let $\left \langle {a_j} \right \rangle_{j \in B}$ be a sequence. Let $\sigma: A \to B$ be a mapping, where $A \subseteq \N$. Then $\left \langle {a_j} \right \rangle \circ \sigma$ is a sequence whose value at each $k \in A$ is $a_{\sigma \left({k}\right)}$. Thus $\left \langle {a_j} \right \rangle \circ \sigma$ is denoted $\left \langle {a_{\sigma \left({k}\right)}} \right \rangle_{k \in A}$.


General Operation from Binary Operation

Let $\left({S, \oplus}\right)$ be a magma. Then there a unique sequence $\left \langle {\oplus_k} \right \rangle_{k \mathop \ge 1}$ such that: :$(1): \quad \forall n \in \N_{>0}: \oplus_n$ is an $n$-ary operation on $S$ such that: :$(2): \quad \forall \left({a_1, \ldots, a_k}\right) \in S^k: \oplus_k \left({a_1, \ldots, a_k}\right) = \begin{cases} a: & k = 1 \\ \oplus_n \left({a_1, \ldots, a_n}\right) \oplus a_{n+1}: & k = n + 1 \end{cases}$ In particular, $\oplus_2$ is the same as the given binary operation $\oplus$. The $n$th term $\oplus_n$ of the sequence $\left \langle {\oplus} \right \rangle$ is called the '''$n$-ary operation defined by $\oplus$'''.


Strictly Increasing Sequence induces Partition

Let $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ be a strictly increasing finite sequence of natural numbers. Let: :$\forall k \in \left[{1 \,.\,.\, n}\right]: A_k := \left[{r_{k-1} + 1 \,.\,.\, r_k}\right]$ Then: :$\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ is a partition of $\left[{r_0 + 1 \,.\,.\, r_n}\right]$.


Fundamental Principle of Counting

Let $A$ be a finite set. Let $\sequence {B_n}$ be a sequence of distinct finite subsets of $A$ which form a partition of $A$. Let $p_k = \size {B_k}$ for each $k \in \closedint 1 n$. Then: :$\displaystyle \size A = \sum_{k \mathop = 1}^n p_k$ That is, the sum of the numbers of elements in the subsets of a partition of a set is equal to the total number of elements in the set.


Odd Number Theorem

:$\displaystyle \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$ That is, the sum of the first $n$ odd numbers is the $n$th square number.


General Associativity Theorem

If an operation is associative on $3$ entities, then it is associative on any number of them.


General Commutativity Theorem

Let $\struct {S, \circ}$ be a semigroup. Let $\family {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of elements of $S$. Suppose that: :$\forall i, j \in \closedint 1 n: a_i \circ a_j = a_j \circ a_i$ Then for every permutation $\sigma: \N_n \to \N_n$: :$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma n} = a_1 \circ \cdots \circ a_n$ where $\N_n$ is used here to denote the initial segment of $\N_{>0}$: :$\N_n = \set {1, 2, \ldots, n}$


General Distributivity Theorem

:$\displaystyle \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$


Associativity on Indexing Set

Let $\left({S, \circ}\right)$ be a commutative semigroup. Let $\left \langle {x_\alpha} \right \rangle_{\alpha \mathop \in A}$ be a family of terms of $S$ indexed by a finite non-empty set $A$. Let $\left \langle {B_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ be a family of distinct subsets of $A$ forming a partition of $A$. Then: : $\displaystyle \prod_{k \mathop = 1}^n \left({\prod_{a \mathop \in B_k} x_\alpha}\right) = \prod_{\alpha \mathop \in A} x_\alpha$


External Direct Product of Groups is Group

Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups whose identity elements are $e_1$ and $e_2$ respectively. Let $\struct {G_1 \times G_2, \circ}$ be the external direct product of $G_1$ and $G_2$. Then $\struct {G_1 \times G_2, \circ}$ is a group whose identity element is $\tuple {e_1, e_2}$.


External Direct Product of Projection with Canonical Injection

Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be algebraic structures with identity elements $e_1$ and $e_2$ respectively. Let $\struct {S_1 \times S_2, \circ}$ be the external direct product of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ Let: :$\pr_1: \struct {S_1 \times S_2, \circ} \to \struct {S_1, \circ_1}$ be the first projection from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_1, \circ_1}$ :$\pr_2: \struct {S_1 \times S_2, \circ} \to \struct {S_2, \circ_2}$ be the second projection from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_2, \circ_2}$. Let: :$\inj_1: \struct {S_1, \circ_1} \to \struct {S_1 \times S_2, \circ}$ be the canonical injection from $\struct {S_1, \circ_1}$ to $\struct {S_1 \times S_2, \circ}$ :$\inj_2: \struct {S_2, \circ_2} \to \struct {S_1 \times S_2, \circ}$ be the canonical injection from $\struct {S_2, \circ_2}$ to $\struct {S_1 \times S_2, \circ}$. Then: :$(1): \quad \pr_1 \circ \inj_1 = I_{S_1}$ :$(2): \quad \pr_2 \circ \inj_2 = I_{S_2}$ where $I_{S_1}$ and $I_{S_2}$ are the identity mappings on $S_1$ and $S_2$ respectively.


Inverse Completion of Natural Numbers

There exists an inverse completion of the natural numbers under addition.


L'Hôpital's Rule

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$. Let: :$\forall x \in \openint a b: \map {g'} x \ne 0$ where $g'$ denotes the derivative of $g$ {{WRT|Differentiation}} $x$. Let: :$\map f a = \map g a = 0$ Then: :$\displaystyle \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$ provided that the second limit exists.


Natural Numbers under Addition form Commutative Monoid

The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is a commutative monoid whose identity is zero.


Natural Numbers form Commutative Semiring

The semiring of natural numbers $\struct {\N, +, \times}$ forms a commutative semiring.


Integer Multiplication is Closed

The set of integers is closed under multiplication: :$\forall a, b \in \Z: a \times b \in \Z$


Integer Multiplication is Commutative

The operation of multiplication on the set of integers $\Z$ is commutative: :$\forall x, y \in \Z: x \times y = y \times x$


Integer Multiplication is Associative

The operation of multiplication on the set of integers $\Z$ is associative: :$\forall x, y, z \in \Z: x \times \paren {y \times z} = \paren {x \times y} \times z$


Integer Multiplication Distributes over Addition

The operation of multiplication on the set of integers $\Z$ is distributive over addition: :$\forall x, y, z \in \Z: x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$ :$\forall x, y, z \in \Z: \paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$


Construction of Inverse Completion

This page consists of a series of linked theorems, each of which builds towards one result. To access the proofs for the individual theorems, click on the links which form the titles of each major section.


Integer Multiplication Identity is One

The identity of integer multiplication is $1$: :$\exists 1 \in \Z: \forall a \in \Z: a \times 1 = a = 1 \times a$


Integer Multiplication has Zero

The set of integers under multiplication $\struct {\Z, \times}$ has a zero element, which is $0$.


Ring of Integers has no Zero Divisors

The integers have no zero divisors: :$\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \lor y = 0$


Integers form Integral Domain

The integers $\Z$ form an integral domain under addition and multiplication.


Natural Numbers are Non-Negative Integers

Let $m \in \Z$. Then: :$(1): \quad m \in \N \iff 0 \le m$ :$(2): \quad m \in \N_{> 0} \iff 0 < m$ :$(3): \quad m \notin \N \iff -m \in \N_{> 0}$ That is, the natural numbers are precisely those integers which are greater than or equal to zero.


Subtraction on Integers is Extension of Natural Numbers

Integer subtraction is an extension of the definition of subtraction on the natural numbers.


Multiplicative Ordering on Integers

Let $x, y, z \in \Z$ such that $z > 0$. Then: :$x < y \iff z x < z y$ :$x \le y \iff z x \le z y$


Invertible Integers under Multiplication

The only invertible elements of $\Z$ for multiplication (that is, units of $\Z$) are $1$ and $-1$.


Index Laws for Monoids

These results are an extension of the results in Index Laws for Semigroup in which the domain of the indices is extended to include all integers. Let $\struct {S, \circ}$ be a monoid whose identity is $e$. Let $a \in S$ be invertible for $\circ$. Let $n \in \N$. Let $a^n$ be the $n$th power of $a$: :$a^n = \begin{cases} e : & n = 0 \\ a^{n - 1} \circ a : & n > 0 \\ \paren {a^{-n}}^{-1} : & n < 0 \end{cases}$ Then we have the following results:


Totally Ordered Abelian Group Isomorphism

Let $\left({\Z', +', \le'}\right)$ be a totally ordered abelian group. Let $0'$ be the identity of $\left({\Z', +', \le'}\right)$. Let $\N' = \left\{{x \in \Z': x \ge' 0'}\right\}$. Let $\Z'$ contain at least two elements. Let $\N'$ be well-ordered for the ordering induced on $\N'$ by $\le'$. Then the mapping $g: \Z \to \Z'$ defined by: :$\forall n \in \Z: g \left({n}\right) = \left({+'}\right)^n 1'$ is an isomorphism from $\left({\Z, +, \le}\right)$ onto $\left({\Z', +', \le'}\right)$, where $1'$ is the smallest element of $\N' \setminus \left\{{0'}\right\}$.


Integers under Addition form Totally Ordered Group

Then the ordered structure $\struct {\Z, +, \le}$ is a totally ordered group.


Integers form Totally Ordered Ring

The structure $\struct {\Z, +, \times, \le}$ is a totally ordered ring.


Congruences on Rational Numbers

There are only two congruence relations on the field of rational numbers $\left({\Q, +, \times}\right)$: :$(1): \quad$ The diagonal relation $\Delta_\Q$ :$(2): \quad$ The trivial relation $\Q \times \Q$.


Number of Elements in Partition

Let $S$ be a set. Let there be a partition on $S$ of $n$ subsets, each of which has $m$ elements. Then: :$\card S = n m$


Cardinality of Complement

Let $T \subseteq S$ such that $\card S = n, \card T = m$. Then: :$\card {\relcomp S T} = \card {S \setminus T} = n - m$ where: :$\relcomp S T$ denotes the complement of $T$ relative to $S$ :$S \setminus T$ denotes the difference between $S$ and $T$.


Cardinality of Cartesian Product

Let $S \times T$ be the cartesian product of two finite sets $S$ and $T$. Then: :$\card {S \times T} = \card S \times \card T$ where $\card S$ denotes cardinality.


Cardinality of Set of All Mappings

Let $S$ and $T$ be sets. The cardinality of the set of all mappings from $S$ to $T$ (that is, the total number of mappings from $S$ to $T$) is: :$\card {T^S} = \card T^{\card S}$


Cardinality of Power Set of Finite Set

Let $S$ be a set such that: :$\card S = n$ where $\card S$ denotes the cardinality of $S$, Then: :$\card {\powerset S} = 2^n$ where $\powerset S$ denotes the power set of $S$.


Cardinality of Set of Injections

Let $S$ and $T$ be finite sets. The number of injections from $S$ to $T$, where $\card S = m, \card T = n$ is often denoted ${}^m P_n$, and is: :${}^m P_n = \begin{cases} \dfrac {n!} {\paren {n - m}!} & : m \le n \\ 0 & : m > n \end{cases}$


Cardinality of Set of Bijections

Let $S$ and $T$ be sets such that $\size S = \size T = n$. Then there are $n!$ bijections from $S$ to $T$.


Cardinality of Set of Subsets

Let $S$ be a set such that $\card S = n$. Let $m \le n$. Then the number of subsets $T$ of $S$ such that $\card T = m$ is: : ${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$


Set of Integers Bounded Below by Integer has Smallest Element

Let $\Z$ be the set of integers. Let $\le$ be the ordering on the integers. Let $\O \subset S \subseteq \Z$ such that $S$ is bounded below in $\struct {\Z, \le}$. Then $S$ has a smallest element.


Set of Integers Bounded Above by Integer has Greatest Element

Let $\Z$ be the set of integers. Let $\le$ be the ordering on the integers. Let $\O \subset S \subseteq \Z$ such that $S$ is bounded above in $\struct {\Z, \le}$. Then $S$ has a greatest element.


Principle of Least Counterexample

Suppose $P \paren n$ is a condition on $n \in \set {x \in \Z: x \ge m \in \Z}$. Suppose next that: $\neg \paren {\forall n \ge m: P \paren n}$. (That is, not all $n \ge m$ satisfy $P \paren n$.) Then there is a '''least counterexample''', that is a smallest integral value of $n$ for which $\neg P \paren n$.


Absolute Value is Bounded Below by Zero

Let $x \in \R$ be a real number. Then the absolute value $\size x$ of $x$ is bounded below by $0$.


Division Theorem

For every pair of integers $a, b$ where $b \ne 0$, there exist unique integers $q, r$ such that $a = q b + r$ and $0 \le r < \size b$: :$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$


Odd Integer 2n + 1

Let $m$ be an odd integer. Then there exists exactly one integer $n$ such that $2 n + 1 = m$.


Integer Divisor Results

Let $m, n \in \Z$ be integers. Let $m \divides n$ denote that $m$ is a divisor of $n$. The following results all hold:


Zero Divides Zero

Let $n \in \Z$ be an integer. Then: :$0 \divides n \implies n = 0$ That is, zero is the only integer divisible by zero.


Absolute Value of Integer is not less than Divisors

A (non-zero) integer is greater than or equal to its divisors in magnitude: :$\forall c \in \Z_{\ne 0}: a \divides c \implies a \le \size a \le \size c$


Divisor Relation on Positive Integers is Partial Ordering

The divisor relation is a partial ordering of $\Z_{>0}$.


Common Divisor in Integral Domain Divides Linear Combination

Let $\struct {D, +, \times}$ be an integral domain. Let $c$ be a common divisor of two elements $a$ and $b$ of $D$. That is: :$a, b, c \in D: c \divides a \land c \divides b$ Then: :$\forall p, q \in D: c \divides \paren {p \times a + q \times b}$


Greatest Common Divisor is at least 1

Let $a, b \in \Z$ be integers. The greatest common divisor of $a$ and $b$ is at least $1$: :$\forall a, b \in \Z_{\ne 0}: \gcd \set {a, b} \ge 1$


GCD of Integer and Divisor

Let $a, b \in \Z_{>0}$, i.e. integers such that $a, b > 0$. Then: : $a \divides b \implies \gcd \set {a, b} = a$


GCD for Negative Integers

:$\gcd \set {a, b} = \gcd \set {\size a, b} = \gcd \set {a, \size b} = \gcd \set {\size a, \size b}$ Alternatively, this can be put: :$\gcd \set {a, b} = \gcd \set {-a, b} = \gcd \set {a, -b} = \gcd \set {-a, -b}$   which follows directly from the above.


GCD with Zero

Let $a \in \Z$ be an integer such that $a \ne 0$. Then: :$\gcd \left\{{a, 0}\right\} = \left\lvert{a}\right\rvert$ where $\gcd$ denotes greatest common divisor (GCD).


Set of Integer Combinations equals Set of Multiples of GCD

The set of all integer combinations of $a$ and $b$ is precisely the set of all integer multiples of the GCD of $a$ and $b$: :$\gcd \set {a, b} \divides c \iff \exists x, y \in \Z: c = x a + y b$


GCD with Remainder

Let $a, b \in \Z$. Let $q, r \in \Z$ such that $a = q b + r$. Then: :$\gcd \set {a, b} = \gcd \set {b, r}$ where $\gcd \set {a, b}$ is the greatest common divisor of $a$ and $b$.


Integer Combination of Coprime Integers

Two integers are coprime {{iff}} there exists an integer combination of them equal to $1$: :$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$


Integers Divided by GCD are Coprime

Any pair of integers, not both zero, can be reduced to a pair of coprime ones by dividing them by their GCD: :$\gcd \set {a, b} = d \iff \dfrac a d, \dfrac b d \in \Z \land \gcd \set {\dfrac a d, \dfrac b d} = 1$ That is: :$\dfrac a {\gcd \set {a, b} } \perp \dfrac b {\gcd \set {a, b} }$


Product of Coprime Factors

Let $a, b, c \in \Z$ such that $a$ and $b$ are coprime. Let both $a$ and $b$ be divisors of $c$. Then $a b$ is also a divisor of $c$. That is: :$a \perp b \land a \divides c \land b \divides c \implies a b \divides c$


Existence of Lowest Common Multiple

Let $a, b \in \Z: a b \ne 0$. The lowest common multiple of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists.


Product of GCD and LCM

:$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$ where: :$\lcm \set {a, b}$ denotes the lowest common multiple of $a$ and $b$ :$\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.


Congruent to Zero if Modulo is Divisor

Let $a, z \in \R$. Then $a$ is congruent to $0$ modulo $z$ {{iff}} $a$ is an integer multiple of $z$. :$\exists k \in \Z: k z = a \iff a \equiv 0 \pmod z$ If $z \in \Z$, then further: :$z \divides a \iff a \equiv 0 \pmod z$


Integer is Congruent Modulo Divisor to Remainder

Let $a \in \Z$. Let $a$ have a remainder $r$ on division by $m$. Then: : $a \equiv r \pmod m$ where the notation denotes that $a$ and $r$ are congruent modulo $m$.


Integer is Congruent to Integer less than Modulus

Let $m \in \Z$. Then each integer is congruent (modulo $m$) to precisely one of the integers $0, 1, \ldots, m - 1$.


Modulo Addition is Well-Defined

Let $m \in \Z$ be an integer. Let $\Z_m$ be the set of integers modulo $m$. The modulo addition operation on $\Z_m$, defined by the rule: :$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$ is a well-defined operation. That is: :If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a + x \equiv b + y \pmod m$.


Modulo Multiplication is Well-Defined

The multiplication modulo $m$ operation on $\Z_m$, the set of integers modulo $m$, defined by the rule: :$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$ is a well-defined operation. That is: :If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a x \equiv b y \pmod m$.


Congruence of Product

Let $a, b, z \in \R$. Let $a$ be congruent to $b$ modulo $z$, that is: :$a \equiv b \pmod z$ Then: :$\forall m \in \Z: m a \equiv m b \pmod z$


Congruence of Powers

Let $a, b \in \R$ and $m \in \Z$. Let $a$ be congruent to $b$ modulo $m$, that is: :$a \equiv b \pmod m$ Then: :$\forall n \in \Z_{\ge 0}: a^n \equiv b^n \pmod m$


Modulo Addition is Associative

Addition modulo $m$ is associative: :$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m +_m \eqclass y m} +_m \eqclass z m = \eqclass x m +_m \paren {\eqclass y m +_m \eqclass z m}$ where $\Z_m$ is the set of integers modulo $m$. That is: :$\forall x, y, z \in \Z: \paren {x + y} + z \equiv x + \paren {y + z} \pmod m$


Modulo Addition is Commutative

Modulo addition is commutative: :$\forall x, y, z \in \Z: x + y \pmod m = y + x \pmod m$


Modulo Addition has Identity

Let $m \in \Z$ be an integer. Then addition modulo $m$ has an identity: :$\forall \eqclass x m \in \Z_m: \eqclass x m +_m \eqclass 0 m = \eqclass x m = \eqclass 0 m +_m \eqclass x m$ That is: :$\forall a \in \Z: a + 0 \equiv a \equiv 0 + a \pmod m$


Modulo Addition has Inverses

Let $m \in \Z$ be an integer. Then addition modulo $m$ has inverses: For each element $\eqclass x m \in \Z_m$, there exists the element $\eqclass {-x} m \in \Z_m$ with the property: :$\eqclass x m +_m \eqclass {-x} m = \eqclass 0 m = \eqclass {-x} m +_m \eqclass x m$ where $\Z_m$ is the set of integers modulo $m$. That is: :$\forall a \in \Z: a + \paren {-a} \equiv 0 \equiv \paren {-a} + a \pmod m$


Modulo Multiplication is Closed

Multiplication modulo $m$ is closed on the set of integers modulo $m$: :$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m \in \Z_m$.


Modulo Multiplication is Associative

Multiplication modulo $m$ is associative: :$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m = \eqclass x m \times_m \paren {\eqclass y m \times_m \eqclass z m}$ That is: :$\forall x, y, z \in \Z_m: \paren {x \cdot_m y} \cdot_m z = x \cdot_m \paren {y \cdot_m z}$


Modulo Multiplication is Commutative

Multiplication modulo $m$ is commutative: :$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m = \eqclass y m \times_m \eqclass x m$


Modulo Multiplication has Identity

Multiplication modulo $m$ has an identity: :$\forall \eqclass x m \in \Z_m: \eqclass x m \times_m \eqclass 1 m = \eqclass x m = \eqclass 1 m \times_m \eqclass x m$


Modulo Multiplication Distributes over Modulo Addition

Multiplication modulo $m$ is distributive over addition modulo $m$: :$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m$: :: $\eqclass x m \times_m \paren {\eqclass y m +_m \eqclass z m} = \paren {\eqclass x m \times_m \eqclass y m} +_m \paren {\eqclass x m \times_m \eqclass z m}$ :: $\paren {\eqclass x m +_m \eqclass y m} \times_m \eqclass z m = \paren {\eqclass x m \times_m \eqclass z m} +_m \paren {\eqclass y m \times_m \eqclass z m}$ where $\Z_m$ is the set of integers modulo $m$. That is, $\forall x, y, z, m \in \Z$: : $x \paren {y + z} \equiv x y + x z \pmod m$ : $\paren {x + y} z \equiv x z + y z \pmod m$


Intersection of Congruence Classes

Let $\mathcal R_m$ denote congruence modulo $m$ on the set of integers $\Z$. Then: :$\mathcal R_m \cap \mathcal R_n = \mathcal R_{\lcm \set {m, n} }$ where $\lcm \set {m, n}$ is the lowest common multiple of $m$ and $n$. In the language of modulo arithmetic, this is equivalent to: :$a \equiv b \pmod m, a \equiv b \pmod n \implies a \equiv b \pmod {\lcm \set {m, n} }$


Mappings Between Residue Classes

Let $\eqclass a m$ be the residue class of $a$ (modulo $m$). Let $\phi: \Z_m \to \Z_n$ be a mapping given by: :$\map \phi {\eqclass x m} = \eqclass x n$ Then $\phi$ is well defined {{iff}} $m$ is a divisor of $n$.


Prime Number has 4 Integral Divisors

Let $p$ be an integer. Then $p$ is a prime number {{iff}} $p$ has exactly four integral divisors: $1, -1, p, -p$.


Prime not Divisor implies Coprime

Let $p, a \in \Z$. If $p$ is a prime number then: : $p \nmid a \implies p \perp a$ where: : $p \nmid a$ denotes that $p$ does not divide $a$ : $p \perp a$ denotes that $p$ and $a$ are coprime. It follows directly that if $p$ and $q$ are primes, then: : $p \divides q \implies p = q$ : $p \ne q \implies p \perp q$.


Composite Number has Two Divisors Less Than It

Let $n \in \Z_{> 1}$ such that $n \notin \mathbb P$. Then: :$\exists a, b \in \Z: 1 < a < n, 1 < b < n: n = a b$ That is, a non-prime number greater than $1$ can be expressed as the product of two positive integers strictly greater than $1$ and less than $n$. Note that these two numbers are not necessarily distinct.


Exponents of Primes in Prime Decomposition are Less iff Divisor

Let $a, b \in \Z_{>0}$. Then $a \divides b$ {{iff}}: :$(1): \quad$ every prime $p_i$ in the prime decomposition of $a$ appears in the prime decomposition of $b$ and: :$(2): \quad$ the exponent of each $p_i$ in $a$ is less than or equal to its exponent in $b$.


Not Coprime means Common Prime Factor

Let $a, b \in \Z$. If $d \divides a$ and $d \divides b$ such that $d > 1$, then $a$ and $b$ have a common divisor which is prime.


Set of Divisors of Integer

Let $n \in \Z_{>1}$. Let $n$ be expressed in its prime decomposition: :$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ where $p_1 < p_2 < \dotsb < p_r$ are distinct primes and $k_1, k_2, \ldots, k_r$ are positive integers. The set of divisors of $n$ is: :$\set {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r}: 0 \le h_i \le k_i, i = 1, 2, \ldots, r}$


Sum Less Minimum is Maximum

For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$: :$a + b - \min \left({a, b}\right) = \max \left({a, b}\right)$


Sum Less Maximum is Minimum

For all numbers $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$: :$a + b - \max \left({a, b}\right) = \min \left({a, b}\right)$.


GCD and LCM from Prime Decomposition

Let $m, n \in \Z$. Let: :$m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ :$n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}$ :$p_i \divides m \lor p_i \divides n, 1 \le i \le r$. That is, the primes given in these prime decompositions may be divisors of ''either'' of the numbers $m$ or $n$. Note that if one of the primes $p_i$ does not appear in the prime decompositions of either one of $m$ or $n$, then its corresponding index $k_i$ or $l_i$ will be zero. Then the following results apply: :$\gcd \set {m, n} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$ :$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$


GCD and LCM Distribute Over Each Other

Let $a, b, c \in \Z$. Then: : $\lcm \set {a, \gcd \set {b, c} } = \gcd \set {\lcm \set {a, b}, \lcm \set {a, c} }$ : $\gcd \set {a, \lcm \set {b, c} } = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }$ That is, greatest common divisor and lowest common multiple are distributive over each other.


Tau Function from Prime Decomposition

Let $n$ be an integer such that $n \ge 2$. Let the prime decomposition of $n$ be: :$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ Let $\map \tau n$ be the tau function of $n$. Then: :$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$


N less than M to the N

:$\forall m, n \in \Z_{>0}: m > 1 \implies n < m^n$


Basis Representation Theorem

Let $b \in \Z: b > 1$. For every $n \in \Z_{> 0}$, there exists one and only one sequence $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that: : $(1): \quad \displaystyle n = \sum_{k \mathop = 0}^t r_k b^k$ : $(2): \quad \displaystyle \forall k \in \closedint 0 t: r_k \in \N_b$ : $(3): \quad r_t \ne 0$ This unique sequence is called the '''representation of $n$ to the base $b$''', or, informally, we can say '''$n$ is (written) in base $b$'''.


Sum of Geometric Sequence

Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$. Let $n \in \N_{>0}$. Then: :$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$


Congruence of Sum of Digits to Base Less 1

Let $x \in \Z$, and $b \in \N, b > 1$. Let $x$ be written in base $b$: :$x = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$ Then: :$\displaystyle \map {s_b} x = \sum_{j \mathop = 0}^m r_j \equiv x \pmod {b - 1}$ where $\map {s_b} x$ is the digit sum of $x$ in base $b$ notation. That is, the digit sum of any integer $x$ in base $b$ notation is congruent to $x$ modulo $b - 1$.


Euler Phi Function of Integer

Let $n \in \Z_{>0}$, that is, a (strictly) positive integer. Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$-function. Then for any $n \in \Z_{>0}$, we have: :$\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \cdots \paren {1 - \dfrac 1 {p_r} }$ where $p_1, p_2, \ldots, p_r$ are the distinct primes dividing $n$. Or, more compactly: :$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$ where $p \divides n$ denotes the primes which divide $n$.


Euler Phi Function is Multiplicative

The Euler $\phi$ function is a multiplicative function: :$m \perp n \implies \map \phi {m n} = \map \phi m \, \map \phi n$ where $m, n \in \Z_{>0}$.


Euler Phi Function of Prime Power

Let $p^n$ be a prime power for some prime number $p > 1$. Then: :$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$ where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function.


Möbius Function is Multiplicative

The Möbius function $\mu$  is a multiplicative function: :$m \perp n \implies \map \mu {m n} = \map \mu m \map \mu n$ where $m, n \in \Z_{>0}$.


Sum of Möbius Function over Divisors

Let $n \in \Z_{>0}$ be a strictly positive integer. Then: :$\displaystyle \sum_{d \mathop \divides n} \map \mu d \frac n d = \map \phi n$ where: :$\displaystyle \sum_{d \mathop \divides n}$ denotes the sum over all of the divisors of $n$ :$\map \phi n$ is the Euler $\phi$ function, the number of integers less than $n$ that are prime to $n$ :$\map \mu d$ is the Möbius function. Equivalently, this says that: :$\phi = \mu * I_{\Z_{>0} }$ where: :$*$ denotes Dirichlet convolution :$I_{\Z_{>0} }$ denotes the identity mapping on $\Z_{>0}$, that is: ::$\forall n \in \Z_{>0}: I_{\Z_{>0} }: n \mapsto n$ {{wtd|Add a link to a page proving this equivalence.}}


Binomial Coefficient of Prime

Let $p$ be a prime number. Then: :$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$ where $\dbinom p k$ is defined as a binomial coefficient.


Prime Power of Sum Modulo Prime

Let $p$ be a prime number. Then: :$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$


Binomial Coefficient involving Power of Prime

: $\dbinom {p^n k} {p^n} \equiv k \pmod p$ where $\dbinom {p^n k} {p^n}$ is a binomial coefficient.


Cassini's Identity

:$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$


Rational Numbers are Countably Infinite

The set $\Q$ of rational numbers is countably infinite.


Real Numbers form Ordered Field

The set of real numbers $\R$ forms an ordered field under addition and multiplication: $\struct {\R, +, \times, \le}$.


Rational Numbers form Subfield of Real Numbers

The (ordered) field $\struct {\Q, +, \times, \le}$ of rational numbers forms a subfield of the field of real numbers $\struct {\R, +, \times, \le}$. That is, the field of real numbers $\struct {\R, +, \times, \le}$ is an extension of the rational numbers $\struct {\Q, +, \times, \le}$.


Definition:Rational Number/Canonical Form

Let $r \in \Q$ be a rational number. The '''canonical form of $r$''' is the expression $\dfrac p q$, where: :$r = \dfrac p q: p \in \Z, q \in \Z_{>0}, p \perp q$ where $p \perp q$ denotes that $p$ and $q$ have no common divisor except $1$.


Ordering Properties of Real Numbers

=== Trichotomy Law === {{:Trichotomy Law for Real Numbers}} === Ordering is Transitive === {{:Real Number Ordering is Transitive}} === Ordering is Compatible with Addition === {{:Real Number Ordering is Compatible with Addition}} === Ordering is Compatible with Multiplication === {{:Real Number Ordering is Compatible with Multiplication}}


Order is Preserved on Positive Reals by Squaring

:$x < y \iff x^2 < y^2$


Real Plus Epsilon

Let $a, b \in \R$, such that: :$\forall \epsilon \in \R_{>0}: a < b + \epsilon$ where $\R_{>0}$ is the set of strictly positive real numbers. That is: :$\epsilon > 0$ Then: :$a \le b$


Difference of Two Squares

Let $\struct {R, +, \circ}$ be a commutative ring whose zero is $0_R$. Let $x, y \in R$. Then: : $x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$ When $R$ is one of the standard sets of numbers, that is $\Z, \Q, \R$, and so on, then this translates into: :$x^2 - y^2 = \paren {x + y} \paren {x - y}$


Mediant is Between

Let $a, b, c, d$ be ''any'' real numbers such that $b > 0, d > 0$. Let $r = \dfrac a b < \dfrac c d = s$. Then: :$r < \dfrac {a + c} {b + d} < s$


Real Numbers Between Epsilons

Let $a, b \in \R$ such that $\forall \epsilon \in \R_{>0}: a - \epsilon < b < a + \epsilon$. Then $a = b$.


Exists Integer Below Any Real Number

Let $x$ be a real number. Then there exists an integer less than $x$: :$\forall x \in \R: \exists n \in \Z: n < x$


Archimedean Principle

Let $x$ be a real number. Then there exists a natural number greater than $x$. :$\forall x \in \R: \exists n \in \N: n > x$ That is, the set of natural numbers is unbounded above.


Real Number is between Floor Functions

:$\forall x \in \R: \floor x \le x < \floor {x + 1}$


Real Number is between Ceiling Functions

:$\forall x \in \R: \left \lceil {x - 1} \right \rceil \le x < \left \lceil {x} \right \rceil$


Real Number minus Floor

:$x - \floor x \in \hointr 0 1$


Ceiling minus Real Number

:$\forall x \in \R: \left \lceil {x} \right \rceil - x \in \left[{0 \,.\,.\, 1}\right)$


Real Number is Floor plus Difference

:There exists an integer $n \in \Z$ such that for some $t \in \hointr 0 1$: ::$x = n + t$ {{iff}}: :$n = \floor x$


Floor plus One

Let $x \in \R$. Then: :$\left \lfloor {x + 1} \right \rfloor = \left \lfloor {x} \right \rfloor + 1$ where $\left \lfloor {x} \right \rfloor$ is the floor function of $x$.


Real Number is Integer iff equals Floor

:$x = \floor x \iff x \in \Z$


Sum of Floor and Floor of Negative

Let $x \in \R$. Then: :$\floor x + \floor {-x} = \begin{cases} 0 & : x \in \Z \\ -1 & : x \notin \Z \end{cases}$ where $\floor x$ denotes the floor of $x$.


Ceiling defines Equivalence Relation

Let $\mathcal R$ be the relation defined on $\R$ such that: :$\forall x, y, \in \R: \left({x, y}\right) \in \mathcal R \iff \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$ where $\left \lceil {x}\right \rceil$ is the ceiling of $x$. Then $\mathcal R$ is an equivalence, and $\forall n \in \Z$, the $\mathcal R$-class of $n$ is the half-open interval $\left({n - 1 \,.\,.\, n}\right]$.


Real Number is Ceiling minus Difference

Let $n$ be a integer. {{TFAE}} :$(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$ :$(2): \quad n = \ceiling x$


Cauchy's Inequality

:$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$


Cancellable Finite Semigroup is Group

Let $\struct {S, \circ}$ be a non-empty finite semigroup in which all elements are cancellable. Then $\struct {S, \circ}$ is a group.


Finite Semigroup Equal Elements for Different Powers

Let $\left({S, \circ}\right)$ be a finite semigroup. Then: : $\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$


Element has Idempotent Power in Finite Semigroup

Let $\struct {S, \circ}$ be a finite semigroup. For every element in $\struct {S, \circ}$, there is a power of that element which is idempotent. That is: :$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$


Powers of Ring Elements

Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$. Let $n \cdot x$ be an integral multiple of $x$: :$n \cdot x = \begin {cases} 0_R & : n = 0 \\ x & : n = 1 \\ \paren {n - 1} \cdot x + x & : n > 1 \end {cases}$ that is $n \cdot x = x + x + \cdots \paren n \cdots x$. For $n < 0$ we use: :$-n \cdot x = n \cdot \paren {-x}$ Then: :$\forall n \in \Z: \forall x \in R: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$


Power of Conjugate equals Conjugate of Power

Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$. That is, let $x$ and $y$ be conjugate. Then: : $\forall n \in \Z: y^n = \paren {a^{-1} \circ x \circ a}^n = a^{-1} \circ x^n \circ a$ It follows directly that: : $\exists b \in G: \forall n \in \Z: y^n = b \circ x^n \circ b^{-1}$ In particular: : $y^{-1} = \paren {a^{-1} \circ x \circ a}^{-1} = a^{-1} \circ x^{-1} \circ a$


Product of Conjugates equals Conjugate of Products

Let $\struct {G, \circ}$ be a group. Then: :$\forall a, x, y \in G: \paren {a \circ x \circ a^{-1} } \circ \paren {a \circ y \circ a^{-1} } = a \circ \paren {x \circ y} \circ a^{-1}$ That is, the product of conjugates is equal to the conjugate of the product.


Power of Product with Inverse

Let $G$ be a group whose identity is $e$. Let $a, b \in G: a b = b a^{-1}$. Then: : $\forall n \in \Z: a^n b = b a^{-n}$


Powers of Elements in Group Direct Product

Let $\left({G, \circ_1}\right)$ and $\left({H, \circ_2}\right)$ be group whose identities are $e_G$ and $e_H$. Let $\left({G \times H, \circ}\right)$ be the group direct product (either external or internal) of $G$ and $H$. Then: : $\forall n \in \Z: \forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$


Powers of Commutative Elements in Semigroups

Let $\left ({S, \circ}\right)$ be a semigroup. Let $a, b \in S$ both be cancellable elements of $S$. Then the following results hold:


Powers of Commutative Elements in Monoids

These results are an extension of the results in Powers of Commutative Elements in Semigroups in which the domain of the indices is extended to include all integers. Let $\left ({S, \circ}\right)$ be a monoid whose identity is $e_S$. Let $a, b \in S$ be invertible elements for $\circ$ that also commute. Then the following results hold.


Powers of Commutative Elements in Groups

Let $\struct {G, \circ}$ be a group. Let $a, b \in G$ such that $a$ and $b$ commute. Then the following results hold:


General Morphism Property for Semigroups

Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups. Let $\phi: S \to T$ be a homomorphism. Then: :$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ Hence it follows that: :$\forall n \in \N_{>0}: \forall s \in S: \map \phi {s^n} = \paren {\map \phi s}^n$


Homomorphism of Power of Group Element

Let $\struct {G, \circ}$ and $\struct {H, \ast}$ be groups. Let $\phi: S \to T$ be a group homomorphism. Then: : $\forall n \in \Z: \forall g \in G: \map \phi {g^n} = \paren {\map \phi g}^n$


Finite Subgroup Test

Let $\struct {G, \circ}$ be a group. Let $H$ be a non-empty finite subset of $G$. Then: :$H$ is a subgroup of $G$ {{iff}}: :$\forall a, b \in H: a \circ b \in H$ That is, a non-empty finite subset of $G$ is a subgroup {{iff}} it is closed.


Powers of Element form Subgroup

Let $\struct {G, \circ}$ be a group. Then: :$\forall a \in G: H = \set {a^n: n \in \Z} \le G$ That is, the subset of $G$ comprising all elements possible as powers of $a \in G$ is a subgroup of $G$.


Homomorphism of Generated Group

Let $\left({G, \circ}\right)$ and $\left({H, \circ}\right)$ be groups. Let $\phi: G \to H$ and $\psi: G \to H$ be homomorphisms.  Let $\left \langle {S} \right \rangle = G$ be the group generated by $S$. Let $\forall x \in S: \phi \left({x}\right) = \psi \left({x}\right)$ Then $\phi = \psi$.


Set of Words Generates Group

Let $S \subseteq G$ where $G$ is a group. Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$. Then $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the set of words of $\hat S$.


Subset Product is Subset of Generator

Let $\struct {G, \circ}$ be a group. Let $X, Y \subseteq \struct {G, \circ}$. Then $X \circ Y \subseteq \gen {X, Y}$ where: :$X \circ Y$ is the Subset Product of $X$ and $Y$ in $G$. :$\gen {X, Y}$ is the subgroup of $G$ generated by $X$ and $Y$.


Order of Subset Product with Singleton

Let $\struct {G, \circ}$ be a group. Let $X, Y \subseteq \struct {G, \circ}$ such that $X$ is a singleton: :$X = \set x$ Then: :$\order {X \circ Y} = \order Y = \order {Y \circ X}$ where $\order S$ is defined as the order of $S$.


Product of Subset with Intersection

Let $\struct {G, \circ}$ be an algebraic structure. Let $X, Y, Z \subseteq G$. Then: :$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$ :$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$ where $X \circ Y$ denotes the subset product of $X$ and $Y$.


Order of Subgroup Product

Let $G$ be a group. Let $H$ and $K$ be subgroups of $G$. Then: :$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$ where: :$H K$ denotes subset product :$\order H$ denotes the order of $H$.


Index of Intersection of Subgroups

Let $G$ be a group. Let $H, K$ be subgroups of finite index of $G$. Then: :$\index G {H \cap K} \le \index G H \index G K$ where $\index G H$ denotes the index of $H$ in $G$. Note that here the symbol $\le$ is being used with its meaning '''less than or equal to'''. Equality holds {{iff}} $H K = \set {h k: h \in H, k \in K} = G$.


Intersection of Subgroups of Prime Order

Let $G$ be a group whose identity is $e$. Let $H$ and $K$ be subsets of $G$ such that: :$\order H = \order K = p$ :$H \ne K$ :$p$ is prime. Then: : $H \cap K = \set e$ That is, the intersection of two unequal subgroups of a group, both of whose order is the same prime, consists solely of the identity.


Tower Law for Subgroups

Let $\struct {G, \circ}$ be a group. Let $H$ be a subgroup of $G$ with finite index. Let $K$ be a subgroup of $H$. Then: :$\index G K = \index G H \index H K$ where $\index G H$ denotes the index of $H$ in $G$.


Morphism from Integers to Group

Let $G$ be a group whose identity is $e$. Let $g \in G$. Let $\phi: \Z \to G$ be the mapping defined as: :$\forall n \in \Z: \map \phi n = g^n$. Then: : If $g$ has infinite order, then $\phi$ is a group isomorphism from $\struct {\Z, +}$ to $\gen g$. : If $g$ has finite order such that $\order g = m$, then $\phi$ is a group epimorphism from $\struct {\Z, +}$ to $\gen g$ whose kernel is the principal ideal $\paren m$. :Thus $\gen g$ is isomorphic to $\struct {\Z, +}$, and $m$ is the smallest (strictly) positive integer such that $g^m = e$.


Identity is Only Group Element of Order 1

In every group, the identity, and only the identity, has order $1$.


Group Element is Self-Inverse iff Order 2

Let $\struct {S, \circ}$ be a group whose identity is $e$. An element $x \in \struct {S, \circ}$ is self-inverse {{iff}}: :$\order x = 2$


Powers of Infinite Order Element

Let $G$ be a group whose identity is $e$. Let $a \in G$ have infinite order in $G$. Then: :$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$


Element of Finite Group is of Finite Order

In any finite group, each element has finite order.


Inverse Element is Power of Order Less 1

Let $G$ be a group whose identity is $e$. Let $g \in G$ be of finite order. Then: : $\order g = n \implies g^{n - 1} = g^{-1}$


Equal Powers of Finite Order Element

:$g^r = g^s \iff k \divides \paren {r - s}$


Order of Element Divides Order of Finite Group

In a finite group, the order of a group element divides the order of its group: :$\forall x \in G: \order x \divides \order G$


Element to Power of Group Order is Identity

Let $G$ be a group whose identity is $e$ and whose order is $n$. Then: :$\forall g \in G: g^n = e$


Boolean Group is Abelian

Let $G$ be a Boolean group. Then $G$ is abelian.


Order of Group Element equals Order of Inverse

Let $G$ be a group whose identity is $e$. Then: : $\forall x \in G: \order x = \order {x^{-1} }$ where $\order x$ denotes the order of $x$.


Order of Conjugate Element equals Order of Element

Let $\struct {G, \circ}$ be a group whose identity is $e$. Then :$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$ where $\order a$ denotes the order of $a$ in $G$.


Order of Homomorphic Image of Group Element

Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively. Let $\phi: G \to H$ be a homomorphism. Let $g \in G$ be of finite order. Then: :$\forall g \in G: \order {\map \phi g} \divides \order g$ where $\divides$ denotes divisibility.


Non-Trivial Group has Non-Trivial Cyclic Subgroup

Let $G$ be a group whose identity element is $e$. Let $g \in G$. If $g$ has infinite order, then $\gen g$ is an infinite cyclic group. If $\order g = n$, then $\gen g$ is a cyclic group with $n$ elements. Thus, every group which is non-trivial has at least one cyclic subgroup which is also non-trivial. In the case that $G$ is itself cyclic, that cyclic subgroup may of course be itself.


Epimorphism from Integers to Cyclic Group

Let $\gen a = \struct {G, \circ}$ be a cyclic group. Let $f: \Z \to G$ be a mapping defined as: $\forall n \in \Z: \map f n = a^n$. Then $f$ is a (group) epimorphism from $\struct {\Z, +}$ onto $\gen a$.


Cyclic Group is Abelian

Let $G$ be a cyclic group. Then $G$ is abelian.


Cyclic Groups of Same Order are Isomorphic

Two cyclic groups of the same order are isomorphic to each other.


Order of Subgroup of Cyclic Group

Let $C_n = \gen g$ be the cyclic group of order $n$ which is generated by $g$ whose identity is $e$. Let $a \in C_n: a = g^i$. Let $H = \gen a$. Then: :$\order H = \dfrac n {\gcd \set {n, i} }$ where: :$\order H$ denotes the order of $H$ :$\gcd \set {n, i}$ denotes the greatest common divisor of $n$ and $i$.


Number of Powers of Cyclic Group Element

Let $G$ be a cyclic group of order $n$, generated by $g$. Let $d \divides n$. Then the element $g^{n/d}$ has $d$ distinct powers.


Subgroup of Finite Cyclic Group is Determined by Order

Let $G = \gen g$ be a cyclic group whose order is $n$ and whose identity is $e$. Let $d \divides n$, where $\divides$ denotes divisibility. Then there exists exactly one subgroup $G_d = \gen {g^{n / d} }$ of $G$ with $d$ elements.


Prime Group is Cyclic

Let $p$ be a prime number. Let $G$ be a group whose order is $p$. Then $G$ is cyclic.


Group of Order less than 6 is Abelian

All groups with less than $6$ elements are abelian.


Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order

Let $C_n$ be the cyclic group of order $n$. Let $C_n = \gen a$, that is, that $C_n$ is generated by $a$. Then: :$C_n = \gen {a^k} \iff k \perp n$ That is, $C_n$ is also generated by $a^k$ {{iff}} $k$ is coprime to $n$.


Order of Conjugate of Subgroup

Let $G$ be a group. Let $H$ be a subgroup of $G$ such that $H$ is of finite order. Then $\order {H^a} = \order H$.


Subgroup of Index 2 is Normal

A subgroup of index $2$ is always normal.


Intersection of Normal Subgroups is Normal

Let $G$ be a group. Let $I$ be an indexing set. Let $\family {N_i}_{i \mathop \in I}$ be a non-empty indexed family of normal subgroups of $G$. Then $\displaystyle \bigcap_{i \mathop \in I} N_i$ is a normal subgroup of $G$.


Union of Conjugacy Classes is Normal

Let $G$ be a group. Let $H \le G$. Then $H$ is normal in $G$ {{iff}} $H$ is a union of conjugacy classes of $G$.


Unique Subgroup of a Given Order is Normal

Let a group $G$ have only one subgroup of a given order. Then that subgroup is normal.


Smallest Normal Subgroup containing Set

Let $S \subseteq G$ where $G$ is a group. Then there exists a unique smallest normal subgroup of $G$ which contains $S$.


Conjugate of Set with Inverse Closed for Inverses

Let $G$ be a group. Let $S \subseteq G$. Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$. Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$. That is, $\tilde S$ is the set containing all the conjugates of the elements of $S$ and all their inverses. Then: : $\forall x \in \tilde S: x^{-1} \in \tilde S$


Conjugate of Set with Inverse is Closed

Let $G$ be a group. Let $S \subseteq G$. Let $\hat S = S \cup S$. Let $\tilde S = \left\{{a s a^{-1}: s \in \hat S, a \in G}\right\}$. Let $W \left({\tilde S}\right)$ be the set of words of $\tilde S$. Then $\forall w \in W \left({\tilde S}\right): \forall a \in G: a w a^{-1} \in W \left({\tilde S}\right)$.


Generator of Normal Subgroup

Let $G$ be a group. Let $S \subseteq G$. Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$. Let $\tilde S = \left\{{a s a^{-1}: s \in \hat S, a \in G}\right\}$. Let $W \left({\tilde S}\right)$ be the set of words of $\tilde S$. Let $N$ be the smallest normal subgroup of $G$ that contains $S$. Then $N = \left \langle {S} \right \rangle = W \left({\tilde S}\right)$.


Subset Product with Normal Subgroup as Generator

Let $G$ be a group whose identity is $e$. Let: :$H$ be a subgroup of $G$ :$N$ be a normal subgroup of $G$. Then: :$N \lhd \gen {N, H} = N H = H N \le G$ where: :$\le$ denotes subgroup :$\lhd$ denotes normal subgroup :$\gen {N, H}$ denotes a subgroup generator :$N H$ denotes subset product.


Subset Product of Normal Subgroups is Normal

Let $\struct {G, \circ}$ be a group. Let $N$ and $N'$ be normal subgroups of $G$. Then $N N'$ is also a normal subgroup of $G$.


Prime Group is Simple

Groups of prime order are simple.


Prime Group has no Proper Subgroups

A nontrivial group $G$ has no proper subgroups except the trivial group {{iff}} $G$ is finite and its order is prime.


Quotient Group of Cyclic Group

Let $G$ be a cyclic group which is generated by $g$. Let $H$ be a subgroup of $G$. Then $g H$ generates $G / H$.


Order of Element in Quotient Group

Let $G$ be a group, and let $H$ be a normal subgroup of $G$. Let $G / H$ be the quotient group of $G$ by $H$. The order of $a H \in G / H$ divides the order of $a \in G$.


Quotient of Group by Center Cyclic implies Abelian

Let $G$ be a group. Let $\map Z G$ be the center of $G$. Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$. Let $G / \map Z G$ be cyclic. Then $G$ is abelian, so $G = \map Z G$. That is, the group $G / \map Z G$ cannot be a cyclic group which is non-trivial.


Centralizer is Normal Subgroup of Normalizer

Let $G$ be a group. Let $H \le G$ be a subgroup of $G$.  Let $\map {C_G} H$ be the centralizer of $H$ in $G$. Let $\map {N_G} H$ be the normalizer of $H$ in $G$. Let $\Aut H$ be the automorphism group of $H$. Then: :$(1): \quad \map {C_G} H \lhd \map {N_G} H$ :$(2): \quad \map {N_G} H / \map {C_G} H \cong K$ where: :$\map {N_G} H / \map {C_G} H$ is the quotient group of $\map {N_G} H$ by $\map {C_G} H$ :$K$ is a subgroup of $\Aut H$.


Number of Distinct Conjugate Subsets is Index of Normalizer

Let $G$ be a group. Let $S$ be a subset of $G$. Let $\map {N_G} S$ be the normalizer of $S$ in $G$. Let $\index G {\map {N_G} S}$ be the index of $\map {N_G} S$ in $G$. The number of distinct subsets of $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$.


Element of Group Not Conjugate to Proper Subgroup

Let $G$ be a finite group. Let $H$ be a proper subgroup of $G$. Then there is at least one element of $G$ not contained in $H$ or in any of its conjugates.


Second Isomorphism Theorem

=== Groups === {{:Second Isomorphism Theorem/Groups}} === Rings === {{:Second Isomorphism Theorem/Rings}} This result is also referred to by some sources as the '''first isomorphism theorem'''.


Third Isomorphism Theorem

=== Groups === {{:Third Isomorphism Theorem/Groups}} === Rings === {{:Third Isomorphism Theorem/Rings}}


Inverse of Inner Automorphism

Let $G$ be a group. Let $x \in G$. Let $\kappa_x$ be the inner automorphism of $G$ given by $x$. Then: : $\paren {\kappa_x}^{-1} = \kappa_{x^{-1} }$


Conjugates of Elements in Centralizer

Let $G$ be a group. Let $\map {C_G} a$ be the centralizer of $a$ in $G$. Then $\forall g, h \in G: g a g^{-1} = h a h^{-1}$ {{iff}} $g$ and $h$ belong to the same left coset of $\map {C_G} a$.


Number of Conjugates is Number of Cosets of Centralizer

Let $G$ be a group. Let $\map {C_G} a$ be the centralizer of $a$ in $G$. Then the number of different conjugates of $a$ in $G$ equals the number of different (left) cosets of $\map {C_G} a$: :$\card {\conjclass a} = \index G {\map {C_G} a}$ where: :$\conjclass a$ is the conjugacy class of $a$ in $G$ :$\index G {\map {C_G} a}$ is the index of $\map {C_G} a$ in $G$. Consequently: :$\card {\conjclass a} \divides \order G$


Size of Conjugacy Class is Index of Normalizer

Let $G$ be a group. Let $x \in G$. Let $\conjclass x$ be the conjugacy class of $x$ in $G$. Let $\map {N_G} x$ be the normalizer of $x$ in $G$. Let $\index G {\map {N_G} x}$ be the index of $\map {N_G} x$ in $G$. The number of elements in $\conjclass x$ is $\index G {\map {N_G} x}$.


Conjugacy Class of Element of Center is Singleton

Let $G$ be a group. Let $\map Z G$ denote the center of $G$. The elements of $\map Z G$ form singleton conjugacy classes, and the elements of $G \setminus \map Z G$ belong to multi-element conjugacy classes.


Conjugacy Class Equation

Let $G$ be a group. Let $\order G$ denote the order of $G$. Let $\map Z G$ denote the center of $G$. Let $x \in G$. Let $\map {N_G} x$ denote the normalizer of $x$ in $G$. Let $\index G {\map {N_G} x}$ denote the index of $\map {N_G} x$ in $G$. Let $m$ be the number of non-singleton conjugacy classes of $G$. Let $x_j: j \in \set {1, 2, \ldots, m}$ be arbitrary elements of those conjugacy classes. Then: :$\displaystyle \order G = \order {\map Z G} + \sum_{j \mathop = 1}^m \index G {\map {N_G} {x_j} }$


Group of Order Prime Squared is Abelian

A group whose order is the square of a prime is abelian.


Center of Group of Prime Power Order is Non-Trivial

Let $G$ be a group whose order is the power of a prime. Then the center of $G$ is non-trivial: :$\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e$


Center of Group of Order Prime Cubed

Let $G$ be a group of order $p^3$, where $p$ is a prime. Let $\map Z G$ be the center of $G$. Then $\order {\map Z G} \ne p^2$.


Prime Power Group has Non-Trivial Proper Normal Subgroup

Let $G$ be a group, whose identity is $e$, such that $\order G = p^n: n > 1, p \in \mathbb P$. Then $G$ has a proper normal subgroup which is non-trivial.


Composition Series of Group of Prime Power Order

Let $G$ be a group whose identity is $e$, and whose order is a prime power: :$\order G = p^n, p \in \mathbb P, n \ge 1$ Then $G$ has a composition series: :$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$ such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is cyclic and of order $p$.


Sum Rule for Counting

Let there be: : $r_1$ different objects in the set $S_1$ : $r_2$ different objects in the set $S_2$ : $\ldots$ : $r_m$ different objects in the set $S_m$. Let $\displaystyle \bigcap_{i \mathop = 1}^m S_i = \varnothing$. Then the number of ways to select an object from one of the $m$ sets is $\displaystyle \sum_{i \mathop = 1}^m r_i$.


Product Rule for Counting

Let it be possible to choose an element $\alpha$ from a given set $S$ in $m$ different ways. Let it be possible to choose an element $\beta$ from a given set $T$ in $n$ different ways. Then the ordered pair $\tuple {\alpha, \beta}$ can be chosen from the cartesian product $S \times T$ in $m n$ different ways.


External Direct Product of Abelian Groups is Abelian Group

Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups. Then the group direct product $\struct {G \times H, \circ}$ is abelian {{iff}} both $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are abelian.


Group Direct Product of Cyclic Groups

Let $G$ and $H$ both be finite cyclic groups with orders $n = \order G$ and $m = \order H$ respectively. Then their group direct product $G \times H$ is cyclic {{iff}} $g$ and $h$ are coprime, that is, $g \perp h$.


Group Direct Product of Infinite Cyclic Groups

The group direct product of two infinite cyclic groups is not cyclic.


Order of Group Element in Group Direct Product

Let $G$ and $H$ be finite groups. Let $g \in G: \order g = m, h \in H: \order h = n$. Then $\order {\tuple {g, h} }$ in $G \times H$ is $\lcm \set {m, n}$.


Subgroup Product is Internal Group Direct Product iff Surjective

Let $G$ be a group. Let $\sequence {H_n}$ be a sequence of subgroups of $G$. Let $\displaystyle \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a mapping defined by: :$\displaystyle \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$ Then $\phi$ is surjective {{iff}}: : $\displaystyle G = \prod_{k \mathop = 1}^n H_k$ That is, {{iff}} $G$ is the internal group direct product of $H_1, H_2, \ldots, H_n$.


Internal Group Direct Product is Injective

Let $G$ be a group whose identity is $e$. Let $H_1, H_2$ be subgroups of $G$. Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by: :$\map \phi {h_1, h_2} = h_1 h_2$ Then $\phi$ is injective {{iff}}: :$H_1 \cap H_2 = \set e$


Internal Group Direct Product Isomorphism

Let $G$ be a group. Let $H_1, H_2$ be subgroups of $G$. Let $\phi: H_1 \times H_2 \to G$ be the mapping defined by $\map \phi {h_1, h_2} := h_1 h_2$. If $\phi$ is a (group) isomorphism, then both $H_1$ and $H_2$ are normal subgroups of $G$.


Internal Group Direct Product of Normal Subgroups

Let $G$ be a group whose identity is $e$. Let $H_1, H_2$ be subgroups of $G$. Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by $\map \phi {h_1, h_2} = h_1 h_2$. Let $H_1$ and $H_2$ be normal subgroups of $G$, and let $H_1 \cap H_2 = \set e$. Then $\phi$ is a (group) homomorphism.


Internal Direct Product Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $H_1, H_2 \le G$. Then $G$ is the internal group direct product of $H_1$ and $H_2$ {{iff}}: :$(1): \quad G = H_1 \circ H_2$ :$(2): \quad H_1 \cap H_2 = \set e$ :$(3): \quad H_1, H_2 \lhd G$ where $H_1 \lhd G$ denotes that $H_1$ is a normal subgroup of $G$.


Inclusion Mapping is Surjection iff Identity

Let $T$ be a set. Let $S\subseteq T$ be a subset. Let $i_S: S \to T$ be the inclusion mapping. Then: :$i_S: S \to T$ is surjective {{iff}} $i_S: S \to T = I_S: S \to S$ where $I_S: S \to S$ denotes the identity mapping on $S$. Alternatively, this theorem can be worded as: :$i_S: S \to S = I_S: S \to S$ It follows directly that from Surjection by Restriction of Codomain, the surjective restriction of $i_S: S \to T$ to $i_S: S \to \Img {i_S}$ is itself the identity mapping.


Internal Direct Product Generated by Subgroups

Let $G$ be a group whose identity is $e$. Let $\sequence {H_n}$ be a sequence of subgroups of $G$. Then: :the subgroup generated by $\displaystyle \bigcup_{k \mathop = 1}^n H_k$ is the internal group direct product of $\sequence {H_n}$ {{iff}}: :$\sequence {H_n}$ is an independent sequence of subgroups such that every element of $H_i$ commutes with every element of $H_j$ whenever $1 \le i < j \le n$.


Internal Group Direct Product Commutativity

Let $\struct {G, \circ}$ be a group whose identity is $e$. Let $H_1, H_2 \le G$. Let $\struct {G, \circ}$ be the internal group direct product of $H_1$ and $H_2$. Then: :$\forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$


Internal and External Group Direct Products are Isomorphic

Let $G$ be a group whose identity is $e$. Then $G$ is the (external) group direct product of $G_1, G_2, \ldots, G_n$ {{iff}} $G$ is the internal group direct product of $N_1, N_2, \ldots, N_n$ such that: :$\forall i \in \N_n: N_i \cong G_i$ where: :$\cong$ denotes (group) isomorphism :$\N_n$ denotes $\set {1, 2, \ldots, n}$


Pullback of Quotient Group Isomorphism is Subgroup

Let $\struct {G, \circ}$ be a group whose identity element is $e_G$. Let $\struct {H, *}$ be a group whose identity element is $e_H$. Let $N \lhd G, K \lhd H$ be normal subgroups of $G$ and $H$ respectively. Let: :$G / N \cong H / K$ where: :$G / N$ denotes the quotient of $G$ by $N$ :$\cong$ denotes group isomorphism. Let $\theta: G / N \to H / K$ be such a group isomorphism. Let $G \times^\theta H$ be the pullback of $G$ and $H$ via $\theta$. Then $G \times^\theta H$ is a subgroup of $G \times H$.


Group/Examples/x+y over 1+xy

Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$. Let $\circ: G \times G \to \R$ be the binary operation defined as: :$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$ The algebraic structure $\struct {G, \circ}$ is a group.


Group/Examples/inv x = 1 - x

Let $S = \set {x \in \R: 0 < x < 1}$. Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$.


Group/Examples/Self-Inverse and Cancellable Elements

Let $S$ be a set with an operation which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that: : $(1): \quad \exists e \in S: a \ast b = e \iff a = b$ : $(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$ Then $\struct {S, \circ}$ is a group, where $\circ$ is defined as $a \circ b = a \ast \paren {e \ast b}$.


Complex Numbers under Addition form Abelian Group

Let $\C$ be the set of complex numbers. The structure $\struct {\C, +}$ is an infinite abelian group.


Non-Zero Complex Numbers under Multiplication form Abelian Group

Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is: :$\C_{\ne 0} = \C \setminus \set 0$ The structure $\struct {\C_{\ne 0}, \times}$ is an infinite abelian group.


Real Numbers under Addition form Abelian Group

Let $\R$ be the set of real numbers. The structure $\struct {\R, +}$ is an infinite abelian group.


Non-Zero Real Numbers under Multiplication form Abelian Group

Let $\R_{\ne 0}$ be the set of real numbers without zero: :$\R_{\ne 0} = \R \setminus \set 0$ The structure $\struct {\R_{\ne 0}, \times}$ is an uncountable abelian group.


Rational Numbers under Addition form Abelian Group

Let $\Q$ be the set of rational numbers. The structure $\struct {\Q, +}$ is a countably infinite abelian group.


Non-Zero Rational Numbers under Multiplication form Abelian Group

Let $\Q_{\ne 0}$ be the set of non-zero rational numbers: :$\Q_{\ne 0} = \Q \setminus \set 0$ The structure $\struct {\Q_{\ne 0}, \times}$ is a countably infinite abelian group.


Integers under Multiplication form Countably Infinite Commutative Monoid

The set of integers under multiplication $\struct {\Z, \times}$ is a countably infinite commutative monoid.


Additive Group of Rationals is Subgroup of Reals

Let $\struct {\Q, +}$ be the additive group of rational numbers. Let $\struct {\R, +}$ be the additive group of real numbers. Then $\struct {\Q, +}$ is a normal subgroup of $\struct {\R, +}$.


Additive Group of Integers is Subgroup of Rationals

Let $\struct {\Z, +}$ be the additive group of integers. Let $\struct {\Q, +}$ be the additive group of rational numbers. Then $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.


Multiplicative Group of Reals is Subgroup of Complex

Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers. Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers. Then $\struct {\R_{\ne 0}, \times}$ is a normal subgroup of $\struct {\C_{\ne 0}, \times}$.


Multiplicative Group of Rationals is Subgroup of Reals

Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers. Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers. Then $\struct {\Q_{\ne 0}, \times}$ is a normal subgroup of $\left({\R_{\ne 0}, \times}\right)$.


Circle Group is Infinite Abelian Group

The circle group $\struct {K, \times}$ is an uncountably infinite abelian group under the operation of complex multiplication.


Homomorphism from Reals to Circle Group

Let $\struct {\R, +}$ be the additive group of real numbers. Let $\struct {K, \times}$ be the circle group. Let $\phi: \struct {\R, +} \to \struct {K, \times}$ be the mapping defined as: :$\forall x \in \R: \map \phi x = e^{i x}$ Then $\phi$ is a (group) homomorphism.


Integers under Addition form Infinite Cyclic Group

The additive group of integers $\struct {\Z, +}$ is an infinite cyclic group which is generated by the element $1 \in \Z$.


Inverse of Generator of Cyclic Group is Generator

Let $\gen g = G$ be a cyclic group. Then: :$G = \gen {g^{-1} }$ where $g^{-1}$ denotes the inverse of $g$. Thus, in general, a generator of a cyclic group is not unique.


Generators of Infinite Cyclic Group

Let $\gen g = G$ be an infinite cyclic group. Then the only other generator of $G$ is $g^{-1}$. Thus an infinite cyclic group has exactly $2$ generators.


Subgroup of Integers is Ideal

Let $\struct {\Z, +}$ be the additive group of integers. Every subgroup of $\struct {\Z, +}$ is an ideal of the ring $\struct {\Z, +, \times}$.


Additive Group of Integers is Subgroup of Reals

Let $\struct {\Z, +}$ be the additive group of integers. Let $\struct {\R, +}$ be the additive group of real numbers. Then $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$.


Quotient Group of Reals by Integers is Circle Group

Let $\struct {\Z, +}$ be the additive group of integers. Let $\struct {\R, +}$ be the additive group of real numbers. Let $K$ be the circle group. Then the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$ is isomorphic to $K$.


Integers Modulo m under Addition form Cyclic Group

Let $\Z_m$ be the set of integers modulo $m$. Let $+_m$ be the operation of addition modulo $m$. Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$. Then $\struct {\Z_m, +_m}$ is a cyclic group of order $m$, generated by the element $\eqclass 1 m \in \Z_m$.


Integers Modulo m under Multiplication form Commutative Monoid

The structure: :$\struct {\Z_m, \times}$ (where $\Z_m$ is the set of integers modulo $m$) is a commutative monoid.


Multiplicative Inverse in Ring of Integers Modulo m

Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$. Then $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$.


Reduced Residue System is Subset of Set of All Residue Classes

Let $\Z_m$ be the set of set of residue classes modulo $m$. Let $\Z'_m$ be the reduced residue system modulo $m$. Then: :$\forall m \in \Z_{> 1}: \O \subset \Z'_m \subset \Z_m$


Reduced Residue System under Multiplication forms Abelian Group

Let $\Z_m$ be the set of set of residue classes modulo $m$. Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$. Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$.


Ring of Integers Modulo Prime is Field

Let $m \in \Z: m \ge 2$. Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$. Then: :$m$ is prime {{iff}}: :$\struct {\Z_m, +, \times}$ is a field.


Subgroups of Additive Group of Integers

Let $\struct {\Z, +}$ be the additive group of integers. Let $n \Z$ be the additive group of integer multiples of $n$. Every non-trivial subgroup of $\struct {\Z, +}$ has the form $n \Z$.


Integer Multiples under Addition form Infinite Cyclic Group

Let $n \Z$ be the set of integer multiples of $n$. Then $\struct {n \Z, +}$ is a countably infinite cyclic group. It is generated by $n$ and $-n$: :$n \Z = \gen n$ :$n \Z = \gen {-n}$ Hence $\struct {n \Z, +}$ can be justifiably referred to as the additive group of integer multiples.


Quotient Group of Integers by Multiples

Let $\struct {\Z, +}$ be the additive group of integers. Let $\struct {m \Z, +}$ be the additive group of integer multiples of $m$. Let $\struct {\Z_m, +_m}$ be the additive group of integers modulo $m$. Then the quotient group of $\struct {\Z, +}$ by $\struct {m \Z, +}$ is $\struct {\Z_m, +_m}$. Thus: :$\index \Z {m \Z} = m$


Euler's Theorem

Let $a, m \in \Z$ be coprime integers: $a \perp m$. Let $\map \phi m$ be the Euler $\phi$ function of $m$. Then: :$a^{\map \phi m} \equiv 1 \pmod m$


Symmetry Group is Group

Let $P$ be a geometric figure. Let $S_P$ be the set of all symmetries of $P$. Let $\circ$ denote composition of mappings. The symmetry group $\struct {S_P, \circ}$ is indeed a group.


Area of Square

A square has an area of $L^2$ where $L$ is the length of a side of the square. Thus we have that the area is a function of the length of the side: :$\forall L \in \R_{\ge 0}: \map \Area L = L^2$ where it is noted that the domain of $L$ is the set of non-negative real numbers.


Order of Symmetric Group

Let $S$ be a finite set of cardinality $n$. Let $\struct {\map \Gamma S, \circ}$ be the symmetric group on $S$. Then $\struct {\map \Gamma S, \circ}$ has $n!$ elements (see factorial).


Powers of Permutation Element

Let $S_n$ denote the symmetric group on $n$ letters. Let $\pi \in S_n$, and let $i \in \N^*_n$. Let $k \in \Z: k > 0$ be the smallest such that: : $\pi^k \left({i}\right) \in \left\{{i, \pi \left({i}\right), \pi^2 \left({i}\right), \ldots, \pi^{k-1} \left({i}\right)}\right\}$ Then $\pi^k \left({i}\right) = i$.


Equality of Cycles

Let $S_n$ denote the symmetric group on $n$ letters, realised as the permutations of $\left\{{1, \ldots, n}\right\}$. Let $\rho = \begin{bmatrix} a_0 & \cdots & a_{k-1} \end{bmatrix}$, $\sigma = \begin{bmatrix} b_0 & \cdots & b_{k-1} \end{bmatrix} \in S_n$ be $k$-cycles of $S_n$. For any $d \in \Z$, by Integer is Congruent to Integer less than Modulus we can associate to $d$ a unique integer $\tilde d \in \left\{{0, \ldots, k-1}\right\}$ such that $d \equiv \tilde d \pmod k$. Define $a_d$ and $b_d$ for any $d \in \Z$ by $a_d = a_{\tilde d}$ and $b_d = b_{\tilde d}$ Choose $i, j \in \left\{{1, \ldots, k}\right\}$ such that: :$\displaystyle a_i = \min\left\{ {a_0, \ldots, a_{k-1} }\right\}$ :$\displaystyle b_j = \min\left\{ {b_0, \ldots, b_{k-1} }\right\}$ Then $\rho = \sigma$ {{iff}} for all $d \in \Z$, $a_{i + d} = b_{j + d}$. That is, $\rho = \sigma$ {{iff}} they are identical when written with the lowest element first.


Identity Permutation is Disjoint from All

Let $S_n$ denote the symmetric group on $n$ letters. Let $e \in S_n$ be the identity permutation on $S_n$. Then $e$ is disjoint from every permutation $\pi$ on $S_n$ (including itself).


Disjoint Permutations Commute

Let $S_n$ denote the symmetric group on $n$ letters. Let $\rho, \sigma \in S_n$ such that $\rho$ and $\sigma$ are disjoint. Then $\rho \sigma = \sigma \rho$.


Permutation Induces Equivalence Relation

Let $S_n$ denote the symmetric group on $n$ letters. Let $\pi \in S_n$. Let $\mathcal R_\pi$ be the relation defined by: :$i \mathrel {\mathcal R_\pi} j \iff \exists k \in \Z: \map {\pi^k} i = j$ Then $\mathcal R_\pi$ is an equivalence relation.


Existence and Uniqueness of Cycle Decomposition

Let $S_n$ denote the symmetric group on $n$ letters. Every element of $S_n$ may be uniquely expressed as a cycle decomposition, up to the order of factors.


Order of Product of Disjoint Permutations

Let $S_n$ denote the symmetric group on $n$ letters. Let $\pi$ be a product of disjoint permutations of orders $k_1, k_2, \ldots, k_r$. Then: :$\order \pi = \lcm \set {k_1, k_2, \ldots, k_r}$ where: :$\order \pi$ denotes the order of $\pi$ in $S_n$ :$\lcm$ denotes lowest common multiple.


Group Action defines Permutation Representation

Let $\map \Gamma X$ be the set of permutations on a set $X$. Let $G$ be a group. Let $\phi: G \times X \to X$ be a group action. For $g \in G$, let $\phi_g: X \to X$ be the mapping defined as: :$\map {\phi_g} x = \map \phi {g, x}$ Let $\tilde \phi: G \to \map \Gamma X$ be the mapping associated to $\phi$, defined by: :$\map {\tilde \phi} g := \phi_g$ Then $\tilde \phi$ is a group homomorphism.


Group Action determines Bijection

Let $*$ be a group action of $G$ on $X$. Then each $g \in G$ determines a bijection $\phi_g: X \to X$ given by: :$\map {\phi_g} x = g * x$ Its inverse is: :$\phi_{g^{-1} }: X \to X$. These bijection are sometimes called '''transformations''' of $X$.


Group Action Induces Equivalence Relation

Let $G$ be a group whose identity is $e$. Let $X$ be a set. Let $*: G \times S \to S$ be a group action. Let $\mathcal R_G$ be the relation induced by $G$, that is: :$x \mathrel {\mathcal R_G} y \iff y \in \Orb x$ where: :$\Orb x$ denotes the orbit of $x \in X$. Then: :$\mathcal R_G$ is an equivalence relation. :The equivalence class of an element is its orbit.


Partition Equation

Let group $G$ act on a finite set $X$. Let the distinct orbits of $X$ under the action of $G$ be: :$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$ Then: :$\card X = \card {\Orb {x_1} } + \card {\Orb {x_2} } + \cdots + \card {\Orb {x_s} }$


Definition:Stabilizer

Let $G$ be a group. Let $X$ be a set. Let $*: G \times X \to X$ be a group action. For each $x \in X$, the '''stabilizer of $x$ by $G$''' is defined as: :$\Stab x := \set {g \in G: g * x = x}$ where $*$ denotes the group action.


Stabilizer is Subgroup

Let $\struct {G, \circ}$ be a group which acts on a set $X$. Let $\Stab x$ be the stabilizer of $x$ by $G$. Then for each $x \in X$, $\Stab x$ is a subgroup of $G$.


Orbit-Stabilizer Theorem

Let $G$ be a group which acts on a finite set $X$. Let $x \in X$. Let $\Orb x$ denote  the orbit of $x$. Let $\Stab x$ denote the stabilizer of $x$ by $G$. Let $\index G {\Stab x}$ denote the index of $\Stab x$ in $G$. Then: :$\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$


Action of Group on Coset Space is Group Action

Let $G$ be a group whose identity is $e$. Let $H$ be a subgroup of $G$. Let $*: G \times G / H \to G / H$ be the action on the (left) coset space: :$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$ Then $G$ is a group action.


Group Action on Sets with k Elements

Let $\struct {G, \circ}$ be a finite group whose identity is $e$. Let $\Bbb S = \set {S \subseteq G: \card S = k}$, that is, the set of all of subsets of $G$ which have exactly $k$ elements. Let $G$ act on $\Bbb S$ by the rule: :$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$ This is a group action, and: : $\forall S \in \Bbb S: \order {\Stab S} \divides \card S$ where $\Stab S$ denotes the stabilizer of $S$ by $G$.