Cosets, Index and Lagrange's Theorem

Given: A group $G$, a subgroup $H\le G$, and elements $a,b,x\in G$.

The left coset $aH$ is $\{ah:h\in H\}$, and the right coset $Ha$ is $\{ha:h\in H\}$ (Left and right cosets $gH$ and $Hg$ of a subgroup).

A subgroup contains the identity and is closed under products and inverses (Subgroup).

In a group, $(uv)^{-1}=v^{-1}u^{-1}$ and $(u^{-1})^{-1}=u$ (In a group $e^{-1} = e$, $(g^{-1})^{-1} = g$ and $(gh)^{-1} = h^{-1}g^{-1}$, the order of the last product being essential).

Left and right cancellation hold in every group (Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$; equivalently left and right translation by $g$ are bijections of $G$, so $gx = h$ and $xg = h$ each have exactly one solution, Group and abelian group).

Given: A group $G$ and a subgroup $H\le G$.

A relation is an equivalence relation when it is reflexive, symmetric and transitive (Equivalence relation, equivalence class, and the quotient set $A/{\sim}$).

The equivalence classes of an equivalence relation on a set are nonempty, cover the set and are pairwise equal or disjoint (The equivalence classes of an equivalence relation are nonempty, cover $A$, and are pairwise equal or disjoint; conversely every such cover arises from exactly one equivalence relation).

For $a,b\in G$, $b\in aH$ if and only if $a^{-1}b\in H$, and $aH=bH$ if and only if $a^{-1}b\in H$ ($x\in aH$ iff $a^{-1}x\in H$, and $aH=bH$ iff $a^{-1}b\in H$).

Because $H\le G$, it contains the identity and is closed under inverses and products (Subgroup).

Given: A group $G$, a subgroup $H\le G$, and $g\in G$.

The cosets are $gH=\{gh:h\in H\}$ and $Hg=\{hg:h\in H\}$ (Left and right cosets $gH$ and $Hg$ of a subgroup).

A map is bijective when it is injective and surjective; two sets are equinumerous when a bijection between them exists (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A \preceq B$).

Left and right cancellation hold in a group (Cancellation in a group: $gx = gy$ or $xg = yg$ forces $x = y$; equivalently left and right translation by $g$ are bijections of $G$, so $gx = h$ and $xg = h$ each have exactly one solution).

Given: A group $G$ and a subgroup $H\le G$.

For left cosets, $gH=kH$ if and only if $g^{-1}k\in H$; for right cosets, $Hg=Hk$ if and only if $gk^{-1}\in H$ ($x\in aH$ iff $a^{-1}x\in H$, and $aH=bH$ iff $a^{-1}b\in H$).

In a group, $(uv)^{-1}=v^{-1}u^{-1}$ and $(u^{-1})^{-1}=u$ (In a group $e^{-1} = e$, $(g^{-1})^{-1} = g$ and $(gh)^{-1} = h^{-1}g^{-1}$, the order of the last product being essential).

A subgroup is closed under inverses (Subgroup).

A map with a two-sided inverse is a bijection (Injection, surjection, bijection, Equinumerous sets, $A \approx B$ and $A \preceq B$).

Given: A finite group $G$ and a subgroup $H\le G$.

Every subset of a finite set is finite (A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$, and equality holds if and only if $B = A$).

The power set of a finite set is finite ($\lvert\mathcal{P}(A)\rvert = 2^{\lvert A\rvert}$ for finite $A$).

Every left or right coset of $H$ is equinumerous with $H$ (Every left or right coset of $H$ is equinumerous with $H$).

A bijection transports finiteness and finite cardinality (The cardinality $\lvert A\rvert$ of a finite set).

The coset set is $G/H=\{gH:g\in G\}$, and its finite cardinality is the index (The coset set $G/H$ and the index $[G:H]$ of a subgroup, The left cosets of a subgroup partition the group).

Given: A finite group $G$ and a subgroup $H\le G$.

The distinct left cosets of $H$ partition $G$ (The left cosets of a subgroup partition the group).

The subgroup, every coset, and $G/H$ are finite; every coset has cardinality $|H|$ and $|G/H|=[G:H]$ (In a finite group, the subgroup, every coset and the set of cosets are finite, Every left or right coset of $H$ is equinumerous with $H$, The coset set $G/H$ and the index $[G:H]$ of a subgroup).

For a finite partition $A=\bigsqcup_{i\in I}A_i$, one has $|A|=\sum_{i\in I}|A_i|$; if every summand has the same value $c$, then $\sum_{i\in I}c=|I|c$ (The sum rule: a finite disjoint union is finite with $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$, and a sum over a finite index set splits along a partition, The sum $\sum_{i \in S} a_i$ over a finite index set, and its product form).

The order $|G|$ of a finite group is the unique natural equinumerous with its underlying set, hence agrees with finite cardinality (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, The cardinality $\lvert A\rvert$ of a finite set).

The embedding $\iota$ preserves multiplication, and $d\mid a$ in $\mathbb Z$ means $a=dq$ for some integer $q$ (The naturals embed in the integers, Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$).

Given: A finite group $G$ and an element $g\in G$.

The generated set $\langle g\rangle$ is a subgroup of $G$ and equals the set of integer powers of $g$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups, $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian).

If $g$ has finite order, then $\langle g\rangle$ is finite and $|\langle g\rangle|=\operatorname{ord}(g)$; every element of a finite group has finite order (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$, The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity).

Lagrange's theorem gives $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group and consequently $\iota(|H|)\mid\iota(|G|)$ (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$, Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$, The naturals embed in the integers).

Given: A finite group $G$ with identity $e$ and an element $g\in G$.

The order $n=\operatorname{ord}(g)$ is finite and $\iota(n)$ divides $\iota(|G|)$ (The order of every element of a finite group divides the order of the group, The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, The naturals embed in the integers).

If $\operatorname{ord}(g)=n$, then for every integer $k$, $g^k=e$ if and only if $n\mid k$ (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$, Powers $g^{n}$: natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$, Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$, and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

Given: A finite group $G$ with identity $e$, with $\iota(|G|)$ prime, and an element $g\in G$ with $g\ne e$.

A prime integer $p$ satisfies $p>1$, and every positive divisor of $p$ is $1$ or $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$).

The natural $\operatorname{ord}(g)$ is positive, equals $1$ exactly when $g=e$, and its image in $\mathbb Z$ divides $\iota(|G|)$; the embedding $\iota:\mathbb N\to\mathbb Z$ is injective and preserves order (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity, The order of every element of a finite group divides the order of the group, The naturals embed in the integers).

The cyclic subgroup $\langle g\rangle$ is a subgroup with $|\langle g\rangle|=\operatorname{ord}(g)$ (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups, $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian, If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$).

If $A\subseteq B$ are finite and $|A|=|B|$, then $A=B$ (A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$, and equality holds if and only if $B = A$).

If a finite set $G$ contains $e$ and $|G|\ne1$, then some element of $G$ differs from $e$: otherwise $G=\{e\}$, whose cardinality is $1$ (The cardinality $\lvert A\rvert$ of a finite set).

Given: A finite group $G$ and subgroups $K\le H\le G$.

Lagrange's theorem gives $|B|=[B:A]|A|$ whenever $A\le B$ and $B$ is finite (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$, The coset set $G/H$ and the index $[G:H]$ of a subgroup, The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity).

Every subgroup contains the identity; hence its underlying set is nonempty. Also, if $K\le H\le G$, then $K\le G$: one has $K\subseteq H\subseteq G$, and the identity, product, and inverse conditions for $K$ are the same inherited operations in $H$ and $G$ (Subgroup).

Every subset of a finite set is finite (A subset of a finite set is finite, with $\lvert B\rvert \le \lvert A\rvert$, and equality holds if and only if $B = A$).

Natural multiplication is associative, and $xz=yz$ with $z\ne0$ implies $x=y$ (Multiplication is associative, Cancellation for multiplication by a nonzero factor).

Given: A group $G$ and a subgroup $H\le G$.

The coset set is $G/H=\{gH:g\in G\}$. Its index is its finite cardinality when the coset set is finite and is the non-natural symbol $\infty$ otherwise; hence $[G:H]=1$ says that $G/H$ is finite of cardinality $1$ (The coset set $G/H$ and the index $[G:H]$ of a subgroup, Left and right cosets $gH$ and $Hg$ of a subgroup).

The cosets partition $G$, and $gH=H$ exactly when $g\in H$ (The left cosets of a subgroup partition the group, $x\in aH$ iff $a^{-1}x\in H$, and $aH=bH$ iff $a^{-1}b\in H$).

A finite set has cardinality $1$ exactly when it is a singleton: a bijection to $1=\{0\}$ has one fibre and hence one element, while the unique map from a singleton to $1$ is a bijection (The cardinality $\lvert A\rvert$ of a finite set).

Given: A positive integer $n$ and an integer $a$ with $\gcd(a,n)=1$.

The unit group $(\mathbb Z/n)^\times$ is finite of order $\varphi(n)$ and has identity $[1]_n$ (The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$).

The class $[a]_n$ is a unit if and only if $\gcd(a,n)=1$ (For $n\ge1$, $[a]_n$ is a unit if and only if $\gcd(a,n)=1$).

Every element $u$ of a finite group satisfies $u^{|G|}=e$ ($g^{|G|}=e$ for every element $g$ of a finite group $G$, Powers $g^{n}$: natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$).

Multiplication of residue classes satisfies $[x]_n[y]_n=[xy]_n$, so natural powers satisfy $[a]_n^k=[a^k]_n$; and $[x]_n=[y]_n$ exactly when $x\equiv y\pmod n$ (Addition and multiplication on $\mathbb{Z}/n$ by $[a]_n+[b]_n=[a+b]_n$ and $[a]_n[b]_n=[ab]_n$, The congruence class $[a]_n$ and the quotient set $\mathbb{Z}/n$, Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$, including the moduli $0$ and $1$, Congruent integers may be added, subtracted and multiplied: representative changes preserve both arithmetic operations).

Given: A prime integer $p$ and an integer $a$.

A prime satisfies $p>1$, and $p\nmid a$ implies $\gcd(p,a)=1$, hence $\gcd(a,p)=1$ by symmetry of the gcd (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$, For a prime $p$ and any integer $a$, $\gcd(p,a)$ is $p$ when $p \mid a$ and $1$ otherwise; so $p \nmid a$ makes $p$ and $a$ coprime, $\gcd$ is symmetric and unchanged by signs: $\gcd(a,b) = \gcd(b,a) = \gcd(|a|,|b|)$; moreover $\gcd(a,0) = |a|$, $\gcd(a,1) = 1$, $\gcd(a,a) = |a|$, and $\gcd(a,b) \ge 1$ unless $a = b = 0$).

Euler's theorem gives $a^{\varphi(p)}\equiv1\pmod p$ when $\gcd(a,p)=1$, and $\varphi(p)=p-1$ for prime $p$ (Euler's theorem: if $n\ge1$ and $\gcd(a,n)=1$, then $a^{\varphi(n)}\equiv1\pmod n$, $\varphi(1)=1$, and $\varphi(p)=p-1$ for every prime $p$).