Cosets and Lagrange's Theorem: Examples and Counterexamples

Given: A positive integer $n$ and the additive group $\mathbb Z$.

The set $n\mathbb Z=\{nk:k\in\mathbb Z\}$ is a subgroup of $(\mathbb Z,+)$ (Every subgroup of $(\mathbb{Z}, +)$ is $\langle n \rangle = n\mathbb{Z}$ for exactly one natural number $n$, The integers form a commutative ring).

In additive notation, the coset represented by $a$ is $a+n\mathbb Z$ (Left and right cosets $gH$ and $Hg$ of a subgroup).

The congruence $x\equiv a\pmod n$ means $n\mid(x-a)$ (Congruence modulo an integer: $a\equiv b\pmod n$ when $n\mid(a-b)$, including the moduli $0$ and $1$).

Every class modulo positive $n$ has exactly one representative $r$ with $0\le r<n$, and there are $n$ classes (For $n\ge 1$, every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$, so $\lvert\mathbb{Z}/n\rvert=n$; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$).

The index is the finite cardinality of the coset set (The coset set $G/H$ and the index $[G:H]$ of a subgroup, The left cosets of a subgroup partition the group).

Given: A group $G$ with identity $e$; for the second assertion, assume $G$ is finite.

The cosets are $gH=\{gh:h\in H\}$, and the index is the cardinality of the coset set when finite (Left and right cosets $gH$ and $Hg$ of a subgroup, The coset set $G/H$ and the index $[G:H]$ of a subgroup).

Lagrange's theorem gives $|G|=[G:H]|H|$ for a subgroup of a finite group (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$, 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).

A subset containing $e$ and closed under products and inverses is a subgroup; moreover $e^{-1}=e$ (Subgroup, 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).

Given: The symmetric group $S_3=\operatorname{Sym}(\{1,2,3\})$ under composition.

$S_3$ is the group of bijections of $\{1,2,3\}$, with cycle notation and composition acting rightmost first (The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

The generated set $\langle g\rangle$ is a subgroup; a nonempty subset closed under products and inverses is a subgroup (The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$, and cyclic groups, Subgroup, One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $gh^{-1} \in H$ for all $g, h \in H$; the identity and the inverses of $H$ are then those of $G$).

The order of a subgroup of a finite group divides the order of the group (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$, 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).

Given: The group $S_3=\operatorname{Sym}(\{1,2,3\})$, the subgroup $H=\{e,(12)\}$, and $g=(123)$.

Cycle products are composites with the rightmost permutation acting first (The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

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

Under rightmost-first composition, if $\tau=(12)$ then $ee=e$, $e\tau=\tau e=\tau$ and $\tau^2=e$; hence $e^{-1}=e$, $\tau^{-1}=\tau$, and $H=\{e,\tau\}$ contains the identity and is closed under products and inverses, so it is a subgroup (The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Subgroup).

Given: A finite cyclic group $G=\langle g\rangle$ of order $n$, and positive integers $d,q$ satisfying $n=dq$.

The generator has $\operatorname{ord}(g)=|G|=n$, and $g^k=e$ exactly when $n\mid k$ (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, 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 $j = k$).

Integer powers satisfy $(g^q)^k=g^{qk}$ (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).

The generated set $\langle g^q\rangle$ is a subgroup, and its cardinality is the order of $g^q$ (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 $j = k$).

The canonical embedding preserves products and order; $\mathbb Z$ is an ordered commutative ring and multiplication by a nonzero integer is cancellative (The naturals embed in the integers, The integers form a commutative ring, The integers form a totally ordered ring, The integers have no zero divisors; multiplicative cancellation, Divisibility in $\mathbb{Z}$: $d \mid a$ when $a = dq$ for some integer $q$).

If a positive natural exponent $r$ satisfies $x^r=e$, then $x$ has finite order, and its order is the least such positive exponent and is at most $r$; if there is no such exponent, its order is $\infty$ (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).

Given: The additive group $\mathbb Z$ and its subgroup $2\mathbb Z$.

The set $2\mathbb Z$ is a subgroup of $(\mathbb Z,+)$ (Every subgroup of $(\mathbb{Z}, +)$ is $\langle n \rangle = n\mathbb{Z}$ for exactly one natural number $n$, The integers form a commutative ring).

The index is the number of cosets, additive cosets have the form $a+2\mathbb Z$, and at modulus $2$ every congruence class has exactly one representative $r$ with $0\le r<2$, hence representative $0$ or $1$ (The coset set $G/H$ and the index $[G:H]$ of a subgroup, Left and right cosets $gH$ and $Hg$ of a subgroup, For $n\ge 1$, every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$, so $\lvert\mathbb{Z}/n\rvert=n$; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$).

Two sets are equinumerous when a bijection between them exists (Equinumerous sets, $A \approx B$ and $A \preceq B$, Injection, surjection, bijection).

Multiplication by a nonzero integer can be cancelled: if $xz=yz$ and $z\ne0$, then $x=y$ (The integers have no zero divisors; multiplicative cancellation).

Given: The group $S_3=\operatorname{Sym}(\{1,2,3\})$, and the subgroups $H=\{e,(12)\}$ and $K=\{e,(23)\}$.

$S_3$ is a group under rightmost-first composition, and $H,K$ are subgroups because each transposition squares to $e$ (The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition, $\operatorname{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements, Subgroup).

If $H$ is a subgroup of a finite group $G$, then under the canonical embedding $\iota:\mathbb N\to\mathbb Z$ one has $\iota(|H|)\mid\iota(|G|)$ (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$, 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).

Given: The group $S_3$, the subgroup $H=\{e,(12)\}$ and the element $g=(123)$ from A left coset that is not the corresponding right coset in $\operatorname{Sym}(\{1,2,3\})$.

The left coset is $gH=\{(123),(13)\}$ (Left and right cosets $gH$ and $Hg$ of a subgroup, A left coset that is not the corresponding right coset in $\operatorname{Sym}(\{1,2,3\})$).

Every subgroup contains the identity element (Subgroup).