coset

Coset

Let $(G,\circ )$ group, $U\le G$ (subgroup).
Left coset: $a\circ U:=\{a\circ u\mid u\in U\},a\in G$
Right coset: $U\circ a:=\{u\circ a\mid u\in U\},a\in G$

A coset is what you get when you take every element of a subgroup and multiply it (from the left or right) by a fixed group element. It represents a kind of "shifted copy" of the subgroup within the larger group.

The direction you multiply from (left or right) can give different results if the group is not commutative.

TLDR: Key properties

• All cosets of a subgroup $U$ in a group $G$ have the same size as $U$: $|U|=|a\circ U|=|U\circ a|$
• Different cosets are either identical or disjoint (no partial overlap).
• The collection of all left cosets (or all right cosets) forms a partition of $G$.
• The number of distinct cosets $[G:U]$ is called the index.
• Any element within a coset can serve as its representative: $b\in a\circ U⟹a\circ U=b\circ U$

Left and right cosets each form a partition of $G$

(1) ${\bigcup }_{a\in G}(a\circ U)=G$
(2) $\exists x\in (a\circ U)\cap (b\circ U)⟹a\circ U=b\circ U$ (two cosets either have nothing in common or are equal)

Proof: ad (1): $a=a\circ e,e\in U⟹a\in a\circ U$ $✓$ ad (2): $x\in (a\circ U)\cap (b\circ U)$, z.z: $a\circ U\subseteq b\circ U\wedge b\circ U\subseteq a\circ U$ $\exists u_1\in U:x=a\circ u_1$ ⇒ $a\circ u_1=b\circ u_2$ $\exists u_2\in U:x=b\circ u_2$ ⇒ $a=b\circ u_2\circ u_1^{-1}$ $y\in a\circ U$ (some $y$) $⟹\exists u\in U:y=a\circ u=b\circ \underset{\in U}{\underbrace{u_2\circ u_1^{-1}\circ u}}\in b\circ U⟹aU\subseteq b\circ U$ $a\sim b⟺a\circ U=b\circ U$ is an equivalence relation.

A representative of a coset $aU$ is any element that generates that coset when multiplied with the subgroup $U$. In other words, if $b\in aU$, then $b$ is a representative of that coset since $bU=aU$.

This means that any element within a coset can serve as its representative. When we write $aU$, we’re using $a$ as one possible representative, but we could have chosen any other element from that coset. This works because cosets partition the group into disjoint equivalence classes - all elements in the same coset generate identical cosets.

Example

For the group ${\mathbb{Z}}_6$ with subgroup $U=\{0,3\}$ the cosets are:
$0+U=\{0,3\}$ where either 0 or 3 can represent this coset
$1+U=\{1,4\}$ where either 1 or 4 can represent this coset
$2+U=\{2,5\}$ where either 2 or 5 can represent this coset
For instance, if we take the coset $1+U$, we can verify that using 4 as a representative gives us the same coset:
$4+U=\{4+0,4+3\}=\{4,1\}=\{1,4\}=1+U$

Transclude of lagrange's-theorem

Example

$S_3=\{e,(12),(13),(23),(123),(132)\}$ (permutations of 3 elements)
$U=\{e,(12)\}$ (some subgroup of $S_3$)
To find all left cosets, we multiply each element of $S_3$ on the left with $U$:
Using $e$:
$eU=\{e\cdot e,e\cdot (12)\}=\{e,(12)\}$
Using $(12)$:
$(12)U=\{(12)\cdot e,(12)\cdot (12)\}=\{(12),e\}=eU$
Using $(13)$:
$(13)U=\{(13)\cdot e,(13)\cdot (12)\}=\{(13),(123)\}$
Using $(23)$:
$(23)U=\{(23)\cdot e,(23)\cdot (12)\}=\{(23),(132)\}$
Using $(123)$:
$(123)U=\{(123)\cdot e,(123)\cdot (12)\}=\{(123),(13)\}=(13)U$
Using $(132)$:
$(132)U=\{(132)\cdot e,(132)\cdot (12)\}=\{(132),(23)\}=(23)U$

Key observations:

We only got 3 distinct left cosets ($[G:U]=3$):
$eU=\{e,(12)\}$
$(13)U=\{(13),(123)\}$
$(23)U=\{(23),(132)\}$

Each coset has size 2 (same as $|U|$)

The cosets partition $S_3$:
Every element appears exactly once, the cosets don’t overlap, together they cover all of $S_3$ → $|S_3|=|U|\cdot [G:U]=2\cdot 3=6$