Group Cosets
Definition
For $H$ a subgroup of $G$, and given some $g \in G$, we define
$gH$ to denote $\{ gh : h \in H \}$
$Hg$ to denote $\{ hg : h \in H \}$
the first term is called a left coset and the second a right coset.
Example
Take the symmetric group $G = S_3$ and subgroup $H = \left\{ \begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} \right\}$ then the cosets of $H$ are: $\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} H = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} H = H$ and $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix} H = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} H = \left\{ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \end{pmatrix} \right\}$ and $\begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix} H = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} H = \left\{ \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix} \right\}$
Properties
Cosets form partition elements. That is, for $H \subseteq G$, the set of all left cosets $\{ gH : g \in G \}$ is a partition of $G$. Likewise for right-cosets.
To prove this, we will show that (1) the collection of left cosets covers $G$, and (2) left cosets are disjoint. 1. For each $g \in G$, we have that $g = g1_H \in gH$. 2. assume $g_1 \neq g_2$. Then for all $h \in H$ we have $g_1h \neq g_2h$, due to the presence of inverses. Further, $g_1H \neq g_2H$.
All left- and right- cosets of $H$ have the same cardinality as $H$. Since left- and right- cosets partition $G$, then there are $\lvert G \rvert / \lvert H \rvert$ left cosets of $H$ and the same number of right cosets.
Coset equality can be found with the following identities:
$g_1H = g_2H$ if and only if $g_2^{-1}g_1 \in H$
$Hg_1 = Hg_2$ if and only if $g_2g_1^{-1} \in H$
I will prove the first identity; the second is proven similarly. $(\Rightarrow)$ Assume that $g_1 H = g_2 H$. Then exists some $h_1, h_2$ where $g_1h_1 = g_2h_2$. Multiplication gives $g_2^{-1}g_1 = h_2h_1^{-1}$. Since $h_2h_1^{-1} \in H$, then also $g_2^{-1}g_1 \in H$. $(\Leftarrow)$ First note that if $h \in H$ then $hH = H$. Now assume $g_2^{-1}g_1 \in H$. Then $g_2H = g_2((g_2^{-1}g_1)H) = \{ g_2(g_2^{-1}g_1h) : h \in H \} = \{ g_1h : h \in H \} = g_1H$ $(\Leftarrow)$, alternative proof. Assume $g_2^{-1}g_1 \in H$. Then $g_1 = g_2(\underbrace{g_2^{-1}g_1}_{\in H}) \in g_2H$. Since $g_1 \in g_2H$ and cosets are disjoint, then $g_1H = g_2H$.

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