Group Cosets
Definition
For HH a subgroup of GG, and given some gGg \in G, we define
gHgH to denote {gh:hH}\{ gh : h \in H \}
HgHg to denote {hg:hH}\{ hg : h \in H \}
the first term is called a left coset and the second a right coset.
Example
Take the symmetric group G=S3G = S_3 and subgroup H={(123123),(123321)}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 HH are: (123123)H=(123321)H=H\begin{pmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \end{pmatrix} H = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix} H = H and (123213)H=(123312)H={(123213),(123312)}\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 (123231)H=(123132)H={(123231),(123132)}\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 HGH \subseteq G, the set of all left cosets {gH:gG}\{ gH : g \in G \} is a partition of GG. Likewise for right-cosets.
To prove this, we will show that (1) the collection of left cosets covers GG, and (2) left cosets are disjoint. 1. For each gGg \in G, we have that g=g1HgHg = g1_H \in gH. 2. assume g1g2g_1 \neq g_2. Then for all hHh \in H we have g1hg2hg_1h \neq g_2h, due to the presence of inverses. Further, g1Hg2Hg_1H \neq g_2H.
All left- and right- cosets of HH have the same cardinality as HH. Since left- and right- cosets partition GG, then there are G/H\lvert G \rvert / \lvert H \rvert left cosets of HH and the same number of right cosets.
Coset equality can be found with the following identities:
g1H=g2Hg_1H = g_2H if and only if g21g1Hg_2^{-1}g_1 \in H
Hg1=Hg2Hg_1 = Hg_2 if and only if g2g11Hg_2g_1^{-1} \in H
I will prove the first identity; the second is proven similarly. ()(\Rightarrow) Assume that g1H=g2Hg_1 H = g_2 H. Then exists some h1,h2h_1, h_2 where g1h1=g2h2g_1h_1 = g_2h_2. Multiplication gives g21g1=h2h11g_2^{-1}g_1 = h_2h_1^{-1}. Since h2h11Hh_2h_1^{-1} \in H, then also g21g1Hg_2^{-1}g_1 \in H. ()(\Leftarrow) First note that if hHh \in H then hH=HhH = H. Now assume g21g1Hg_2^{-1}g_1 \in H. Then g2H=g2((g21g1)H)={g2(g21g1h):hH}={g1h:hH}=g1Hg_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 g21g1Hg_2^{-1}g_1 \in H. Then g1=g2(g21g1H)g2Hg_1 = g_2(\underbrace{g_2^{-1}g_1}_{\in H}) \in g_2H. Since g1g2Hg_1 \in g_2H and cosets are disjoint, then g1H=g2Hg_1H = g_2H.



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