Let \(G\) be a group and \(H\) a subgroup of \(G\). Define a *left coset* of \(H\) with *representative* \(g \in G\) to be the set
\begin{equation*}gH = \{ gh : h \in H \}.\end{equation*}
*Right cosets* can be defined similarly by
\begin{equation*}Hg = \{ hg : h \in H \}.\end{equation*}
If left and right cosets coincide or if it is clear from the context to which type of coset that we are referring, we will use the word *coset* without specifying left or right.

#####
Example6.1

Let \(H\) be the subgroup of \({\mathbb Z}_6\) consisting of the elements 0 and 3. The cosets are
\begin{gather*}
0 + H = 3 + H = \{ 0, 3 \}\\
1 + H = 4 + H = \{ 1, 4 \}\\
2 + H = 5 + H = \{ 2, 5 \}.
\end{gather*}
We will always write the cosets of subgroups of \({\mathbb Z}\) and \({\mathbb Z}_n\) with the additive notation we have used for cosets here. In a commutative group, left and right cosets are always identical.

#####
Example6.2

Let \(H\) be the subgroup of \(S_3\) defined by the permutations \(\{(1), (123), (132) \}\). The left cosets of \(H\) are
\begin{gather*}
(1)H = (1 2 3)H = (132)H = \{(1), (1 23), (132) \}\\
(1 2)H = (1 3)H = (2 3)H = \{ (1 2), (1 3), (2 3) \}.
\end{gather*}
The right cosets of \(H\) are exactly the same as the left cosets:
\begin{gather*}
H(1) = H(1 2 3) = H(132) = \{(1), (1 23), (132) \}\\
H(1 2) = H(1 3) = H(2 3) = \{ (1 2), (1 3), (2 3) \}.
\end{gather*}

It is not always the case that a left coset is the same as a right coset. Let \(K\) be the subgroup of \(S_3\) defined by the permutations \(\{(1), (1 2)\}\). Then the left cosets of \(K\) are
\begin{gather*}
(1)K = (1 2)K = \{(1), (1 2)\}\\
(1 3)K = (1 2 3)K = \{(1 3), (1 2 3)\}\\
(2 3)K = (1 3 2)K = \{(2 3), (1 3 2)\};
\end{gather*}
however, the right cosets of \(K\) are
\begin{gather*}
K(1) = K(1 2) = \{(1), (1 2)\}\\
K(1 3) = K(1 3 2) = \{(1 3), (1 3 2)\}\\
K(2 3) = K(1 2 3) = \{(2 3), (1 2 3)\}.
\end{gather*}

The following lemma is quite useful when dealing with cosets. (We leave its proof as an exercise.)

#####
Lemma6.3

Let \(H\) be a subgroup of a group \(G\) and suppose that \(g_1, g_2 \in G\). The following conditions are equivalent.

\(g_1 H = g_2 H\);

\(H g_1^{-1} = H g_2^{-1}\);

\(g_1 H \subseteq g_2 H\);

\(g_2 \in g_1 H\);

\(g_1^{-1} g_2 \in H\).

In all of our examples the cosets of a subgroup \(H\) partition the larger group \(G\). The following theorem proclaims that this will always be the case.

#####
Theorem6.4

Let \(H\) be a subgroup of a group \(G\). Then the left cosets of \(H\) in \(G\) partition \(G\). That is, the group \(G\) is the disjoint union of the left cosets of \(H\) in \(G\).

##### Proof

Let \(g_1 H\) and \(g_2 H\) be two cosets of \(H\) in \(G\). We must show that either \(g_1 H \cap g_2 H = \emptyset\) or \(g_1 H = g_2 H\). Suppose that \(g_1 H \cap g_2 H \neq \emptyset\) and \(a \in g_1 H \cap g_2 H\). Then by the definition of a left coset, \(a = g_1 h_1 = g_2 h_2\) for some elements \(h_1\) and \(h_2\) in \(H\). Hence, \(g_1 = g_2 h_2 h_1^{-1}\) or \(g_1 \in g_2 H\). By Lemma 6.3, \(g_1 H = g_2 H\).

Let \(G\) be a group and \(H\) be a subgroup of \(G\). Define the *index* of \(H\) in \(G\) to be the number of left cosets of \(H\) in \(G\). We will denote the index by \([G:H]\).

#####
Example6.6

Let \(G= {\mathbb Z}_6\) and \(H = \{ 0, 3 \}\). Then \([G:H] = 3\).

#####
Example6.7

Suppose that \(G= S_3\), \(H = \{ (1),(123), (132) \}\), and \(K= \{ (1), (12) \}\). Then \([G:H] = 2\) and \([G:K] = 3\).

#####
Theorem6.8

Let \(H\) be a subgroup of a group \(G\). The number of left cosets of \(H\) in \(G\) is the same as the number of right cosets of \(H\) in \(G\).

##### Proof

Let \({\mathcal L}_H\) and \({\mathcal R}_H\) denote the set of left and right cosets of \(H\) in \(G\), respectively. If we can define a bijective map \(\phi : {\mathcal L}_H \rightarrow {\mathcal R}_H\), then the theorem will be proved. If \(gH \in {\mathcal L}_H\), let \(\phi( gH ) = Hg^{-1}\). By Lemma 6.3, the map \(\phi\) is well-defined; that is, if \(g_1 H = g_2 H\), then \(H g_1^{-1} = H g_2^{-1}\). To show that \(\phi\) is one-to-one, suppose that
\begin{equation*}H g_1^{-1} = \phi( g_1 H ) = \phi( g_2 H ) = H g_2^{-1}.\end{equation*}
Again by Lemma 6.3, \(g_1 H = g_2 H\). The map \(\phi\) is onto since \(\phi(g^{-1} H ) = H g\).