# Cosets

If $G$ is a group, $H$ is a subgroup and $a$ is any element in $G$, then the set
$\left\{ {ha:h \in H} \right\}$
is called the right coset generated by $a$ and $H$ is denoted by $Ha$.

Similarly, the set
$\left\{ {ah:h \in H} \right\}$
is called the left coset generated by $a$ and $H$ is denoted by $aH$.

Since $eH = He = H$, we see that $H$ itself is a right as well as a left coset. Moreover, since $e \in H$, it is evident that $a \in aH$.

If the group operation is addition, we define the right coset of $H$ in $G$ by
$H + a = \left\{ {h + a:h \in H} \right\}$

Similarly, the left coset of $H$ in $G$ is defined by
$a + H = \left\{ {a + h:h \in H} \right\}$

It must be noted that cosets are not necessarily subgroups of $G$. Rather, they are only special types of complexes which are sometimes called residue classes modulo subgroup.

In general $aH \ne Ha$. In the case of an abelian group, each right coset coincides with the corresponding left coset.

Example: Let $G = \left\{ {a,{a^2},{a^3},{a^4} = 1} \right\}$, and $o\left( G \right) = 4$, $H = \left\{ {1,{a^2}} \right\}$ is a group of $G$. Find all the cosets of $H$ in $G$ and prove that $G$ is equal to the union of all these cosets and also establish that any two cosets are either disjoint or identical.

Solution: We have
$1 \cdot H = \left\{ {1,{a^2}} \right\} = H$
$a \cdot H = \left\{ {a,{a^3}} \right\}$
${a^2} \cdot H = \left\{ {{a^2},{a^4}} \right\} = \left\{ {{a^2},1} \right\} = H$
${a^3} \cdot H = \left\{ {{a^3},{a^5}} \right\} = \left\{ {{a^3},a} \right\} = \left\{ {a,{a^3}} \right\} = a \cdot H$

Thus there are only distinct cosets, namely $H$ and $aH$ which is disjoint. Also $H$ is identical to ${a^2}H$ and $aH$ is identical to ${a^3}H$.

Again
$\begin{gathered} H \cup aH = \left\{ {1,{a^2}} \right\} \cup \left\{ {a,{a^3}} \right\} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {1,a,{a^2},{a^3}} \right\} = G \\ \end{gathered}$