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 $${a^3}H$$.

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