# Coset Decomposition

Let $$H$$ be a subgroup of group $$G$$. We know that no right coset of $$H$$ in $$G$$ is empty and any two right cosets of $$H$$ in $$G$$ are either disjoint or identical. The union of all right cosets of $$H$$ in $$G$$ is equal to $$G$$. Hence the set of all right cosets of $$H$$ in $$G$$ gives a partition of $$G$$.

This partition is called the right coset decomposition of $$G$$. The procedure to obtain distinct members of this partition is given below:

$$H$$ itself is a right coset. Now suppose $$a \in G$$ and $$a \notin H$$ then $$Ha$$ will be another distinct right coset. Again let $$b$$ be another such element that $$b \in G$$ and $$b \notin H$$ and also $$b \notin Ha$$, then $$Hb$$ will be another distinct right coset. Proceeding in this way, all distinct right cosets of $$H$$ in $$G$$ will be obtained.

Thus $$G = H \cup Ha \cup Hb \cup Hc \ldots $$ where $$a,b,c$$ are elements of $$G$$ so chosen that all right cosets are distinct. In the same way, left coset decomposition of $$G$$ can be obtained.