Given a group and a subgroup $H$ , the left cosets of $H$ in $G$ partition $G$ , in the sense that every element of $g$ is in precisely one coset.

Proof

Firstly, every element is in a coset: since $g \in g H$ for any $g$ . So we must show that no element is in more than one coset.

Suppose $c$ is in both $a H$ and $b H$ . Then we claim that $a H = c H = b H$ , so in fact the two cosets $a H$ and $b H$ were the same. Indeed, $c \in a H$ , so there is $k \in H$ such that $c = a k$ . Therefore $c H = {c h : h \in H} = {a k h : h \in H}$ .

Exercise: ${a k h : h \in H} = {a r : r \in H}$ .

Show solution

Suppose $a k h$ is in the left-hand side. Then it is in the right-hand side immediately: letting $r = k h$ .

Conversely, suppose $a r$ is in the right-hand side. Then we may write $r = k k^{- 1} r$ , so $a k k^{- 1} r$ is in the right-hand side; but then $k^{- 1} r$ is in $H$ so this is exactly an object which lies in the left-hand side.

But that is just $a H$ .

By repeating the reasoning with $a$ and $b$ interchanged, we have $c H = b H$ ; this completes the proof.

Why is this interesting?

The fact that the left cosets partition the group means that we can, in some sense, "compress" the group $G$ with respect to $H$ . If we are only interested in $G$ "up to" $H$ , we can deal with the partition rather than the individual elements, throwing away the information we're not interested in.

This concept is most importantly used in defining the quotient group. To do this, the subgroup must be normal (proof). In this case, the collection of cosets itself inherits a group structure from the parent group $G$ , and the structure of the quotient group can often tell us a lot about the parent group.