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

Proof

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

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

Exercise: $\{akh:h\in H\}=\{ar:r\in H\}$.

But that is just $aH$.

By repeating the reasoning with $a$ and $b$ interchanged, we have $cH=bH$; 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.