Below is my attempt at a proof of the following. I hope somebody will grade my work.
Let $G$ be a group and $H\leq G$. Define an equivalence relation $\sim$ on $G$ by:
$a\sim b$ if and only if $a^{-1}b\in H$.
Prove that: $[a]=aH$.
First, I note that $[a] = \{x\in G : x\sim a\}$ and $aH = \{ah : h\in H\}$.
Let $x\in [a]$. Then $x\sim a$ i.e. $x^{-1}a \in H$. If we write that $h=x^{-1}a \in H$ and since we know $H\leq G$, then $x=ah^{-1}$. But $h^{-1}\in H$ which means $x\in aH$.
Let $x\in aH$. Then there exists $h\in H$ such that $x=ah$. Since $H\leq G$, we have inverse so $a^{-1}x = h$ and $h\in H$ so $a\sim x$. $\sim$ is an symmetric so $x \sim a$. Thus, $x\in [a]$.
It's mostly right. The only time you really need the fact that $H$ it's a subgroup of $G$ it's when you say that $h^{-1} \in H$ (and when you say that the relation is an equivalence relation). All the other times you are just using the fact that $G$ it's a group.