Let $H$ be a subgroup of a group $G$ and let $a,b \in G$. Define the relation $\equiv$ on $G$ by $a\equiv b$ if and only if $ab^{-1} \in H$. Show that $\equiv$ is an equivalence relation on $G$.
Well my only question is in the symmetry part. This is what I did.
NTS. If $a\equiv b$ then $b \equiv a$.
$a\equiv b \Rightarrow ab^{-1} \in H$
$\Rightarrow (ab^{-1})^{-1}\in H$ since inverse exist if H is a subgroup
$\Rightarrow a^{-1}b \in H$
$\Rightarrow ba^{-1} \in H$ *Can I do this? Can I commute an inverse?
No, you can't assume commutativity -and you don't need to: $ab^{-1}\equiv h\in H\Longleftrightarrow a=hb\Longleftrightarrow e=hba^{-1}\Longleftrightarrow h^{-1}=ba^{-1}\in H$