Suppose $U$ and $V$ are subgroups of a group $G$. Show that if $V$ is normal in $G$, then $UV = \{\,uv\,|\,u \in U, v \in V\,\}$ is a subgroup of $G$.
I have shown the identity axiom. For the associativity axiom, associativity follows from $G$, right?
But I'm not sure about the inverse axiom. Here's what I have -
$u \in U$ and so $u^{-1} \in U$
$v \in V$ and so $v^{-1} \in V$
So take $uv$ to be $u^{-1}v^{-1}$
$u^{-1}v^{-1} = (uv)^{-1}$
So an inverse exists for all $uv \in UV$. Is that correct?