A subgroup $H$ is normal in $G$ if and only if... Prove that $H$ is a normal subgroup of $G$ if and only if $\{g^{-1}h^{-1}gh|g \in G, h\in H\}$ is a subgroup of $H$.
Attempt:
For the necessary condition, assume $H$ is normal in $G$. Then $g^{-1}h^{-1}g \in H$ for all $h \in H$. Then elements of the form $g^{-1}h^{-1}gh$ make a subgroup of $H$ by closure.
 A: Let $H$ be a normal subgroup of $G$.  Then $g^{-1}h^{-1}gh=(g^{-1}h^{-1}g)h$ is the product of two elements of $H$ and thus is contained in $H$.  Therefore, the set $\{g^{-1}h^{-1}gh|g\in G, h\in H\}$ is contained in $H$.  We want to show that the subgroup generated by this set is a subgroup of $H$.  To do this, it suffices to consider the product of two elements of the form $g^{-1}h^{-1}gh$, for $g\in G$ and $h\in H$: $$\begin{eqnarray*}
g_1^{-1}h_1^{-1}g_1h_1 g_2^{-1}h_2^{-1}g_2h_2&=&
(g_1^{-1}h_1^{-1}g_1)h_1 (g_2^{-1}h_2^{-1}g_2)h_2\\
\end{eqnarray*}$$
We have that $g_1^{-1}h_1^{-1}g_1$ and $g_2^{-1}h_2^{-1}g_2$ are elements of $H$ by normality, and of course $h_1$ and $h_2$ are elements of $H$, so $(g_1^{-1}h_1^{-1}g_1)h_1 (g_2^{-1}h_2^{-1}g_2)h_2\in H$.  Therefore, $\langle g^{-1}h^{-1}gh|g\in G, h\in H\rangle$ is contained in $H$, so it is a subgroup of $H$.
Conversely, if the subgroup generated by the set $\{g^{-1}h^{-1}gh|g\in G, h\in H\}$ is a subgroup of $H$, then $g^{-1}h^{-1}g\in H$ by closure for every $g\in G, h\in H$, so $H$ is normal in $G$.
