## A subgroup is normal precisely when it contains its commutator by the whole group

Prove that a subgroup $H \leq G$ is normal if and only if $[G,H] \leq H$.

$(\Rightarrow)$ Suppose $H \leq G$ is normal. Let $g \in G$ and $h \in H$; then $g^{-1}h^{-1}g \in H$, so that $g^{-1}h^{-1}gh = [g,h] \in H$. Thus $[G,H] \leq H$.

$(\Leftarrow)$ Suppose $[G,H] \leq H$. In particular, for all $g \in G$ and $h \in H$, $[g,h] = g^{-1}h^{-1}gh \in H$. Thus $g^{-1}h^{-1}g \in H$. Since $h^{-1}$ is arbitrary in $H$, $H$ is normal in $G$.

I think the point of this exercise was to use the result in exercise 5.4.1, as well as the second result in proposition 7 (in section 5.4): $H \trianglelefteq G$ if and only if $[H,G] \leq H$.