Prove $H \lhd G$ and $G/H$ is abelian $\iff [G,G] \leq H$

Question

Let $H \leq G $, then

$H \lhd G$ and $G/H$ is abelian $\iff$ $[G,G] \leq H$

I have proved the $\Longrightarrow$

but I fail to prove $\Longleftarrow$

Can someone provide me details?

Thanks in advance.

Answer 1

Note that if $[G,G] \leq H$, then also $[G,H] \leq H$. So for all $g \in G, h \in H$, we have that $g^{-1}h^{-1}gh \in H$, thus also $g^{-1}h^{-1}g \in H$. Because $h^{-1}$ is arbitrary, $H$ is a normal subgroup of $G$.

As to why $G/H$ is abelian: take elements $g_1H, g_2H \in G/H$. Then: $$(g_1H)(g_2H) = g_1g_2H = g_2g_1[g_1,g_2]H = g_2g_1H = (g_2H)(g_1H)$$ since $[g_1,g_2] \in H$.