Is $[G,H]\unlhd G$ for any $H\leq G$? Possible counterexample? Suppose $[G,H]$ is the subgroup of $G$ generated by commutators $[g,h]=ghg^{-1}h^{-1}$ where, $g\in G$ and $h\in H$, where $H$ is an arbitrary subgroup of $G$.
I'm having a hard time seeing why $[G,H]\unlhd G$. I've derived a few identities, such as
$$
g[a,h]g^{-1}=[gag^{-1},ghg^{-1}]
$$
and
$$
[g,h]^{-1}=[h,g].
$$
I'm just trying to see that the conjugate by arbitrary $g\in G$ of a commutator of form $[a,h]$ or $[h,a]$ (where $a\in G$) must be of the form $[a,h]$ or $[h,a]$? However, if $H$ is not normal, I don't see any reason why this would be true. Is there some little trick I'm missing?

To prove normality, you can use the commutator identity $[ab, c] = a[b,c]a^{-1}[a,c]$. This gives
$$g[g_0, h_0]g^{-1} = [gg_0, h_0][g, h_0]^{-1}$$
for all $g, g_0 \in G$ and $h \in H$. Hence $[G, H]$ is a normal subgroup.
With the same method, you can prove that $[H, K]$ is always a normal subgroup of $\langle H, K \rangle$.
