Is $H$ the Commutator Subgroup of $G$ if $H\neq \{1\}, H\leq G$ and $[H, G] = [H, H]$?

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I'm working on the following problem obtained when trying to prove in a different way that the alternating group $A_n$ is the commutator subgroup of the symmetric group $S_n$ if $n\geq 5$.

(Again, I have proved the original problem, and the following is just part of a different attempt)

Gievn a group $G$ and a nontrivial subgroup $H\leq G$, prove or disprove that if $[H, G] = [H, H]$, then $H$ is, or contains, the commutator subgroup of $G$.

Here $[A, B]=<[a, b] | a\in A, b\in B>$ denotes the group generated by commutators of elements from $A$ and from $B$ ($A, B$ are subgroups of a certain group).

Any idea is appreciated. Thanks!

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This is not true. If $H = Z(G)$ is the center then $[H, G] = [H, H] = e$ but $Z(G)$ need not be or contain the commutator subgroup.