Retract implies derived subgroup equals intersection with whole derived subgroup

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., retract) must also satisfy the second subgroup property (i.e., subgroup whose derived subgroup equals its intersection with whole derived subgroup)
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Statement

Statement with symbols

Suppose $H$ is a retract of a group $G$: in other words, $H$ is a subgroup of $G$ such that there is a retraction from $G$ to $H$: a surjective homomorphism $\alpha :G\to H$ such that the restriction of $\alpha$ to $H$ is the identity map. Then, $H\cap [G,G]=[H,H]$.

Proof

Given: A group $G$, a subgroup $H$ with retraction $\alpha :G\to H$.

To prove: $H\cap [G,G]=[H,H]$.

Proof: By definition, we have $[H,H]\leq H$ and $[H,H]\leq [G,G]$. Thus, we have $[H,H]\leq H\cap [G,G]$. We need to prove the reverse inclusion.

Note that a generic element of $[G,G]$ is of the form:

$g=[a_{1},b_{1}][a_{1},b_{2}]\dots [a_{n},b_{n}]$

where $a_{i},b_{i}\in G$. Note that we do not need negative powers, since the inverse of a commutator is again a commutator.

Since $\alpha$ is a homomorphism, we have:

$\alpha (g)=\alpha ([a_{1},b_{1}][a_{2},b_{2}]\dots [a_{n},b_{n}])=[\alpha (a_{1}),\alpha (b_{1})][\alpha (a_{2}),\alpha (b_{2})]\dots [\alpha (a_{n}),\alpha (b_{n})]$.

We want to show that if $g\in H$, then $g\in [H,H]$.

Suppose $g\in H$. Then, since $\alpha$ is a retraction, $\alpha (g)=g$, so we get that $g$ is a product of commutators $[\alpha (a_{i}),\alpha (b_{i})]$ between elements of $H$. Thus, $g\in [H,H]$.