Conjugacy-closed and Hall implies commutator subgroup equals intersection with whole commutator subgroup

Statement
Suppose $$G$$ is a finite group and $$H$$ is a conjugacy-closed Hall subgroup of $$G$$. Then, we have:

$$H \cap [G,G] = [H,H]$$.

Facts used

 * 1) uses::Analogue of focal subgroup theorem for Hall subgroups

Proof
Given: A finite group $$G$$, a conjugacy-closed Hall subgroup $$H$$ of $$G$$.

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

Proof: By fact (1):

$$H \cap [G,G] = H_0$$

where $$H_0$$ is the focal subgroup of $$H$$ in $$G$$: the subgroup generated by $$xy^{-1}$$ with $$x,y \in H$$ conjugate in $$G$$.

Since $$H$$ is given to be conjugacy-closed in $$G$$, $$x,y \in H$$ are conjugate in $$G$$ if and only if they are conjugate in $$H$$. Thus, $$H \cap [G,G]$$ is the subgroup generated by $$xy^{-1}$$ for $$x,y \in H$$ conjugate in $$H$$. This is the same as $$[H,H]$$, so the proof is completed.