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Let $G$ be a group, and let $K$ be a normal subgroup of $G$ which is a direct product of simple non-abelian groups.

I wanted to prove that $K=C_K(H)[K,\, H]$ for every subgroup $H$ of $G$.

Is this result out there? How can I start the proof?

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$[H,K]$ is a normal subgroup of $K$, so it must be the direct product of some subset of the nonabelian simple direct factors of $K$.

Let $S$ be a simple direct factor of $K$ that is not centralized by $H$. So, there exists $h \in H$, $k \in S$ with $[h,k] = (h^{-1}k^{-1}h)k \ne 1$. Now $h^{-1}kh$ is in the simple direct factor $k^{-1}Sk$ of $K$, which may or may not be equal to $S$, but in any case the projection of $[h,k]$ onto $S$ is nontrivial and hence $S$ must be one of the direct factors that are contained in $[H,K]$.

It follows that $K=C_K(H)[H,K]$.

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