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Let $G$ be a finite perfect group. Let $H\le G^n$ be a subgroup which surjects onto each factor of $G$ (ie, is a subdirect product). Must $H$ be perfect?

EDIT: If $G$ is nonabelian simple, then any subdirect product $H\le G^n$ should be perfect right? (if I'm understanding Goursat's lemma correctly, the image should be isomorphic to $G^m$ for some $m\le n$).

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    No not necessarily. Try two copies of $G = {\rm SL}(2,5)$, with $H$ having structure $2^2.A_5$. – 2017-01-19
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    @DerekHolt How are you embedding $H$ in $G^2$ here? – 2017-01-20
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    There is essentially only one subdirect product of ${\rm SL}(2,5)^2$ with structure $2^2.A_5$ and it cannot be perfect because the Schur Multiplier of $A_5$ has order $2$. If $G$ is simple then any subdirect prodcut of $G^n$ is isomorphic to $G^m$ for some $m$. – 2017-01-20

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Let $G$ be any group with nontrivial center $Z$. Let $H = \{(a,b)\in G^2\;|\;ab^{-1}\in Z\}$, equivalently, let $H$ be the subgroup of $G^2$ generated by the diagonal copy of $G$ and the subgroup $Z^2$.

Claim 1. $H$ is a subdirect product of copies of $G$. [$H$ contains the diagonal copy of $G$, $D\leq G^2$, which is itself subdirect.]

Claim 2. $H$ is not perfect. [The diagonal subgroup $D\leq H$ is normal in $H$, and $H/D\cong Z$ is abelian. Thus $H'\leq D \lt H$.]

The above shows how to produce nonperfect subdirect products. To complete the answer to the question, we need to apply the construction when $G$ is perfect, so we need a group $G$ that is perfect and has a nontrivial center. $G=\textrm{SL}_n(q)$ has these properties if $n,q>2$ or if $n=2$, $q>3$.