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Is the class of finite perfect groups closed under fiber products? In other words if $G$ is a finite group, $N_1$, $N_2$ normal subgroups with trivial intersection such that $G/N_1$ and $G/N_2$ are perfect, must $G$ be perfect as well?

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  • $\begingroup$ Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ – Zev Chonoles Aug 11 '13 at 4:55
  • $\begingroup$ The error in my answer was that $\bar N_1 \cap \bar N_2 \geq \overline{N_1 \cap N_2} = \bar 1$, but one does not necessarily have equality as in Derek Holt's example. $\endgroup$ – Jack Schmidt Aug 11 '13 at 17:47
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Let $X$ be the direct product of two copies of ${\rm SL}(2,5)$ and $G$ the inverse image in $X$ of a diagonal subgroup of $A_5 \times A_5$. Then $G$ is the fiber (fibre?) product of two copies of ${\rm SL}(2,5)$ with $|N_1|=|N_2|=2$. Since $G$ is generated by the diagonal subgroup $D$ of $X$ and $Z(G) = N_1 \times N_2$, we have $G' = D' = D \cong {\rm SL}(2,5)$. So $|G'|=120$ but $|G|=240$, so $G$ is not perfect.

Of course there are also many examples when the fiber product is perfect. For example, if you do the same construction with $X=2^4:A_5$ with a nontrivial irreducible action of $A_5$ on the normal subgroup of order $2^4$ (there are actually two such modules for $A_5$), then the corresponding fiber product $2^{4+4}:A_5$ is perfect.

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  • $\begingroup$ Sorry, I did not write what I meant to write! I was referring to the intersection of the derived subgroups of $G$ with $Z(G)$. I have corrected my post now. $\endgroup$ – Derek Holt Aug 11 '13 at 20:08
  • $\begingroup$ Yeah, I just figured that out a few minutes ago, was just about to write a clarification (classic example of term confusion). Thank you for the nice solution! $\endgroup$ – walcher Aug 11 '13 at 20:10

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