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Let $G=N_1\times N_2\dots \times N_n$. Suppose that $H$ is a simple subgroup of $G$. Is $H$ isomorphic to a subgroup of $N_i$, for some $N_i$?

This is a weaker version of this question, which turned out to be false:

If $G$ is direct product of simple normal subgroups, then is every simple subgroup isomorphic to some factor

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  • $\begingroup$ This is the same question as math.stackexchange.com/questions/1622808 $\endgroup$ – Derek Holt Jan 22 '16 at 21:57
  • $\begingroup$ @DerekHolt I thought it was different. $\endgroup$ – Jorge Fernández Hidalgo Jan 22 '16 at 22:02
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    $\begingroup$ Well the other question asks whether it is possible for a simple subgroup not to be isomorphic to a subgroup of some $N_i$, which is not completely dissimilar :} $\endgroup$ – Derek Holt Jan 22 '16 at 22:28
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Let $\pi_i:G\to N_i$ be the projection onto the $i$th factor. Then $\pi_i(H)\neq \{1\}$ for some $i$ and, for this $i$, we must have $\ker\pi_i=\{1\}$. Hence, $H$ is isomorphic to a subgroup of $N_i$.

This is false if you require equality. Take $G=A_6\times A_6$ and $A_5\cong H=\Delta(A_5)\leq G$, where $\Delta(A_5)=\{(\sigma,\sigma)\mid \sigma\in A_5\}$.

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  • $\begingroup$ Lol, I need to refresh my group theory skills, thank you very much. $\endgroup$ – Jorge Fernández Hidalgo Jan 22 '16 at 22:04

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