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Jun 8, 2015 at 23:18 comment added The Masked Avenger I think A_4^n being big means the answer to the question is no.
Jun 8, 2015 at 22:02 comment added YCor @GerhardPaseman Yes it's easy to check that if $G_i$ are finitely many groups and $M_i$ are non-normal maximal subgroups, then any subgroup of $\prod G_i$ containing $\prod M_i$, has the form $\prod H_i$, where $H_i\in\{M_i,G_i\}$ for all $i$.
Jun 8, 2015 at 21:33 comment added Gerhard Paseman I suspect ${A_4}^n$ is big in ${A_5}^n$, but do not know this. Gerhard "Not Fully Remembering Subdirectly Irreducible" Paseman, 2015.06.08
Jun 8, 2015 at 21:27 history asked Pablo CC BY-SA 3.0