$G$ is a finite group. If $G$ has, say, $n$ conjugacy classes of maximal subgroups, can we say that each subgroup of $G$ has at most $n$ conjugacy classes of maximal subgroups?

I tried some small groups, $S_4$ for example. Is it true for all finite groups?


No. For example, the group $C_p\wr C_p$, the Sylow $p$-subgroup of $S_{p^2}$, has $p+1$ maximal subgroups. It contains the group $C_p\times C_p\times C_p$, if $p\geq 3$, which has $p^2+p+1$ maximal subgroups.

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  • $\begingroup$ Thanks for your answer. +1 from me. Now I know that this is not true for every subgroup, but is it also not true for commutator subgroup $G’$? I wonder if you could give a counterexample $G$ whose commutator subgroup $G’$ has more conjugacy classes of maximal subgroups that $G$. Any help is appreciated. $\endgroup$ – Benjamin Aug 1 at 14:10
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    $\begingroup$ The commutator subgroup of this group has order $p^{p-1}$ and is elementary abelian, so is also a counterexample to that statement as well. $\endgroup$ – David A. Craven Aug 1 at 14:13

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