Are the maximal subgroups of a finite group always conjugate? Suppose that $H$ is a maximal subgroup of $G$. Consider a subgroup $H$ conjugate to any element of $G$, this subgroup has the same order as $H$, hence max is maximal and all maximal subgroups are conjugate. Is this reasoning correct?


closed as off-topic by Shaun, Leucippus, JMP, Isaac Browne, Claude Leibovici May 7 '18 at 10:08

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    $\begingroup$ Consider the group $S_4$. Then, $A_4$ and $S_3$ are maximal subgroups of $S_4$ but they have different orders. $\endgroup$ – Moritz May 6 '18 at 20:00
  • $\begingroup$ Smallest counterexample is $S_3$; the maximal subgroups are the proper subgroups, one of order $3$ and three of order $2$; the latter cannot be conjugate to the former. In addition, if all maximal subgroups were conjugate, then you would never be able to have a normal maximal subgroup unless the maximal subgroup were a unique maximal subgroup... and the only finite groups with a unique maximal subgroup are cyclic of prime power order. Finally, maximality is not determined by the order. So, “not just no, but hell no.” $\endgroup$ – Arturo Magidin May 7 '18 at 0:53

All conjugates of maximal subgroups are maximal, but not all maximal subgroups are conjugate of each other. They don't even need to be the same order: take a product of two cyclic groups of orders two different primes.


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