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Suppose that for every pair of distinct maximal subgroups of $G$, these two maximal subgroups have distinct indices. Then $G$ is cyclic (for example, all such maximal subgroups have prime index, and the Frattini quotient embeds into a product of cyclic groups of distinct prime orders). In particular, $G$ has unique subgroups of every index.

I'm interested in a generalization of this claim: fix $m$, and suppose $G$ has at most one maximal subgroup of index $d$, for every $d$ dividing $m$. Does it follow that $G$ has at most one subgroup (not necessarily maximal) of index $d$, for every $d$ dividing $m$? (We recover the above by taking $m=|G|$.)

bm0525
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    $S_3$ is a counterexample to your assertion in the first paragraph. – Nicky Hekster May 25 '25 at 15:46
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    I suspect that by "$G$ has maximal subgroups with distinct indexes", it is meant that the indices are pairwise distinct. – sTertooy May 25 '25 at 16:41
  • Yes, that is clearly the intent. I’ve edited the question. – bm0525 May 25 '25 at 16:49
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    The question is still unclear. $S_3$ has maximal subgroups with index $2$ and $3$ and these are pairwise distinct. – Derek Holt May 25 '25 at 17:20
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    @DerekHolt But it has three subgroups with index $3$, so the indices are not pairwise distinct in the sense that for any two different maximal subgroups the indices are different. I think that is the intended meaning. (Though it would be good to make the intended meaning quite clear.) – Dermot Craddock May 25 '25 at 17:28
  • @DerekHolt is pairwise distinct a way of saying intersect trivially? – suckling pig May 25 '25 at 20:44
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    @sucklingpig In my answer I just meant that $2$ and $3$ are pairwise distinct: i.e. $2 \ne 3$. I was just pointing out that the question is badly worded. – Derek Holt May 26 '25 at 08:05
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    Is the claim in the first paragraph: if $G$ is such that every pair of maximal subgroups have distinct indexes, then $G$ is cyclic? – user1729 May 26 '25 at 08:19
  • You might check out https://math.stackexchange.com/questions/4108229/if-a-finite-group-has-exactly-2-maximal-subgroups-then-it-is-3-cyclic/4108259#4108259 – Nicky Hekster May 26 '25 at 10:15
  • @user1729. Yes, this is the claim, and it is true. – bm0525 May 26 '25 at 13:32
  • I am voting to reopen the question, as the OP has addressed all feedback. The question is now clear and the preamble correct. – user1729 May 26 '25 at 15:03

1 Answers1

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Let $A_4$ be the alternating group on $4$ elements. It has the following proper, non-trivial subgroups:

  • Three subgroups of index $6$, all non-maximal,
  • Four subgroups of index $4$, all maximal,
  • One subgroup of index $3$, maximal.

So taking $m = 6$ gives us a counterexample: the only maximal subgroup of index dividing $m$ is the maximal subgroup of index $3$; yet there are three (non-maximal) subgroups of index $6$.

sTertooy
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