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I thought this question up myself. It's somewhat similar to a question I asked previously Progress on a conjecture of Burnside...

For the relevant definitions consult basic texts on group theory. (I like Fraleigh, but there are plenty of others.)

Maximal subgroups are invariant under automorphisms. (This can be used to see that the Frattini subgroup is characteristic. And in fact that's what got me to thinking about it.)

Now, is every automorphism fixing the maximal subgroups the identity? By fixing I mean: maps each maximal subgroup to itself.

Checking in $\mathscr S_3$, for example, all the automorphisms are inner. So we have $6$ automorphisms to check.

There's work left to do. It's just a question of how much. I doubt this will be as difficult as the classification of finite simple groups, or Fermat's last theorem. But I don't actually know.

Any ideas?

This is of course naive of me. I haven't checked around that much. $\mathscr S_4$ is already more difficult, with $24$ automorphisms and maximal subgroups of orders $6,8$ and $12$.

So we can continue to look around for a counterexample; or try to prove a general result.

The question has been answered in the negative with @reuns's help. Cyclic groups provide a counter example, since there is only one subgroup of each order.

Don't know why I chose symmetric groups first. I appear to have overshot. No reason why an answer couldn't come from abelian groups (as it turns out to have).

  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Dec 20 '21 at 20:59
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    Have you tried to look for small counterexamples? – Derek Holt Dec 20 '21 at 21:34
  • An automorphism will of course preserve the order of a finite subgroup. So I'd look for a finite group with more than one maximal subgroup of the same order, and see if there isn't an automorphism that, instead of fixing them, swaps one to the other. – hardmath Dec 20 '21 at 22:39
  • @hardmath There are such examples, but that doesn't provide a counter example. –  Dec 21 '21 at 04:03
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    Can you clarify "fixing the maximal subgroups" ? – reuns Dec 21 '21 at 04:12
  • Yes. @reuns I mean by that mapping each such to itself again. So if $a$ is the automorphism and $M$ any maximal subgroup, $a(M)=M$. Of course this doesn't mean pointwise. Not sure about that case either btw. –  Dec 21 '21 at 04:33
  • Ok. Then try with a cyclic group! – reuns Dec 21 '21 at 04:34
  • The maximal subgroups of $S_n$ includes the $n$ copies of $S_{n-1}$ and an automorphism sending each of those to themselves will be trivial. – reuns Dec 21 '21 at 04:37
  • Every automorphism of $\Bbb{Z}/n\Bbb{Z}$ is given by multiplication by an integer, it sends any subgroup to itself. Also I think ${0}$ is a maximal subgroup of $\Bbb{Z}/3\Bbb{Z}$. – reuns Dec 21 '21 at 04:40
  • So the answer is no. Thanks. @reuns –  Dec 21 '21 at 05:05

2 Answers2

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Let $(A, +)$ be an abelian group which contains maximal subgroups*. Consider the inversion map $x\mapsto -x$.

As $A$ is abelian, this is an automorphism. This map fixes all subgroups, not just the maximal ones. It is non-trivial if $A$ contains a non-trivial element of order other than $2$ (e.g. $A$ cyclic of order $\geq3$ works). Therefore, it satisfies all the properties you wish.

*e.g. not $\mathbb{Q}$, but any finitely generated abelian group works.

user1729
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For the cyclic group $\mathbb Z/n\mathbb Z=\mathbb Z_n$, we have that the automorphism group is $(\mathbb Z_n)^×$.

Thus we just need to choose $n$ large enough that $\varphi(n)\gt1$.

To finish we use that cyclic groups have only one subgroup of each order dividing $n$. Therefore any automorphism fixes any subgroup.