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This question came up in the discussion over here

My first thought was that then it fixes the Frattini subgroup. Any help?

For reference we found that the answer is no when each maximal subgroup is merely mapped back to itself.

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    With questions like this, one of the first things you should do is to look for small counterexamples. Did you do that? You don't have to look far. The cyclic group of order $3$ is a counterexample. You might try and prove that, for finite groups, the only counterexamples are cyclic groups of prime power order. – Derek Holt Dec 22 '21 at 08:24
  • To be honest I used most of my RAM posing the question. What you propose looks feasible. I'll think about it. @Derek Holt –  Dec 22 '21 at 08:32

2 Answers2

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The additive group of the rational numbers has no maximal subgroups, yet it admits many non-trivial automorphisms $x \mapsto kx$ for fixed $k \ne 0$.

Alon Amit
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  • Very sneaky.... –  Dec 22 '21 at 06:41
  • @Cpc Not at all. This should always be your first thought: your condition says that all borogoves are mimsy… what if there aren’t any? Or just one? Take things to extreme. – Alon Amit Dec 22 '21 at 07:13
  • Right. Maybe I could also add the condition that $G$ isn't its Frattini subgroup. –  Dec 22 '21 at 07:17
  • @Cpc I wouldn’t expect this to change things. The Frattini can still be large, and still admit non-trivial automorphisms which fix the rest of the group. – Alon Amit Dec 22 '21 at 07:19
  • I'm not convinced that this answers the question. The automorphisms here don't fix maximal subgroups pointwise (as there are no maximal subgroups to fix). – user1729 Dec 23 '21 at 09:05
  • Since there are no maximal subgroups, of course these automorphisms fix their elements pointwise (“vacuously”). – Alon Amit Dec 23 '21 at 15:29
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Small counterexample: $G=C_4$, the cyclic group of order 4, say generated by $a$. The automorphism $a \rightarrow a^{-1}$, fixes the maximal subgroup $ \langle a^2 \rangle$ pointwise.

Nicky Hekster
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  • Yeah, here the Frattini is large as can be. –  Dec 22 '21 at 08:17
  • $G = C_3$ also works. – Derek Holt Dec 22 '21 at 08:24
  • @Derek Holt vacuously for any $C_p,,p\gt2$. Or, the Frattini is $0$, I should say. –  Dec 22 '21 at 08:39
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    Yes, on purpose I did not take that example to avoid an example like Alon Amit's below (which one is perfectly OK of course). – Nicky Hekster Dec 22 '21 at 08:51
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    @Cpc To be more precise, in $C_p$, the trivial subgroup, which contains just the identity element, is a maximal subgroup, and is also equal to the Frattini subgroup. In $({\mathbb Q},+)$ on the other hand, there are no maximal subgroups, and so the Frattini subgroup is the intersection of an empty set of subgroups, which is (by definition) equal to the whole group. – Derek Holt Dec 22 '21 at 11:48