Example of an outer automorphism that maps all elements to conjugates.

Question

While answering this question my mind wandered a bit, and it got me wondering about the following question:

Does there exist a (finite) group $G$ with an outer automorphism $\varphi$ such that $g$ and $\varphi(g)$ are conjugate for all $g\in G$?

The word 'finite' is in parentheses because I'd be interested in any example of such a group, but I'd prefer an example with a finite group. I'm convinced such an example must exists, as otherwise this would give a seemingly very weak sufficient condition for an automorphism to be inner, which I would find hard to believe. But I don't mind being surprised.

I've checked a few very non-abelian groups and a few small groups, but I haven't been able to find any such outer automorphisms. If you know of an example, or have some good ideas of where to look for one, I'd be glad to hear.

@Adelafif https://en.wikipedia.org/wiki/Inner_automorphism#Inner_and_outer_automorphism_groups — Mario Carneiro, Oct 24 '15 at 18:59
This might be helpful: (Part of the proof that $S_n$ has no other exceptional outer automorphisms) "First, show that any automorphism that preserves the conjugacy class of transpositions is an inner automorphism." Perhaps the proof of that could shed some light on your problem. — Mario Carneiro, Oct 24 '15 at 19:12

score 5 · Accepted Answer · answered Oct 24 '15 at 19:19

5

Examples of such group of order $p^6$ for primes $p \equiv \pm 3 \bmod 8$ were found by Burnside in 1913. See here for exampes of $2$-groups with this property.

answered Oct 24 '15 at 19:19

Derek Holt

90,008

Mario Carneiro · Answer 2 · 2015-10-25T05:42:35.727

4

The infinite alternating group $A_\infty$ of even permutations on a countable set (fixing all but finitely many elements) has an outer automorphism of conjugation by some odd permutation $g$, but for any fixed $x\in A_\infty$ we can find some transposition $t$ in the set of fixed points of both $x$ and $g$, and then $tgxg^{-1}t^{-1}=gxg^{-1}$, so $x$ is conjugate (via the even permutation $tg$) to $gxg^{-1}$, even though $g$ is not in $A_\infty$.

edited Oct 25 '15 at 05:42

answered Oct 24 '15 at 19:31

Mario Carneiro

27,438

1

That's a nice example! – hmakholm left over Monica Oct 24 '15 at 21:56
"of odd" $: \mapsto :$ "of even" $;;;$ ? $;;;;;;;;;$ – Oct 25 '15 at 00:44
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One could also look at the group of all finitely supported permutations of a countable set, and let $g$ be a non-finitely supported permutation. That way one doesn't need to worry about parity. – hmakholm left over Monica Oct 25 '15 at 02:25
@RickyDemer fixed – Mario Carneiro Oct 25 '15 at 05:42

Example of an outer automorphism that maps all elements to conjugates.

2 Answers2