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For a finitely generated group $G$, is it always the case that if two elements $g_1, g_2$ have the same order then there is an automorphism that sends one to the other (i.e. $\phi$ such that $\phi(g_1)=g_2$)?

I can't seem to prove it, but playing around with several groups, abelian or not (to include $D_3$; a cyclic group of prime order and $\mathbb{Z}_4$), I'm unable to refute this.

How could one go about proving this? (assuming indeed it can be proven).

Shaun
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Anon
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  • My first thought is to look at semidirect products as a potential source of counterexamples; however, I'm not sure how to justify that idea. – Shaun Feb 03 '23 at 20:27
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    @Shaun I actually thought of that (!) so I'm glad you're saying this. I wasn't able to think of how to find a counter example though. – Anon Feb 03 '23 at 20:28
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    Take any finite group with two conjugacy classes of elements of the same order, but the classes have different sizes. – David A. Craven Feb 03 '23 at 20:30
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    The smallest counterexample is $\mathbb{Z}_2 \times \mathbb{Z}_4$. There are three elements of order $2$ there, but when you take $g$ of order four, then $g+g$ has just one possible value. – radekzak Feb 03 '23 at 20:30
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    One of the simplest examples is the infinite cyclic group. – Moishe Kohan Feb 03 '23 at 22:01

3 Answers3

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Take $G=S_3\times\mathbb{Z_2}$. It contains an element of order $2$ which is in the center, and an element of order $2$ not in the center. So clearly there is no automorphism which maps one of these elements to the other one.

Mark
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  • Is there another counterexample with an abelian finite group? – mathematico Apr 29 '24 at 15:32
  • @mathematico There is. For example, $G=\mathbb{Z_4}\times\mathbb{Z_2}$. The elements $(2,0)$ and $(0,1)$ are both of order $2$. But there can't be an automorphism that sends one of the elements to the other, because the quotient groups $G/\langle (2,0)\rangle$ and $G/\langle (0,1)\rangle$ are not isomorphic. – Mark Apr 29 '24 at 15:45
  • Thank you! Actually, I missed some important detail in my question. Is there another counterexample with an abelian finite group and two elements with maximal order? – mathematico Apr 30 '24 at 08:59
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In $S_4$ there is no automorphism that takes a two cycle like $(12)$ to a product of two cycles like $(12)(34)$.

In general, $S_n$ has only inner automorphisms when $n \ne 2,6$ and those preserve cycle structure, so you can construct lots of similar examples.

David A. Craven
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Ethan Bolker
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    Can I ask - how did you think of this? Was it just trial and error or something that popped into mind or is there some guiding principle you used? – Anon Feb 03 '23 at 20:30
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    From the start I was pretty sure the conjecture was false, so I looked for a counterexample. The smallest decently compicated group is $S_4$. There are two different kinds of permutations of order $2$ there. If there's a guiding principle it's that it's useful to be familiar with lots of examples, particularly the symmetric groups. – Ethan Bolker Feb 03 '23 at 20:56
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Automorphisms send conjugacy classes to conjugacy classes, while preserving elements' order. In $S_4$, the only distinct conjugacy classes of elements of the same order are the class of the transpositions and the class of the double transpositions. But the former has size $6$, whereas the latter has size $3$, so no one automorphism of $S_4$ sends transpositions to double transpositions, despite having the elements of both classes order $2$. Yet another affordable case where this argument works, verbatim, is $S_5$. (Incidentally, this proves also that in these two groups the automorphisms are class-preserving, and hence inner.)

Kan't
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  • Not all automorphisms are class preserving. See this old post of mine: https://math.stackexchange.com/q/4075551/1070376. Inner automorphisms do though. So when $n\neq2,6,$ you are good in $S_n.$ Also I missed the argument that the automorphisms in $S_4,S_5$ are all inner. – suckling pig Feb 03 '23 at 22:58
  • You even acknowledged this in your answer there. – suckling pig Feb 03 '23 at 23:07