# Finitely generated group has automorphism mapping between two elements of the same order?

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).

• 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. Feb 3 at 20:27
• @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 3 at 20:28
• Take any finite group with two conjugacy classes of elements of the same order, but the classes have different sizes. Feb 3 at 20:30
• 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. Feb 3 at 20:30
• One of the simplest examples is the infinite cyclic group. Feb 3 at 22:01

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.

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.

• 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 3 at 20:30
• 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. Feb 3 at 20:56

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.)

• Not all automorphisms are class preserving. See this old post of mine: 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. Feb 3 at 22:58
• You even acknowledged this in your answer there. Feb 3 at 23:07