Consider a finite group $G$, elements $u,v \in G$ are said to be $\text{Aut}$ equivalent if $\exists \phi \in Aut(G) | \phi(u) = v$. You can then consider Aut-equivalence to be an equivalence relation leading to Aut-equivalent classes.

A restriction of this idea is to consider the "conjugacy" classes of a group. Which are defined by the conjugacy relations, said to be true for a pair $u,v \in G$ if $\exists g \in G \ s.t. \ gug^{-1} = v$.

Claim: The Aut-equivalent classes of a group are the conjugacy classes of a group

My work

  1. Let $\phi$ be an arbitrary automorphism. Then if $u,v$ are conjugate meaning $ \exists g \ gug^{-1} = v$ it follows that $\phi(g)\phi(u)\phi(g)^{-1} = \phi(v)$ that is $\phi(u), \phi(v)$ are conjugate.

  2. Suppose there doesn't exist $t$ such that $tut^{-1} = v$ but there if there exists $t$ such that $t \phi(u) t^{-1} = \phi(v)$ Then it must be the case that $\phi^{-1}(t) u \phi^{-1}(t)^{-1} = v$. That is if 2 elements are conjugate post automorphism, then they must have been conjugate prior.

This tells me that under any automorphism, conjugacy is respected. Which is a start... but doesn't really tell me anything more useful.

Another aspect is that for every Aut-equivalent class $A$ there exists one more entire conjugacy classes $C_i$ tha $C_i \subseteq A$, in other words Aut-equivalent classes can only be larger than a union of some finite number conjugacy classes.

Furthermore conjugation is an automorphism, so the "larger" statement can be made stronger into

Every Aut-equivalent classes is a finite union of some number of conjugacy classes.

But going from finite to 1... feels like a jump

  • $\begingroup$ Have you tried looking at what happens in examples? For instance the cyclic group of order $3$? $\endgroup$ – Sebastian Schoennenbeck Mar 2 '17 at 7:51
  • $\begingroup$ oh... {0}, {1}, {2} form their own conjugacy classes but here there's clearly an automorphism that combines them. I'm silly $\endgroup$ – frogeyedpeas Mar 2 '17 at 7:53
  • $\begingroup$ On the positive side: You learned something about groups, and everything else you did is perfectly fine, it just doesn't lead to the desired conclusion. $\endgroup$ – Sebastian Schoennenbeck Mar 2 '17 at 7:57
  • $\begingroup$ And, hopefully, you've also learned that you have to consider examples. $\endgroup$ – Mariano Suárez-Álvarez Mar 2 '17 at 8:09

The topic you are looking for is the outer automorphism group.

For each $g \in G $ the map $G \to G$ of the form $x \mapsto gxg^{-1}$ is called an inner automorphism of $G$. It is a simple exercise to check this map is indeed an automorphism. Every inner automorphism is uniquely determined by the element $g$.

If an automorphism is not inner we call it an outer automorphism. (This terminology is not quite right but is correct in spirit and precise enough for the purpose of this question).

Observe that all inner automorphisms on an Abelian group are trivial, and indeed the conjugacy classes are trivial. So our best bet is to find an Abelian group $A$ with a nontrivial automorphism $h \colon A \to A$. Then any pair $a$ and $h(a)$ will be Aut-equivalent but not conjugate unless $a = h(a)$.


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