I was asked this question in an interview-

Let G be an non-abelian group then what about it's automorphism group? Is it abelian or non-abelian?

I could not think about a single example at that time to contradict the proposition nor I know any general result to support it. Now I am thinking over it but still I am unable to find a suitable argument. Can anybody help me to solve this? Apology if this is too trivial. Thanks.

  • $\begingroup$ What is the automorphism group of $S_3$? $\endgroup$ – Mathmo123 Aug 24 '16 at 11:51

Conjugations are part of the automorphism group, and are relatively easy to study. Given $a, b \in G$, we look at the automorphisms $g\mapsto aga^{-1}$ and $g\mapsto bgb^{-1}$.

Assume that the two automorphisms do commute. That means that for any $g\in G$, we have $abgb^{-1}a^{-1} = baga^{-1}b^{-1}$. Specifically, this holds for $g = ab$. We get $$ ababb^{-1}a^{-1} = baaba^{-1}b^{-1}\\ ab = ba^2ba^{-1}b^{-1}\\ ab^2a = ba^2b $$ or, in other words, $ab$ commutes with $ba$. So, if you want to be certain that you have a non-abelian automorphism group, all you have to do is find a group with two elements $a, b$ such that $ab$ and $ba$ are unequal, and also do not commute.

Example: In $S_3$, the symmetric group on three objects, take $a = (1\,2\,3)$ and $b=(1\,2)$. In that case, $ab = (1\,3)$ while $ba = (2\,3)$. This makes $abba = (1\,3\,2)$ and $baab = (1\,2\,3)$. Therefore, the automorphisms given by conjugation by $a$ and by conjugation by $b$ do not commute.

Some non-abelian groups has abelian automorphism groups. However, they are not easy to describe. For instance, according to this mathoverflow question there were no concrete examples known until about 1975.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.