Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a group and let $H$ be a factor group of $G$. Is there any result that relates $\operatorname{Aut}(G)$ (the automorphism group of $G$) and $\operatorname{Aut}(H)$?

As a very special case of the question, let $F_2$ be the free group with two generators $x$ and $y$. Let $G_2$ be the factor group of $F_2$ by adding relations such that $[x,[x,y]]=[y,[x,y]]=1$; that is, $G_2$ is the discrete Heisenberg group. Then is there any relation between $\operatorname{Aut}(F_2)$ and $\operatorname{Aut}(G_2)$? Hence or otherwise, how to find out $\operatorname{Aut}(G_2)$?

share|improve this question

1 Answer 1

up vote 8 down vote accepted

If $H=G/N$ and $N$ is a characteristic subgroup, then there is a natural homomorphism $\phi$ from $Aut(G)$ to $Aut(H)$. If $G=F_n$, the image $\phi(Aut(F_n))$ consists of tame automorphisms. In general $\phi(Aut(F_n))$ may not coincide with $Aut(H)$ and there is a lot of work on this topic, especially for the case when $N$ is a fully invariant subgroup, hence $H$ is relatively free. In particular, the Heisenberg group is the relatively free group in the variety of nilpotent groups of class 2 (of rank 2). In that case all automorphiss of $H$ are tame. See Bachmuth, S. Induced automorphisms of free groups and free metabelian groups. Trans. Amer. Math. Soc. 122 1966 1–17 and S. Andreadakis, Automorphisms of free groups and free nilpotent groups, Proc. London Math. Soc. 15 (1965), 239-268. But if you consider the free nilpotent group of class 4, then not all automorphisms are tame. One of the latest papers on the subject is Papistas, Athanassios I. On automorphisms of free nilpotent-by-abelian groups. Internat. J. Algebra Comput. 16 (2006), no. 5, 827–837. Look also at the list of references there.

share|improve this answer
@Mark Sapir, Thank you for your answer! Just to clarify: did you mean that in the free group with generators $x$ and $y$, the normal subgroup generated by $[x,[x,y]]$ and $[y,[x,y]]$ is a fully invariant group? If yes, why is it so? –  Zuriel May 21 '12 at 7:34
$\gamma_3(F_2)=\langle [x,[x,y]], [y,[x,y]] \rangle ^G$, and is fully invariant –  Wei Zhou May 21 '12 at 8:50
Yes, Wei is correct. The Heisenberg group $H$ is nilpotent of class 2, hence the second member $F_2^{(2)}$ of the lower central series of $F_2$ is inside $N$. On the other hand every relation of $H$ is inside $F_2^{(2)}$, so $N\subseteq F_2^{(2)}$, whence $N=F_2^{(2)}$ and is fully invariant. –  Mark Sapir May 21 '12 at 15:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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