# Example of a finitely generated infinite group with a non-inner automorphism of finite order

Does there exist an infinite finitely generated group $G$ together with a finite group $B$ of automorphisms of $G$ such that

1. The non-identity elements of $B$ are not inner automorphisms of $G$;

2. For every element $g \in G, g\neq 1,$ the finite set $g^B$ is a generating set for $G$?

It looks like a Tarski Monster $p$-group $T_p$ (an infinite group in which every non-identity element has order $p$) might be a contender for $G$. If $T_p$ has an automorphism of finite order coprime to $p$, then this would work perfectly. Unfortunately I'm not sure that it does.

Another, more straightforward, way to construct an example is the following. Fix the free group $F_2$ and its obvious automorphism $\alpha$ of order 2 that switches the generators. Now consider the procedure of creating the Tarski monster quotient of $F$ described in the book by Olshanskii "Geometry of defining relations...". At each step together with a new relation $r_i=1$, impose the relation $\alpha(r_i)=1$. It is not difficult to see that $\alpha$ will be an outer automorphism of the resulting Tarski monster without non-trivial fixed points.
• Thanks so much Mark. Your post was very helpful indeed. I've been giving this some serious thought, and I'm not sure if we have a full answer yet. Obrazstov's paper that you kindly pointed me towards gets us close, but not all the way I think. The monsters that he constructs aren't quasi-finite which causes problems. Indeed, his results (specifically Theorem C") can be used to prove the existence of a simple group $L$ that has, say a group of non-inner automorphisms $H$, where $H$ is cyclic of order $3$. – Simon Smith Apr 18 '11 at 2:50
• In addition, we can require that $L$ contains only nice subgroups (of our choosing), together with infinite cyclic subgroups. Unfortunately, we're not able to exclude the possibility that $L$ has proper subgroups that are infinite and cyclic using Obrazstov's results alone. And if $g$ is a generator for some infinite cyclic subgroup of $L$, I don't think it's necessarily true that $g^H$ is a set of generators for $L$, no matter which nice subgroups we require $L$ to possess. – Simon Smith Apr 18 '11 at 2:51
• Your more straightforward" way to construct an example sounds promising, but I'm struggling (because of my own ignorance) to see that it always works. You suggest that we fix the free group $F_2$ with generators $a$ and $b$, and let $\alpha$ be the obvious automorphism of $F_2$ that interchanges the two generators. At each stage in the construction of the Tarski monster in Olshanskii's book (page 297 - 298 I assume), every time the book introduces a new relation $r_i = 1$, you suggest that we impose the relation $\alpha(r_i) = 1$. – Simon Smith Apr 18 '11 at 2:51
• I agree that doing so will result in a group for which $\alpha$ is an automorphism. However, I don't see why such a group would still have to be a Tarski monster. The defining relations (2) and (2') used in the book aren't symmetric in the two generators and I don't see how we are guaranteed that the construction will still result in an infinite group whose subgroups are all finite. Is this easy to see? If so, any pointers would be most welcome! Thanks again for your help. – Simon Smith Apr 18 '11 at 2:51
• The resulting group will be a Tarski monster because every two elements will either generate the whole group or will commute and hence generate a cyclic group of finite order. The "only" thing to prove here is that the resulting group is infinite, i.e. at every step you get a non-elementary hyperbolic group. That fact is true and is proved the same way as in Olshanskii's paper. In fact you should look not at the original paper but at the result that every hyperbolic torsion-free group has a Tarski monster factor (a paper about $G$-subgroups of hyperbolic groups). – Mark Sapir Apr 18 '11 at 3:03