2
$\begingroup$

I would like to ask a question about automorphisms of free products of groups. More specifically, let $G = G_1 \ast ... \ G_n \ast F_r$ where $F_r$ is free group on r generators. We can define the group $Out(G)$ of outer automorphisms of $G$, as the quotient of automorphisms with the inner automorphisms.

Then we can see as a subgroup of $Out(G)$, the automorphisms that have a representative $\phi$ from $G$ to $G$, which induce the identity on $G_i$'s and restricted to $F_r$ is an automorphism from $F_r$ to $F_r$. My question is if this subgroup shares any properties with $Out(F_r)$, for example can we say that this subgroup is virtually torsion free or maybe every torsion subgroup is finite?

Thanks a lot in advance for your time.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

EDITED: Thanks to @Mathieu for finding a bug.

The group you describe is actually isomorphic to $\text{Aut}(F_r)$ (which, of course, has a well known relation to $\text{Out}(F_r)$).

To see why, you have defined a subgroup of $\text{Aut}(G)$ which I will denote $A_0$, consisting of all automorhisms which take each $G_i$ to itself by the identity and which take $F_r$ to itself by some automorphism. Restricting each element of $A_0$ to $F_r$ defines an isomorphism $A_0 \mapsto \text{Aut}(F_r)$, so we just need to show that the induced map $A_0 \mapsto \text{Out}(G)$ is injective. For this it suffices to show that for each $g \in G$ with corresponding inner automorphism $i_g(h) = ghg^{-1}$, and for each $\phi \in A_0$, if $i_g \circ h \in A_0$ then $g$ is the identity.

For each nontrivial free factor $B \in \{G_1,...,G_n,F_r\}$, if $g \not\in B$ then for any non-identity element $h \in B$ we have $ghg^{-1} \not\in B$, contradicting invariance of $B$. Therefore $g$ is an element of each of the free factors $B$, and this forces $g$ to be the identity (assuming that at least one of the $G_i$'s is nontrivial and that $r \ge 1$).

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks a lot for your answer. I suspected that it was related a lot with $Out(F_r)$, but I wasn't sure for the details. It is very helpful for me, thanks again. $\endgroup$
    – user75691
    Dec 16, 2014 at 14:05
  • $\begingroup$ @user75691: It is my pleasure. $\endgroup$
    – Lee Mosher
    Dec 16, 2014 at 14:08
  • 2
    $\begingroup$ Even though this question is old, I thought I'd nitpick on the accepted answer: assuming some $G_i$ is non-trivial, isn't the described subgroup isomorphic to $Aut(F_r)$ instead of $Out(F_r)$? Indeed, $Aut(F_r) \subset Aut(G)$ intersects $Inn(G)$ trivially: for any non-trivial $g \in F_r$ the inner automorphism $i_g(h) = ghg^{-1}$ does not act as the identity on any non-trivial $G_i$. $\endgroup$
    – Mathieu
    Jun 29, 2018 at 8:37
  • $\begingroup$ Ah, that's a good point; having checked $g \in F_r$, I should also have checked $g \in G_i$ for each $i$. I'll fix it. $\endgroup$
    – Lee Mosher
    Jun 29, 2018 at 14:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .