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It is well known that if $F_n$ is the free group on $n\in \mathbb{N}$ generators then $\operatorname{Out}(F_n)$ is residually finite. However, I cannot seem to find a reference for what happens when we have $F_\infty$. So,

Is $\operatorname{Out}(F_\infty)$ residually finite?

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Can't you embed the symmetric group on the naturals by looking at automorphisms of $F_\infty$ induced by permutations of the generating set? –  user83827 Aug 15 '11 at 21:29
    
@ccc: You can, but how does that answer the question? –  Grumpy Parsnip Aug 16 '11 at 0:40
    
@ccc: Are you saying $S_\infty$ is not residually finite, implying that the larger group $Out(F_\infty)$ can't be? That makes sense, I just wasn't aware of the theorem for $S_\infty$. –  Grumpy Parsnip Aug 16 '11 at 1:06
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@Jim: (Good thing I refreshed before posting my comment!) Yes, that's what I mean. In fact, $S_\infty$ contains an infinite simple subgroup: the (finitely supported) even permutations. This more than precludes residual finiteness. –  user83827 Aug 16 '11 at 1:09
    
Thanks, that's quite a neat proof. If you post it as an answer I'll accept it. –  user1729 Aug 16 '11 at 10:13

1 Answer 1

up vote 7 down vote accepted

Set-theoretic disclaimer: this answer assumes that $F_\infty$ is the free group on a countably infinite set of generators. Under a very mild choice principle, every infinite set contains a countably infinite set, so in that context this argument works for any infinite generating set. Who knows what happens if, for example, the set of generators is an infinite, Dedekind finite set, but certainly such investigations are at best tangential to the intent of the question.


We consider the free group $F_\infty$ with free generating set $S = \{s_0, s_1, \ldots\}$. Its outer automorphism group $\operatorname{Out}(F_\infty)$ is not residually finite, and in fact contains an infinite simple group. For a self-contained answer, it is easiest to proceed in two steps.

First, we embed the symmetric group on the natural numbers (denoted by $S_\infty$) into $\operatorname{Out}(F_\infty)$. With each permutation $\sigma: \mathbb{N} \to \mathbb{N}$ we associate the unique automorphism $T_\sigma$ of $F_\infty$ extending the map $s_i \mapsto s_{\sigma(i)}$ (explicitly, $T_\sigma: s_{i_0}^{\epsilon_0} \cdots s_{i_n}^{\epsilon_n} \mapsto s_{\sigma(i_0)}^{\epsilon_0} \cdots s_{\sigma(i_n)}^{\epsilon_n}$). This map is evidently an embedding of $S_\infty$ into $\operatorname{Aut}(F_\infty)$, so it suffices to check that that $T_\sigma$ is outer for each nonidentity $\sigma$. But if $w \in F_\infty$ is not the identity and $s \in S$ is a generator not occurring in (the reduced) $w$, then $w^{-1} s w$ is not in $S$, and in particular is not equal to $T_\sigma(s)$. So $\operatorname{Out}(F_\infty)$ contains a copy of $S_\infty$.

Next, we exhibit an infinite simple subgroup of $S_\infty$. In analogy with the finite setting, we define the alternating group $A\subset F_\infty$ to consist of the products of an even number of transpositions (in particular, each element of $A$ has finite support). One can establish the simplicity of $A$ directly, by mimicking one's favorite argument for the simplicity of $A_n$ for sufficiently large $n$. Alternatively, the simplicity of large $A_n$ abstractly implies that of $A$ by considering the canonical copy of $A_n$ as the set of permutations in $A$ with support contained in $\{0, \ldots, n-1\}$. Then any nontrivial normal subgroup $G \unlhd A$ would contain a nonidentity element of some $A_n$, thus by simplicity $G \supseteq A_n$ for sufficiently large $n$. But that implies that $G \supseteq A$, since $A = \bigcup_n A_n$ (and the union is increasing). So $A$ is simple in $S_\infty$, and finally we can realize it as an infinite simple subgroup of $F_\infty$ as we wanted.

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