The automorphism group of a free group (of finite rank) is known (see this). The group is infinite, if the number of generators is at least $2$, since there are automorphisms of the form $x_i\mapsto x_ix_j$ and other $x_k$'s fixed (here $\{x_i\}_{i=1}^n$ is a minimal generating set). Also note that $S_n$ is a subgroup of automorphism group of free group of rank $n$.

Question: Consider the automorphisms of a free group of rank $n$ which are of finite order; is there upper bound on this order in terms of $n$?


Yes, there is such a bound, namely $n! 2^n$. You can look this up in Vogtman's survey paper entitled "Automorphisms of free groups and outer space". Here's an outline.

Let's denote the rank $n$ free group by $F_n$ and its automorphism group by $\text{Aut}(F_n)$. Also, denote its normal subgroup of inner automorphisms by $\text{Inn}(F_n)$, and its outer automorphism group $\text{Out}(F_n) = \text{Aut}(F_n) / \text{Inn}(F_n)$. Since $\text{Inn}(F_n)$ is isomorphic to $F_n$ itself which is torsion-free, for any finite subgroup $G \subset \text{Aut}(F_n)$ the quotient homomorphism $\text{Aut}(F_n) \mapsto \text{Out}(F_n)$ is injective on $G$. It therefore suffices to bound the orders of finite subgroups of $\text{Out}(F_n)$.

The theorem you need is a realization theorem (proved independently by Culler, by Khramtsov, and by Zimmerman). This theorem says that for every finite subgroup $G < \text{Out}(F_n)$ there exists a connected, finite graph $\Gamma$ whose vertices all have valence $\ge 3$ and whose fundamental group is free of rank $n$, such that $G$ is isomorphic to a subgroup of the simplicial automorphism group of $\Gamma$. Since the number of vertices and edges of $\Gamma$ is bounded in terms of $n$ (by a simple Euler characteristic calculation), one obtains a concrete bound on the order of the simplicial automorphism group of $\Gamma$ and therefore on the order of $G$.

The largest finite order subgroup of $\text{Out}(F_n)$ is the simplicial automorphism group of the "rose with $n$ petals", the graph with one vertex and $n$ edges each having both ends attached to the vertex. The automorphism group of this graph has order $n! 2^n$. It can be understood quite concretely in terms of a free basis $F_n = \langle a_1,...,a_n \rangle$, namely the group of automorphisms of the form $a_i \mapsto a^{\epsilon(i)}_{\sigma(i)}$ where $\epsilon : \{1,...,n\} \to \{-1,+1\}$ is any function and $\sigma : \{1,...,n\} \to \{1,...,n\}$ is any permutation. Up to isomorphism this is just the wreath product of $\mathbb{Z}/2\mathbb{Z}$ and the symmetric group $S_n$.


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