# Why does every finite subgroup of $\mathrm{Aut}(F_n)$ acts on a graph of Euler characteristic $n-1$?

My question is the following:

In a paper I read that:

Any finite subgroup of $\mathrm{Aut}(F_n)$ can be realised as agroup of baspoint-preserving isometries of a graph of Euler characteristic $1-n$. Why is this fact true?

Thanks for help.

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you might look at Karen Vogtmann's survey paper on $Out(F_n)$ (which also discusses $Aut(F_n)$). –  Grumpy Parsnip Jan 16 '12 at 13:19

## 1 Answer

This statement is called The Realization Theorem by Vogtmann in her survey paper. She gives the following references:

1. M. Culler, Finite groups of outer automorphisms of a free group, Contribu- tions to group theory, 197–207, Contemp. Math., 33, Amer. Math. Soc., Providence, R.I., 1984.

2. D. G. Khramtsov, Finite groups of automorphisms of free groups, Mat. Za- metki 38 (1985), no. 3, 386–392, 476.

3. B. Zimmermann, Uber Homomorphismen n-dimensionaler Henkelkörper und endliche Erweiterungen von Schottky-Gruppen, Comment. Math. Helv. 56 (1981), no. 3, 474–486.

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None of these articles is easily available on the internet (that I could determine), which is ashame. Please post if you know a good internet reference. –  Grumpy Parsnip Jan 17 '12 at 2:09