It is known that the mapping class group of a closed (orientable) surface is generated by elements of finite order. Is this also known to be true for $Out(F_n)?$ A related question is the following: The mapping class group is known to be a perfect group for $g\geq 3,$ that is, it is equal to its commutator. Is this known for $Out(F_n)?$ I can find references to the effect that the rational homology is trivial in low dimensions, but that's not quite the same...
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asked Oct 11, 2011 at 14:30
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$\begingroup$ The abelianization is almost trivial: it's $\mathbb Z_2$. $\endgroup$– Jim ConantCommented Oct 11, 2011 at 14:52
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$\begingroup$ @Jim: how do you show this? Is there a reference? $\endgroup$– Igor RivinCommented Oct 11, 2011 at 15:10
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$\begingroup$ It follows directly from the Nielsen presentation for $Out(F_n)$. Vogtmann has a survey, "Automorphisms of free groups and outer space," which covers this point. $\endgroup$– Jim ConantCommented Oct 11, 2011 at 15:22
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$\begingroup$ And I should have mentioned $n>2$. $\endgroup$– Jim ConantCommented Oct 11, 2011 at 15:23
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$\begingroup$ Cool, will check out KV's paper, tnx! $\endgroup$– Igor RivinCommented Oct 11, 2011 at 15:30
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4 Answers
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$Aut(F_2)$ is generated by torsion, and $Aut(F_n)$ is normally generated by $Aut(F_2)$, so $Aut(F_n)$ is generated by torsion, hence $Out(F_n)$.
answered Oct 11, 2011 at 15:45
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In Bridson's "A condition that prevents groups from acting nontrivially on trees" it is shown that for $n\geq 3$, $Aut (F_n)$ is generated by subgroups $A_1,A_2,A_3$ such that $\langle A_i,A_j \rangle$ is finite for $i,j=1,2,3$. This answers your question and moreover implies (via Helly's theorem) that for $n\geq 3$, any action of $Aut (F_n)$ on a tree fixes a point.
Beautiful stuff!
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For your second point, the map $$d : \mathrm{Out}(F_n) \longrightarrow GL_n(\mathbb{Z}) \overset{\mathrm{det}}\longrightarrow \mathbb{Z}^\times$$ (given by the action on $\mathbb{Z}^n = H_1(F_n;\mathbb{Z})$) is surjective, and so $Out(F_n)$ is not perfect. I do not know whether $SOut(F_n) := \mathrm{Ker}(d)$ is perfect, but I would guess so.
answered Oct 11, 2011 at 14:52
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You can cook up lots of normal subgroups by looking at any characteristic subgroup of the free group. For example, if $F^{(k)}$ is the $k$th term of the lower central series, there is a surjection $$Out(F_n)\twoheadrightarrow Out(F_n/F_n^{(k)}).$$ The kernel of this surjection is an interesting normal subgroup. Oscar's construction is a special case of this.
(Edit: I misread "perfect" as "simple," so this is not an answer to the question.)
answered Oct 11, 2011 at 15:04