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Let $G$ be a finitely-generated, torsion-free group. Under what circumstances is $Inn(G)$, the group of inner automorphisms $\phi(x)=g^{-1}xg$ of $G$, torsion-free?

Since $Inn(G)\cong G/Z(G)$, the group $G$ modulo its centre, this is certainly true if $G$ is abelian, or, at the other extreme, centreless. But when else?

Apologies if this is too elementary, I am a topologist who does not know the group theory literature as well as he would like.

flag	asked Apr 1 2011 at 15:36 Mark Grant 9,061●1●13●33

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I am no group theorist either, but I think your question is way too general. BTW in the abelian case $Inn(G)$ is trivial, which I would not call torsion-free. – GH Apr 1 2011 at 15:49

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The trivial group is torsion free. – Fernando Muro Apr 1 2011 at 16:13

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True for any torsion-free nilpotent group as well. – Steve D Apr 1 2011 at 16:29

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An example in which $G/Z(G)$ is not torsion-free is $\langle x,y \mid y^{-1}xy=x^{-1} \rangle$, because $y^2 \in Z(G)$. – Derek Holt Apr 1 2011 at 17:33

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It is 1.2.20 in Lennox and Robinson's "The Theory of Infinite Soluble Groups". – Steve D Apr 1 2011 at 21:12

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Jon Bannon · Answer 1 · 2011-04-01 18:56:22Z UTC

Here's a one-relator ("large") example of a torsion free group such that the group of inner automorphisms has torsion:

$$\langle a_{1},a_{2},a_{3} : a_{1}^{2}=a_{2}^{3}, a_{2}^{5}=a_{3}^{7}\rangle$$.

This group is one-relator on $a_{1}$ and $a_{3}$, and by a result of Murasugi is torsion-free. The center of the group is generated by $a_{1}^{10}$.

This fact is mentioned in the first paragraph of this readily googlable paper of James McCool: A class of one-relator groups with centre, Bulletin of the Australian Mathematical Society, Volume 44, Issue 2, 2009.

The result is proved in this other readily googlable paper of Medskin, Pietrowski and Steinberg: One relator groups with center, Journal of the Australian Mathematical Society, Volume 16, 1973.

The result of Murasugi can be found in: The center of a group with a single defining relation, Math. Ann., 155, 1964.

When is $Inn(G)$ torsion-free?

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