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It is well known that a torsion-free group which is virtually free must be free, by works of Serre, Stallings, Swan...

Is there a simple cohomological proof of the fact that a torsion-free group which is virtually-$\mathbb{Z}$ must be isomorphic to $\mathbb{Z}$?

Thanks.

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1 Answer 1

Here's a proof that requires no cohomology at all.

Let $H$ be the infinite cyclic subgroup of finite index. Your group $G$ will be finitely generated, say by $g_1,\ldots, g_n$. Now the centralizer of any of these, say $C_G(g_1)$, will intersect $H$ non-trivially, and hence $C_G(g_1)\cap H$ is of finite index in $H$; hence is of finite index in $G$. But then $Z(G)=\cap C_G(g_i)$ is of finite index in $G$; this implies that $G'$ is finite, and since $G$ is torsion-free, $G'$ is actually trivial, so $G$ is abelian. But the only f.g. (torsion-)free abelian group which is virtually cyclic is $\mathbb{Z}$.

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Nice! I'm being stupid now, but why $H\cap C_G(g_i)$ is non-trivial? Is your last sentence really obvious? –  P M Oct 7 '11 at 2:08
    
Of course, there really is some cohomology hidden in there somewhere, since I have used Schur's theorem on finite index centers. –  user641 Oct 7 '11 at 2:08
    
@PM: $[C_G(g_i) : H\cap C_G(g_i)]\le [G:H]$, but $C_G(g_i)$ is torsion free... As for the last sentence, we're talking about $\mathbb{Z}^n$ groups; what values of $n$ could give us virtually cyclic? –  user641 Oct 7 '11 at 2:10
    
Ok... Sure, f.g. torsion-free abelian is free abelian. Yes, I saw Schur. Thanks! –  P M Oct 7 '11 at 2:15

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