Let $G$ and $H$ be two finitely generated groups, where $H$ is abelian. I'm curious in which cases $Hom(G,H)$ turns out to be cyclic or virtually cyclic.

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    $\begingroup$ Note that $Hom(G,H)$ has in general no natural group structure, only if $H$ is abelian. $\endgroup$ – Ralph Feb 25 '12 at 18:12
  • $\begingroup$ $\text{Hom}(G, H)$ doesn't carry a natural group structure unless $H$ (and therefore $G$ WLOG) is abelian or $G$ is free (this is a theorem of Kan). In the first case you can finish using the structure theorem and the second case ought to be undecidable even when $G = \mathbb{Z}$. $\endgroup$ – Qiaochu Yuan Feb 25 '12 at 18:13
  • $\begingroup$ Sorry, I missed one assumption. Of course I require $H$ to be abelian. $\endgroup$ – Sebastian Feb 25 '12 at 18:22
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    $\begingroup$ Qiaochu - in your second case, $Hom(F_r,H)\cong H^r$. So, undecidable or not, it's never cyclic unless $H$ is, in which case we're back in the abelian case anyway. $\endgroup$ – HJRW Feb 27 '12 at 10:01

If $H$ is abelian, any homomorphism $G \to H$ factors through the abelianization $G/[G, G] \to H$, so we may assume WLOG that $G$ is also abelian, so we can apply the structure theorem to both $G$ and $H$. Then $\text{Hom}(G, H)$ is virtually cyclic if and only if it has rank at most $1$, hence if and only if both $G$ and $H$ themselves have rank at most $1$, and this is easy to test (it reduces to linear algebra over $\mathbb{Q}$ given a presentation of $G$ and $H$).

To see when $\text{Hom}(G, H)$ is cyclic, write $G_p, H_p$ for the $p$-parts of $G, H$ respectively and $\mathbb{Z}^a, \mathbb{Z}^b$ for the torsion-free parts. Then $$\text{Hom}(G, H) \cong H^a \oplus \bigoplus_p \text{Hom}(G_p, H_p).$$

In particular, it has rank $ab \le 1$. If $ab = 1$, then $H \cong \mathbb{Z}$ and $G$ has rank $1$ so that $\text{Hom}(G, H) \cong \mathbb{Z}$.

If $ab = 0$, then $\text{Hom}(G, H)$ is finite, so it is cyclic if and only if its $p$-parts $$\text{Hom}(G, H)_p \cong H_p^a \oplus \text{Hom}(G_p, H_p)$$

are cyclic. This can occur in a few different ways; either $\text{Hom}(G_p, H_p) = 0, a = 1$, and $H_p$ is cyclic, or $H_p = 0$, or $a = 0$ and $G_p, H_p$ are both cyclic.

  • $\begingroup$ Why is it that H=Z if ab=1? $\endgroup$ – Hiro Feb 25 '12 at 19:54
  • $\begingroup$ @Hiro: if $\text{Hom}(G, H)$ has rank $1$ then it is cyclic if and only if it is torsion-free. Since $G$ has a free summand, $H$ cannot itself have any torsion. $\endgroup$ – Qiaochu Yuan Feb 25 '12 at 19:59
  • $\begingroup$ @Qiaochu Yuan: I understood clearly, thanks! $\endgroup$ – Hiro Feb 25 '12 at 20:04

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