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If $A$ is a finite group and $D$ is a divisible abelian group, then will $Hom(A,D)$ be a finite group?. My thought are that if $D$ is torsion free then this would be true, but I'm worried if $D$ was something like $\mathbb{Q}/\mathbb{Z}$.

Thank you.

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$\operatorname{Hom}(\mathbb Z/2\mathbb Z, (\mathbb Q/\mathbb Z)^{\mathbb N})$ is not finite. –  Hagen von Eitzen Dec 19 '12 at 16:07

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up vote 1 down vote accepted

Any morphism from a group $G$ to an abelian group factors through the abelianized $G^{\text{ab}}$. Moreover $G$ finite implies that $G^{\text{ab}}$ is aproduct of cyclic groups and since $$\hom(C_1\times C_2,-)\simeq\hom(C_1,-)\times\hom(C_2,-)$$ we may assume that $G$ is finite cyclic. On the other hand $$ \hom(\Bbb Z/n\Bbb Z,\Bbb Q/\Bbb Z)\simeq\Bbb Z/n\Bbb Z $$ as one can see by examining what can be the image of $\bar 1$ under a map.

On the other hand, if we take as target group a product of infinitely many copies of $\Bbb Q/\Bbb Z$, we sure have infinitely many morphisms.

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Thank you, this is what I was worried about, so i thought I'd ask. I think $Hom(A,D)$ would be finite if the torsion sungroup of D was finite, does this sound reasonable? –  Chris Birkbeck Dec 19 '12 at 16:27
    
Very much so, because if $G$ is finite (or more generally, torsion) the image of a morphism $G\rightarrow A$ ($A$ abelian) is contained in the torsion subgroup of $A$. –  Andrea Mori Dec 19 '12 at 16:29
    
Now I'm trying to find conditions for $D$ under which $Hom(X,D$ is finite, I thought $D$ having finite torsion group would be good, but I know this is true for $D=\mathbb{C}^{\times}$ so is there a better restriction to put on $D$ to make this true? –  Chris Birkbeck Dec 24 '12 at 14:42
    
I believe that the following two are equivalent: (1)$\text{Hom}(G,D)$ is finite for all finite groups $G$; (2) for all $n$, D has finitely many elements of order $n$. Is there a "simpler" way to express (2)? –  Andrea Mori Dec 25 '12 at 11:36

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