2
$\begingroup$

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.

  • 1
    $\operatorname{Hom}(\mathbb Z/2\mathbb Z, (\mathbb Q/\mathbb Z)^{\mathbb N})$ is not finite.2012-12-19

1 Answers 1

1

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.

  • 0
    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)?2012-12-25