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Let $G,H$ be groups and let $Hom (G,H)$ be the set of group homomorphisms from $G$ to $H$. Prove that $Hom (G,H)$ is a group. Find the order of $Hom (\mathbb{Z}/ m \mathbb{Z}, \mathbb{Z}/ n \mathbb{Z})$ in terms of $m$ and $n$.

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Have you tried something? –  Najib Idrissi Jun 4 '13 at 12:07
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...please...? ...some self effort there...? –  DonAntonio Jun 4 '13 at 12:09
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I don't think you'll find that $\operatorname{Hom}(G,H)$ is a group unless $H$ is abelian. Take $G = H$ and $1 \in \operatorname{Hom}(G,H)$. What is the product of $1$ and $1$ in the group on $\operatorname{Hom}(G,H)$? –  Jack Schmidt Jun 4 '13 at 12:24
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Questions posted here should come with citations of the source, and indication of having put more effort into the question than just copy-pasting it in. Also, some idea of what you know about the topic would be helpful to anyone trying to compose an answer. –  Gerry Myerson Jun 4 '13 at 12:38
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Here is what I get: I think the group operation is $(f+g)(x)= f(x) + g(x)$? And the identity is $Id: G \rightarrow H$ which sends everything to $e_H$? And the inverse of $f$ is $(-f)(x)= -f(x) $? But then I still don't have any clue on the order of $Hom (\mathbb{Z}/m\mathbb{Z}, \mathbb{Z}/n\mathbb{Z})$... –  rosa Jun 4 '13 at 14:24
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