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Consider $H$ abelian (so $\operatorname{Hom}(G,H)$ has a group structurep) and the group operation of $\operatorname{Hom}(G,H)$ is defined by $(\phi_{1}\phi_{2})(g):=\phi_{1}(g)\phi_{2}(g)$.

Take $G=S_{3}$ and $H=<e>$. Then $\operatorname{Hom}(G,H)$ has only the trivial homomorphism $\phi(g)=e$, for all $g\in G$. Hence, $|\operatorname{Hom}(G,H)|=1$ and $\operatorname{Hom}(G,H)\simeq <e>=H$. Thus $\operatorname{Hom}(G,H)$ is abelian, but $G=S_{3}$ does not abelian.

My question is: Do you know another examples? Thanks.

Bernard
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Margot Viènna
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1 Answers1

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Take for $G$ an arbitrary finite non-abelian group an for $H$ and arbitrary group whose order satisfies for all primes $p$ $$p \mid |G| \Rightarrow p \nmid |H|.$$ In other words $|G|$ and $|H|$ are coprime. This condition ensures that there is only one homomorphism $\varphi : G \to H$, namely the trivial one. A special case of this is $|H|=1$ which Chris Custer pointed out in the comments.

By the way: If $H$ is abelian $\operatorname{Hom}(G,H)$ is abelian as well: $$(\phi_1\phi_2)(g)=\phi_1(g)\phi_2(g)=\phi_2(g)\phi_1(g)=(\phi_2\phi_1)(g).$$ So since you demand $H$ to be abelian there is no counter-example!

principal-ideal-domain
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