# Hom and tensor are isomorphic for finite cyclic groups

Something I noticed today: $\mathsf{Ab}(\mathbb{Z}/m, \mathbb{Z}/n) \cong \mathbb{Z}/m \mathop{\otimes}\limits_\mathbb{Z} \mathbb{Z}/n \cong \mathbb{Z}/\operatorname{gcd}(m,n)$.

I'm not sure there's much deeper going on there, like for instance a nice consequence of the hom/tensor adjunction. I'd say the upshot is that that cyclic groups are really rigid. (Two important things here are that $\mathbb{Z}/m \to \mathbb{Z}/m \mathop{\otimes}\limits_\mathbb{Z} \mathbb{Z}/n$ is surjective, and everything in sight has to be cyclic, so same order implies isomorphic.

Does anyone have something to add to this? Does this reveal anything cool? Or have connections to anything?

• Ab being the abelianization? – Zelos Malum Dec 28 '16 at 6:56
• @ZelosMalum Ab(A,B) meaning the mapping space. – Eric Auld Dec 28 '16 at 7:00
• Do you mean Hom functor then? – Zelos Malum Dec 28 '16 at 7:21
• @ZelosMalum Yes. In a category $\mathscr{C}$, another notation for $\operatorname{Hom}_\mathscr{C}(a,b)$ is $\mathscr{C}(a,b)$. – Eric Auld Dec 28 '16 at 17:21
• I was not familiar with it, only seen "Ab" for Abelianization – Zelos Malum Dec 28 '16 at 17:22

I know the question is old, but was wondering the same thing. A partial answer is that the Hom/tensor adjunction reduces the problem of showing, say, $$\mathrm{Hom}_{\mathbb Z} (\mathbb Z/m ,\mathbb Z/n)\cong \mathbb Z /(m,n)$$ if we know the corresponding statement about tensor. For short, let $$d = (m,n)$$. Then \begin{align*} \mathrm{Hom}_{\mathbb Z}(\mathbb Z /m,\mathbb Z/n)&\cong \mathrm{Hom}_{\mathbb Z}(\mathbb Z/m,\mathrm{Hom}_{\mathbb Z}(\mathbb Z/n,\mathbb Z/n)) \\ &\cong \mathrm{Hom}_{\mathbb Z}(\mathbb Z /m\otimes_{\mathbb Z}\mathbb Z/n,\mathbb Z/n)\\ &\cong \mathrm{Hom}_{\mathbb Z}(\mathbb Z/d,\mathbb Z/n)\\ &\cong \mathbb Z/d. \end{align*} The first isomorphism is trivial and the second is the Hom/tensor adjunction. The final isomorphism is certainly not as bad in the general case (since here $$d$$ divides $$n$$).

Similar tricks probably work in the opposite direction, i.e. going from the statement about Hom to the one about tensor.