Show $\operatorname{Hom}_{\mathbb{Z}}\left ( \mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z} \right )\cong \mathbb{Z}/\left ( n,m \right )\mathbb{Z}$

Question

Show that $\operatorname{Hom}_{\mathbb{Z}}\left ( \mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z} \right )\cong \mathbb{Z}/\left ( n,m \right )\mathbb{Z}$

I think that the hom-set (of $\mathbb{Z}$ module homomorphisms ) is isomorphic to $\left \{ a\in \mathbb{Z}/m\mathbb{Z},na=m\mathbb{Z} \right \}=\left \{ k+m\mathbb{Z} \right \}$ where $m\mid (nk)$ but I can't show that it's isomorphic to $\mathbb{Z}/\left ( n,m \right )\mathbb{Z}$

See also here and here. – Dietrich Burde May 27 '20 at 10:06 — Dietrich Burde, May 27 '20 at 10:06

Hanul Jeon · Answer 1 · 2020-05-27T09:19:00.297

2

Hint. Let $f:\Bbb{Z}/n\Bbb{Z}\to\Bbb{Z}/m\Bbb{Z}$ be a homomorphism then you can check that $$0=f(0)=f(n\cdot 1)=nf(1)$$ so $m\mid nf(1)$ and $m/(m,n)\mid f(1)$. Also, $f(1)$ determines whole $f$.

edited May 27 '20 at 09:19

answered Feb 11 '15 at 15:53

Hanul Jeon

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How are you getting $(m,n) \mid f(1)$? – user5826 May 27 '20 at 07:46
@aljebr I was wrong, it should be $m/(m,n)\mid f(1)$ – Hanul Jeon May 27 '20 at 09:18
Got it. So, we have that $f(1)$ must be a multiple of $m/(m,n)$, i.e., $f(1) \in \frac{m}{(m,n)}\mathbb Z$. Can you help me proceed from here? – user5826 May 27 '20 at 17:54
More specifically, how do we know that $f(1)$ cannot be a number small than $m/(m,n)$? – user5826 May 27 '20 at 18:43
@AlJebr f(1) can be 0, so it can be smaller. The gist of the proof is that multiples of $m/(m,n) \pmod n$ are the only possible values of $f(1)$. – Hanul Jeon May 27 '20 at 19:37

score 2 · Answer 2 · answered Feb 11 '15 at 16:08

the seq. $Z\to Z\to Z_m \to 0$ is exact, when the 1st homomorphism, $f$ is defined as $f(a)=ma$ and the 2nd, $g$ as $g(x)=\bar x$.
so we have the exact seq.: $$0\to Hom_Z(Z_m,Z_n)\to Hom_Z(Z,Z_n)\to Hom_Z(Z,Z_n)$$ here the 1st homomorphism, is $hom(g,1)$ (which is injective map), and the 2nd is $hom(f,1)$.

so $ Hom_Z(Z_m,Z_n)/ker (hom(g,1)) \equiv ker(hom(f,1))$.
now can you prove that $ker(hom(f,1))\equiv Z/\left ( n,m \right )Z$ ?

Show $\operatorname{Hom}_{\mathbb{Z}}\left ( \mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z} \right )\cong \mathbb{Z}/\left ( n,m \right )\mathbb{Z}$

2 Answers2

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