\operatorname {gcd}(m,n).$

Asked Dec 02 '23 at 15:30

Active Dec 02 '23 at 15:42

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I'm having a hard time proving the statement in the title.

In one proof I read, they used the isomorphism $\phi:\operatorname {Hom}_{{{\mathbb {Z}}}}({\mathbb {Z}}/m,{\mathbb {Z}}/n)\to{\mathbb {Z}}/\operatorname {gcd}(m,n).$ that sends $ \phi(\varphi)= \varphi([1]_m)$.

However, in the case that $m=3,n=6$ a homomorphism $\varphi$ can map $\varphi([1]_m)\in\{0,2,4\}$. But if $\varphi([1]_n)=4$ so $\varphi([1]_m)\notin {\mathbb {Z}}/\operatorname {gcd}(m,n)$. So, how is this isomorphism even well-defined?

Please, if anyone can help me prove the statement (without exact sequences), thanks!

edited Dec 02 '23 at 15:32

asked Dec 02 '23 at 15:30

yoyok3

The definition of $\phi$ is wrong. If $\varphi(1 \bmod m)=k \bmod n$, we have $\phi(\varphi)=k \bmod \mathrm{gcd}(m,n)$. More details can be found in the linked duplicates. – Martin Brandenburg Dec 02 '23 at 15:43
thanks! so, with the new def is this indeed isomorphism? – yoyok3 Dec 02 '23 at 15:48

How to prove that $\operatorname {Hom}_{{{\mathbb {Z}}}}({\mathbb {Z}}/m,{\mathbb {Z}}/n)\cong{\mathbb {Z}}/\operatorname {gcd}(m,n).$

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