Prove that the total number of distinct group homomorphism between $\mathbb{Z}_m$ and $\mathbb{Z}_n$ is $gcd(m,n)$.
Proof:
Let $x\in\mathbb{Z}_m$, and let $f:\mathbb{Z}_m\to\mathbb{Z}_n$ be a homomorphism. Then $f(x)=x\cdot f(1)$. Thus we just have to determine the possible values of $f(1)$.
We show that $f(x)=a\cdot x$ for some fixed $a\in\mathbb{Z}_n$ is a homomorphism iff $ma\equiv0$ (mod $n$). For that, assume that $f(x)=a\cdot x$ is a homomorphism for some fixed $a\in\mathbb{Z}_n$. Then $0=f(0)=f(m)=a\cdot m$. Since our codomain is $\mathbb{Z}_n$, $ma\equiv0$ (mod $n$). Conversely assume that $ma\equiv0$ (mod $n$). Let $x,y\in\mathbb{Z}$. Since $\mathbb{Z}$ is an Euclidean ring, there are integers $q,r$ such that $x+y=mq+r$ where $0\leq r<m$. Now using our assumption, we get $f(x+y)=f(mq+r)=f(r)=a\cdot r=a\cdot (x+y-mq)=a\cdot x+a\cdot y-a\cdot mq=a\cdot x+a\cdot y=f(x)+f(y)$. Hence $f:\mathbb{Z}_m\to\mathbb{Z}_n$ defined by $f(x)=a\cdot x$ for some fixed $a\in\mathbb{Z}_n$ is $indeed$ homomorphism provided $ma\equiv0$ (mod $n$).
Now by the virtue of techniques developed in Number Theory, We know that the congruence $ma\equiv0$ (mod $n$) has $gcd(m,n)$ distinct solutions.
Here proof is completely understood by me but I want to prove the theorem by using $only$ Group Theoretic techniques. Can someone suggest me an another proof ?