Let $m,n,d$ be positive integers.find the sufficient and necessary condition that $f:\mathbb{Z}_n=\{0,1,\ldots,n-1\}\rightarrow\mathbb{Z}_m$ such that $f(x)=dx$ is a group homomorphism.
i tried to solve this with noticing that $f$ is uniquely determined just with $f(1)$ and we know that $O(f(1)=d)|O(1)=m$ so we must have $\frac{n}{(n,d)}|m$.
i think it is sufficient condition,but how to prove that?
thanks a lot
At first we should explicit the definition of $f$ . You say $f(x)=dx$ this means multiplication by $d$ act on the set $ \{0,1,...,n-1\}$ and then you consider the answers$\mod m$.
For example if $n=5,m=10,d=3$ then $f(4)=3*4 \pmod {10} = 2$ .
For checking homomorphism of $f$ we should show : $$f(x +_n y)=f(x) +_m f(y)$$
I wrote $+_n$ and $+_m$ to mention the difference between operations of $\mathbb Z_n$ and $\mathbb Z_m$.
It's equal to say $$d(x +_n y) \equiv d(x + y) \pmod m\quad \forall x,y \in \mathbb Z_n$$ and by congruence theorems if $\gcd(m,d)=k$ we can cancel $d$ from both sides and get $$x +_n y \equiv x + y \pmod {m/k}\quad \forall x,y \in \mathbb Z_n$$
Set $x=1 , y=n-1$ hence we should have $\frac{m}{k}|n$ .
And if $\frac{m}{k}|n$ there is no difference between $x +_n y \mod \frac{m}{k}$ and $x+y \mod \frac{m}{k}$ .
Therefore $f$ is a homomorphism if and only if $\frac{m}{\gcd(m,d)}|n$ .