Problem:
Let $k,n \in \mathbb{N}$ We define the following group homomorphism.
$$\phi_k: (\mathbb{Z}/n\mathbb{Z}) \to (\mathbb{Z}/n\mathbb{Z}), [a] \mapsto [ka]$$
Let $n$ be prime. For which $k \in \mathbb{N}$ is $\phi_k$ a group isomorphism.
Questions:
Right now from what i know is that $\phi_k$ is a homomorphism. So what i need to prove for the group isomorphism, is that $\phi_k$ is either injective or surjective. Though right now i am coming to a little bit of a blockade involving the use of the fact that n is prime. I would require either a comprehensive solution to this relatively probably simple problem. Or some way of helping me get onto the right track of using the properties of n and $\phi_k$. I thank you for your help in advance.
For a finite set $X$ a function $X\to X$ is injective if and only if it is surjective. So it is enough to check when $\phi_k$ is injective. And for homomorphisms it is enough to check when the kernel is trivial.
So assume that $\phi_k([x])=[0]$ which is if and only if $[kx]=[0]$ which is if and only if $n$ divides $kx$. Now if $gcd(k,n)=1$ then the only possibility is that $n$ divides $x$, i.e. $[x]=[0]$. So we get that when $gcd(k,n)=1$ then $\phi_k$ is an isomorphism.
On the other hand, assume that $g=gcd(k,n)\neq 1$. Put $k=k'g$. Then for $x=n/g$ we get $$\phi_k([x])=[kx]=[k'g\frac{n}{g}]=[k'n]=[0]$$ Note that when $g\neq 1$ then $1\leq x <n$ and thus $[x]\neq[0]$. Therefore $\phi_k$ is not an isomorphism.
All in all we get: $\phi_k$ is an isomorphism if and only if $gcd(k,n)=1$. Which for prime $n$ means: if and only if $n$ does not divide $k$.