More formally, I want to prove that for any cyclic groups $C_p$ and $C_q$ where $p, q$ are distinct primes, the only homomorphism $\phi:C_{p} \rightarrow C_{q}$ is $\phi(g) = e$ for all $g \in C_{p}$.
I think I have a proof, but I want to make sure it is correct. I also want to know if there is a nicer way to prove it that doesn't involve splitting the problem into cases. Here is my solution:
Case 1: $|\ker(\phi)| = p$. This is produces the trivial homomorphism since it must be that $\phi(g) = e$ for all $g \in C_{p}$.
Case 2: $1 < |\ker(\phi)| < p$. Then $\ker(\phi)$ is not a subgroup of $C_p$ By Lagrange's theorem, since $|\ker(\phi)|$ doesn't divide $|C_p|$. Thus I have a contradiction since the kernel must be a subgroup of the domain.
Case 3: $|\ker(\phi)| = 1$. Then $C_p \subset C_q$ and furthermore $C_p$ is a subgroup of $C_q$. I haven't proved this yet but it seems true. But $|C_p|$ doesn't divide $|C_q|$ so using Lagrange's theorem again I obtain a contradiction.
Then the only possible case is the first: the only homomorphism is the trivial one.
Best Answer
Your proof works.
The key idea is that every non-identity element of $C_q$ is a generator, since $q$ is prime. So if $C_p$ is generated by $g$, then if a homomorphism $\theta$ does not match $g \mapsto e$ (i.e. if the homomorphism is not trivial) then it must map $g$ to a generator of $C_q$. But then $$e=\theta(e) = \theta(g^p) = \theta(g)^p = h^p$$where $h$ is a non-zero element of $C_q$. But since $q\nmid p$, we can't have $h^p = e$, so this is a contradiction.
More generally, if $C_n$ and $C_m$ are cyclic groups of orders not necessarily prime, then the image of any homomorphism $C_n\to C_m$ will be isomorphic to a subgroup of $C_{\mathrm{gcd}(m,n)}$