Prove that the only homomorphism between two cyclic groups with distinct prime orders is the trivial one 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.
 A: 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 $C_{\mathrm{gcd}(m,n)}$
A: Suppose $\phi(1)\not=0$.  Then $q\times\phi(1)=0$  So $\phi(q)=0$.  But $q\not=0$ in a cyclic group of size $p\not=q$ (since $p$ and $q$ are distinct primes).  So the kernel of $\phi$ is not trivial. The kernel of $\phi$ is a subgroup, so it must be the whole group.
A: Indeed Lagrange's theorem is enough to prove this. We have $\left|\ker\phi\right|\in\{1,p\}$ and $\left|\phi(C_p)\right|\in\{1,q\}$. Assuming $\left|\ker\phi\right|=1$, we know that $\phi$ is injective, so we would have $\left|\phi(C_p)\right|=\left|C_p\right|=p$, but $p$ doesn't divide $q$, so we arrive at a contradiction. Hence, $\left|\ker\phi\right|=p$ and $\phi$ is trivial.
A: All good answers. Note that $\text{im }\phi$ is a subgroup of $C_q$. However, $C_q$ has only two subgroups: $\{e\}$ and $C_q$ itself. The first possibility is the trivial map, so we need only show the second possibility doesn't occur. If it did, the fundamental isomorphism theorem implies that:
$q = \dfrac{p}{|\text{ker }\phi|}$
which is not possible with two distinct primes, no matter what positive integer we might have for $|\text{ker }\phi|$.
A: Assume $\phi$ is not trivial, if $C_p=\langle x\rangle$, $C_q=\langle y\rangle$, then $\phi(x)=y^k$ for some $k$. But $q$ is a prime, so $|y^k|=|y|$ and $C_q=\langle y^k\rangle$. So w.l.o.g. we just say $\phi(x)=y$. 
Now $\phi(C_p)=\phi(\langle x\rangle)=\{1,y,\dots,y^{p-1}\}$ is a subgroup of $C_q$, which has to be $C_q$ itself since otherwise $p|q$. But this is impossible since $p\neq q$.
