Group homomorphism from $\mathbb{Z}_p$ to $\mathbb{Z}_q$ By $\mathbb{Z}_p$ of course I mean $p$-adic integers.

I am fairly certain that there is no non-trivial group homomorphism between $\mathbb{Z}_p$ and $\mathbb{Z}_q$ for $p\neq q$ as additive groups but I can't come up with a simple proof.
Note that $q$ is invertible in $\mathbb{Z}_p$.  So in particular for any $a$ in $\mathbb{Z}_p$ there is a unique $a'$ such that $qa'=a,$ a unique $a''$ such that $q^2a''=a$, and so on.

However in $\mathbb{Z}_q$, $0$ is the only element infinitely divisible by $q$ in this sense. Therefore everything must get sent to $0$.
For prime $p$ and $q$ you already have the answer.

I would however adjust Nate's reasoning as follows: there is no need for doing "infinite" division by $q$. Simply, since in $\mathbb Z_p$ you can divide by $q$ then for some $a$ we have $qa = 1$. In case homomorphism $h: \mathbb Z_p\rightarrow \mathbb Z_q$ exists we have $h(a+a+...+a) = h(a)+...+h(a)$  ($q$ summands in each) and therefore $h(1) = qh(a) = 0$, so homomrphism is trivial.

Same proof if $p$ and $q$ are just co-prime.