I'm working on a problem that asks me to show all the homomorphisms from $\Bbb Z_{10}$ to $\Bbb Z_{15}$.
So far, my attempt is as follows:
Since $\Bbb Z_{10}$ and $\Bbb Z_{15}$ are both cyclic, I can just define a homomorphism on $1$ in $\Bbb Z_{10}$, and extend to the other elements (since $1$ is a generator of $\Bbb Z_{10}$).
$|1| = 10$, so $|f(1)$| has to be either $1$, $2$, $5$, or $10$. $|f(1)| \mid 15$, so $|f(1)| = 1, 3, 5$ or $15$.
The two statements combined give $|f(1)| = 1$ or $5$.
If $ f(1)$ has order $1$, then $f(1) = 0$ and for all $g$ in $\Bbb Z_{10}$, $f(g) = 0$ in $\Bbb Z_{15}$.
I'm have trouble with the case where $f(1)$ has order $5$. If $f(1)$ has order $5$, how do I determine what values $f(1)$ can actually have?
Your approach is great, and in the comments it looks like you may have your answer.
Alternatively, you could view this as a question about going from $\mathbb{Z}_2\times\mathbb{Z}_5$ to $\mathbb{Z}_3\times\mathbb{Z}_5$. Instead of one generator, you have two. But $(1,0)$ has to go to $(0,0)$, since nothing has order $2$ in $\mathbb{Z}_3\times\mathbb{Z}_5$.
You are left asking where $(0,1)$ can go, and of course it can only go to any element of the form $(0,a)$. If $a=0$, the whole map is trivial. Otherwise, $a$ is a generator of $\mathbb{Z}_5$.
So the image of $(0,1)$ is $(0,a)$. Since $(1,1)$ is a generator of $\mathbb{Z}_2\times\mathbb{Z}_5$, we focus on the fact that $(1,1)$ goes to $(0,a)\in\mathbb{Z}_3\times\mathbb{Z}_5$. In general in $\mathbb{Z}_3\times\mathbb{Z}_5$, $(x,y)\leftrightarrow5x+3y\in\mathbb{Z}_{15}$. So the generator of $\mathbb{Z}_{10}$ is being mapped to $3a\in\mathbb{Z}{15}$.