Given finite groups $G:|G|=70,H:|H|=91$, how many distinct homomorphisms $f:G\to H$ exist?
Noting that $|G|=7\cdot5\cdot2,|H|=7\cdot13$, and that $|f(G)| \big\vert|G|$ and $|f(G)|\big\vert|H|$, we see that $|f(G)|\in\{1,7\}$. But how do I proceed from here...?
$G$ has a subgroup of order $7 \cong C_7$. Say $x$ is the generator. $x$ and his powers are the only elements that can hope not to be in $\ker f$, because a homomorphism conserves the order of an element, and there aren't elements of order 2 or 5 or multiples of 2 and 5 in $H$. Since $H$ has a subgroup of order $7$ too (call $y$ the generator), the homomorphisms $x\mapsto y^i, \quad i=0,\dots, 6$ are the only possibilities.
So $7$ is the right answer.
Note that $G$ and $H$ have only one subgroup of order $7$, and that descends from Sylow's theorems.