I cannot seem to make any progress at all with this proof. Suggestions and hints. No full proofs, please. Thanks!
Suppose $\phi:G\to G'$ is a homomorphism of finite groups, and define $K=\ker(\phi)$. If $|K|=k$ and $N$ is a subgroup of $G'$, show that $$ \left|\phi^{-1}(N)\right|=k\cdot|N| $$ That is, prove that for every $n\in N$, there are precisely $k$ elements in $G$ that map to $n$. One might say in this case that $\phi$ is a $k$-to-one homomorphism.
This was 2015 and the question shows up in search results, so it's time for a full proof.
The statement is not true (in general) if $\phi$ is not surjective. Consider the embedding. $\mathbb Z_2\to \mathbb Z_2 \times \mathbb Z_2$ of finite abelian groups. $\mathbb Z_2\times \mathbb Z_2$ has order $4$, its preimage has order $2$, and the kernel has order $1$.
In the case of a canonical quotient $\pi: G \to G/N$ the statement is clearly true. $\pi$ sends cosets of size $|N|$ to points, so for each subset $S$ of $G/N$ one gets $|\pi^{–1}(S)| = |S|\, |N|$. Let $\phi: G \to H$ be a surjective morphism of finite groups. The first isomorphism theorem states that $\phi$ is, up to composition with an isomorphism, a canonical quotient map as above, and the result follows immediately.
Consider using first theorem of isomorphism and Lagrange theorem. Notice that $K$ is a subgroup of $(\phi)^{-1}(N)$
