$|\phi(G):\phi(H)|$ divides $|G:H|$ 
Let $\phi$ be a homomorphism defined on a finite group $G$, and let $H\subseteq G$. Show that $|\phi(G):\phi(H)|$ divides $|G:H|$.

Not quite sure where to start on this one. We have a theorem saying that $\phi(H)$ is a subgroup of $\phi(G)$, but what is $|\phi(H)|$? If $H$ contains the kernel $N$ of $\phi$, then I think everything behaves nicely and $|\phi(G):\phi(H)|=|G:H|$. But if $N\not\subseteq H$, I don't know what I can conclude.
 A: Hint: $\phi(H) = \phi(NH)$ and $|\phi(G):\phi(NH)| = |G: NH|$.
Note that finiteness of $G$ is not necessary, we only need $|G: H|$ to be finite.
A: The important thing is finiteness, then we have that
$$|G:H|=|G|/|H|\text{,}$$
$$|\phi(G):\phi(H)|=|\phi(G)|/|\phi(H)|$$
and
$$|H:H\cap\ker\phi|=|H|/|H\cap\ker\phi|\text{.}$$
Here, remember that by the first isomorphism theorem and finiteness
$$|\phi(G)|=|G:\ker\phi|=|G|/|\ker \phi|$$
and
$$|\phi(H)|=|H:H\cap\ker\phi|=|H|/|H\cap \ker \phi|\text{.}$$
Now, do you see that everything is reduced to divide and rearrange terms in order to obtain the wanted divisibility?
A: In one answer it is claimed that $G$ must be finite, in one answer it is claimed that we only need that $|G:H|$ is finite. In fact, no finiteness assumptions are necessary (on the other hand, in the infinite case, the statement is rather boring due to cardinal arithmetic).
Here is an alternative bijective proof, which is perhaps more conceptual and works without any finiteness assumptions: The map $G/H \to \phi(G)/\phi(H), [g] \mapsto [\phi(g)]$ is a well-defined surjective homomorphism of $G$-sets. Thus it suffices to prove:

Lemma. If $f : X \to Y$ is a surjective homomorphism of $G$-sets and $X$ is transitive, then $|Y|$ divides $|X|$. Namely, if $y_0 \in Y$, then $|X|=|Y| \cdot |f^{-1}(y_0)|$.

Proof: Since $f^{-1}(gy)=g f^{-1}(y)$, every fiber of $f$ has the same cardinality $N$. Then $X = \coprod\limits_{y \in Y} f^{-1}(y)$ (as a set) witnesses $|X| = |Y| \cdot N$. $\square$
By the way, applying the Lemma to $G/K \twoheadrightarrow G/H$ for subgroups $K \subseteq H \subseteq G$, we get Lagrange's Theorem $|G:K|=|G:H| \cdot |H:K|$. It is also useful in other contexts.
