$|\phi^{-1}(\phi(H))| = |\ker \phi||\phi(H)|$ ? where $H \leq G$ Suppose $G$ is a group of finite order, let $\phi: G\to G'$ be a homomorphism and $H$ be a subgroup of $G$ then is it always true that $|\phi^{-1}(\phi(H))| = |\ker \phi||\phi(H)|$ ?
I was reading group homomorphisms and its properties and here's my observation
 A: This is just the statement that fibers of $\phi$ are disjoint and have the same number of elements.
Concretely, if $h \in H$, then $\phi(x)=\phi(h)$ iff $xh^{-1} \in \ker\phi$ and so $$\phi^{-1}(\phi(h))=\{ x \in G : \phi(x) = \phi(h) \} = (\ker\phi) h$$
Apply this to $\phi^{-1}(\phi(H))= \bigcup_{h \in H} \phi^{-1}(\phi(h))$.
A: Let $H'\le {\rm im}(\phi)\le G'$ be an arbitrary subgroup.
Then apply the first isomorphism theorem for $\phi|_{\phi^{-1}(H')}:\phi^{-1}(H')\to H'$, and note that $\ker\phi\subseteq\phi^{-1}(H')$.
A: First let's see $\phi^{-1}(\phi(H))$ is a subgroup. Of course $0\in\phi^{-1}(\phi(H))$. Now, if $x,y\in\phi^{-1}(\phi(H))$ then $\phi(x),\phi(y)\in\phi(H)$, which is a subgroup, so $\phi(x*y^{-1})=\phi(x)\phi(y)^{-1}\in\phi(H)$, so $x*y^{-1}\in\phi^{-1}(\phi(H))$ and it is a subgroup indeed.
Then we can consider the homomorphism $\varphi={\phi}_{\mid\phi^{-1}(\phi(H))}$ (the restriction). Then, by the first isomorphism theorem, $\phi^{-1}(\phi(H))/\ker(\varphi)\simeq\varphi(\phi^{-1}(\phi (H)))=\phi(\phi^{-1}(\phi(H)))=\phi(H)$.
Since $\ker(\varphi)$ is a normal subgroup of $\phi^{-1}(\phi(H))$ then $\left|\phi^{-1}(\phi(H))/\ker(\varphi)\right|=\left|\phi^{-1}(\phi(H))\right|/\left|\ker(\varphi)\right|$.
Now, clearly $\ker(\varphi)\subset\ker(\phi)$. But since $0\in\phi(H)$ we have that $\ker(\phi)=\phi^{-1}(0)\subset\phi^{-1}(\phi(H))$, so we can do $\varphi(\ker(\phi))=\phi(\ker(\phi))=0$ and then $\ker(\phi)\subset\ker(\varphi)$. Therefore $\ker(\phi)=\ker(\varphi)$.
We conclude $\left|\phi^{-1}(\phi(H))\right|/\left|\ker(\phi)\right|=\left|\phi^{-1}(\phi(H))/\ker(\phi)\right|=\left|\phi^{-1}(\phi(H))/\ker(\varphi)\right|=$
$=\left|\phi^{-1}(\phi(H))\right|/\left|\ker(\varphi)\right|=\left|\phi(H)\right|$.
