Let $\phi:G\to\mathcal{G}$ be a surjective homomorphism with kernel $K$. There is a bijective correspondence between subgroups of $\mathcal{G}$ (where we denote as $\mathcal{H}=\phi(H)$) and subgroups of $G$ that contain $K$ (where we denote as $H$).
I am trying to show that if $H$ and $\mathcal{H}$ are corresponding subgroups, then $|H|=|\mathcal{H}||K|$.
I tried using the counting formula, we have $|G|=|\text{ker }\phi|\cdot|\text{im }\phi|$. We can then restrict the map $\phi$ to $H$, and we get $\phi|_H:H\to\mathcal{G}$. And by definition, we get $\text{ker }(\phi|_H)=(\text{ker }\phi)\cap H$. So we have, $|H|=|\text{ker }\phi\cap H|\cdot|\phi(H)|=|K\cap H|\cdot|\mathcal{H}|$.
This is where I got stuck, unless $|K\cap H|=|K|$, but I don't think we can show that.
Is there any mistake in my proof? Could somebody give me some hints on how to fix my proof?
Thanks!
In fact what you are trying to prove is a direct consequence of the first isomorphism thm. You can prove it by taking $g: G/K \rightarrow$ image$(\phi)$: $g(x + K) = \phi(x)$