Show that $\pi^{-1}(\tilde{H})$ is a subgroup of order $|\tilde{H}|\cdot |N|$ where $\tilde{H}$ is a subgroup of $G/N$ Since the title does not explain the problem well enough, I'll elaborate in the following lines:
Let $G$ be a finite group with $N\lhd G$. If $\tilde{H}$ is a subgroup of $G/N$, show that $\pi^{-1}(\tilde{H})$ is a subgroup of order $|\tilde{H}|\cdot |N|$. Here, the map $\pi$ is the canonical map $G\to G/N$ given by $g\mapsto gN$.
I think that we should use the canonical homomorphism theorem, which is that $\pi$ is a surjective homomorphism with $\text{ker }\pi = N$. But I can't figure out how to go further. Can you help me to prove this problem?

Since $\pi(1)=N\in \tilde H$, $1\in \pi^{-1}(\tilde H)$.
Let $x,y\in \pi^{-1}(\tilde H)$.
Then $\pi(x),\pi(y)\in \tilde H$, which means $xN,yN\in \tilde H$.
Since $\tilde H$ is a group,
$$\pi(xy^{-1})=(xy^{-1})N=xN(yN)^{-1}\in \tilde H.$$
Thus $xy^{-1}\in \pi^{-1}(\tilde H)$.
We conclude that $\pi^{-1}(\tilde H)$ is a subgroup of $G$.
Next we want to prove that $\tilde H=\pi^{-1}(\tilde H)/N$.
For $n\in N$, $\pi(n)=N\in \tilde H$. Hence $n\in\pi^{-1}(\tilde H)$.
We have $N\subseteq \pi^{-1}(\tilde H)$.
Let $gN\in\tilde H$, that is $\pi(g)\in \tilde H$. Hence $g\in \pi^{-1}(\tilde H)$.
This means that $\tilde H\subseteq \pi^{-1}(\tilde H)/N$.
Conversely, let $gN\in \pi^{-1}(\tilde H)/N$.
Then $g\in \pi^{-1}(\tilde H)$, which means $gN\in \tilde H$.
So we have $\pi^{-1}(\tilde H)/N \subseteq \tilde H$
By Lagrange's Theorem,
$$|\tilde H|=|\pi^{-1}(\tilde H)/N|=|\pi^{-1}(\tilde H)|/|N|.$$
Therefore we have
$$|\pi^{-1}(\tilde H)|=|\tilde H||N|.$$
