I have read this statement, but I don't find why it's true:
Let $G$ be a group and $H\leq G$ with $[G:H]=m$. Let $N$ be the largest normal subgroup of $G$ contained in $H$. Then there is a group homomorphism $\varphi\colon G\longrightarrow \text{S}_m$ with kernel $N$. In particular $G/N$ is isomorphic to a subgroup of $\text{S}_m$.
I understand that we can see $G/N$ as a set of cosets, but I don't see why this lets me make the homomorphism. Can anyone explain me why is true?
For any group $G$ and subgroup $H$ there is a natural action of $G$ on the set of cosets $G/H$: $g\cdot xH= gxH$. This action is clearly transitive.
So we have a homomorphism of $\phi\colon G\to Perm(G/H)$, (here $Perm(X)$ denotes group of permutations of a given set $X$).
Coming to the scenario of your question for this action elements of $N$ act trivially: $n\cdot xH = xn'H$ (for some $n'\in N$ as $N$ is a normal subgroup).
But $n'H = H$ by the hypothesis that $N\subset H$. So $n\cdot xH= xH$.
That is $N$ acts trivially and so is in the kernel of $G\to Perm(G/H)$
This shows that $G/N$ is a subgroup of $S_m$ as $m=[G:H]$