I was reading about groups of order 12 and found this:
Let $G$ be a non-abelian group of order $12$. By Cauchy's theorem, it has an element, hence subgroup, $H$, of order $3$
H is not normal in G, then as $[G:H]=4$, there is a homomorphism $ϕ:G→S_4$
I don't understand why there is such homomorphism and what would it be.
I understand that since $[G:H]=4$, I could define the set of the left cosets of $H$ in $G$, $A=\{H, g_1H, g_2H, g_3H\}$ and the group of permutations of $A$ would be isomorphic to $S_4$. But I still don't know how I could define the homomorphism $\phi$. Any help would be greatly appreciated.
A basic fact about group actions is that if a group $G$ acts on a set $X$, then that is equivalent to having a group homomorphism $\phi:G\to\mathrm{Sym}(X)$. This is proved by noting that, for each $g\in G$, the map $$\sigma_g:X\to X$$ given by $\sigma_g(x)=g.x$ is a bijection (with inverse $\sigma_g^{-1}=\sigma_{g^{-1}}$). The assignment $\phi(g)=\sigma_g$ is the desired homomorphism.
Examples:
$G$ acts on itself by left multiplication, so there is a homomorphism $G\to \mathrm{Sym}(G)$ (Cayley's theorem)
The dihedral group $D_{2n}$ acts on the set of vertices of a regular $n$-gon. This yields a homomorphism $D_{2n}\to S_n$
If $H\leq G$ is a subgroup, then $G$ acts on the cosets $G/H$ by left multiplication. This yields a homomorphism $G\to \mathrm{Sym}(G/H)$.
In your question, $[G:H]=4$, so example 3 applies and you get a homomorphism $G\to S_4\cong\mathrm{Sym}(G/H)$.