Finite index non-normal subgroups of free groups. Consider a free group on some variable set $A=\{a_1,a_2,\cdots,a_n\}$
$n$ is dependent on us, we can choose it.
I want to find some finite indexed non-normal subgroup of this free group on $A$.
For example how can we construct a subgroup $H$ of Free group on $\{a,b\}$ which is non-normal and index of $H$ is $3$? I know some methods to find normal subgroups but for non-normal subgroups I can't construct.
For normal subgroups
Consider a map $$\phi:F_2\to \mathbb Z_n\\a_1\to \bar 0\\ a_2\to \bar 1\\ F_2/_{<ker\phi>}\simeq \mathbb Z_n$$
And yes we know $\{(1),(1,2)\}\subset S_3$ is non normal subgroup but how to relate this fact to free groups? How can I find index 3 non normal subgroup?
 A: For normal subgroups we can use homomorphisms, and the kernels themselves will be our normal subgroups as suggested in the OP. For non-normal subgroups we can still use homomorphisms, but finding the subgroups requires an extra observation:

Let $\phi:G\twoheadrightarrow H$ be a surjective group homomorphism. If $K$ is a non-normal subgroup of $H$ then $\phi^{-1}(K)$ is a non-normal subgroup of $G$. Moreover, $|H:K|=|G:\phi^{-1}(K)|$.

This result is pretty standard and not too hard to prove. For example, it follows from the Correspondence Theorem.
So, writing $F_n$ for the free group of rank $n$, to find a non-normal subgroup $L$ of $F_n$ with index $i=|F_n:L|$ we simply have to find a finite group $H$ which can be generated by $n$ elements and which contains a non-normal subgroup $K$ of index $i$; as $H$ can be generated by $n$ elements there exists a map $\phi:F_n\twoheadrightarrow H$, and then $L:=\phi^{-1}(K)$ has the required properties by the above observation.
For $i=3$, take $H$ to be the symmetric group $S_3$ (this can be generated by $2$ elements, and hence by $n$ elements for all $n\geq 2$ as we don't really care about repetition), while it contains a non-normal subgroup of index $3$, say $K=\{id, (12)\}$.
For general index $i$, take $H$ to be the symmetric group $S_i$ and take $K$ to be canonical image of $S_{i-1}$ in $S_i$ (all permutations of $\{1, \ldots, i\}$ which fix $i$), so here $|S_i:K|=i$. Moreover, $K$ is not normal in $S_i$ because for example $(12\cdots i-1)$ and $(23\cdots i)$ are conjugate in $S_i$, but $(12\cdots i-1)\in K$ as it fixes $i$ while $(23\cdots i)\not\in K$ because it does not fix $i$.
