# finite index, self-normalizing subgroup of $F_2$ [closed]

Denote $F_2=\langle a, b\rangle$ to be the free group on two generators $a, b$.

Let $H\leq F_2$ to be a subgroup with finite index $n$, so $H\cong F_{n+1}$ by Nielsen–Schreier theorem, recall that $H$ is called self-normalizing if the normalizer of $H$ inside $F_2$ to equal to $H$,

Question:

Can anyone give me a subgroup $H\leq F_2$ with finite index and self-normalizing?

Note that $\langle a\rangle$ is a normalizing subgroup but with infinite index. Also $n$ should be odd.

• Just take a non-normal (maximal subgroup) in a 2-generated finite group (it exists in the smallest non-abelian group, of order 6), and pull it back to $F_2$. – YCor Feb 8 at 23:46

Take any finite group generated by two elements with a subgroup of odd index that is self-normalizing, and lift it to $F_2$.
For example, take $S_3$, generated by $(1,2,3)$ and $(1,2)$, and consider the subgroup generated by $(1,2)$. It is self-normalizing. Now consider the induced homomorphism $F_2\to S_3$ mapping $a$ to $(1,2,3)$ and $b$ to $(1,2)$; by the Lattice Isomorphism Theorem, the subgroup $\langle (1,2)\rangle$ of $S_3$ lifts to a subgroup $H$ of $F_2$ of index $3$, and the image of the normalizer of $H$ in $F_2$ is equal to the normalizer of the image of $H$ in $S_3$; hence $H$ is self-normalizing in $F_2$, giving you an example. Plenty of others can be constructed this way.