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.



finite index, self-normalizing subgroup of $F_2$
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.


