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 $H$,


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.


closed as off-topic by YCor, Derek Holt, Andrey Rekalo, Jiang, David White Dec 3 '13 at 13:34

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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.

  • $\begingroup$ Thanks, I did not realize this simple observation. $\endgroup$ – Jiang Dec 3 '13 at 5:41

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