-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $F_2$. $\endgroup$ – YCor Feb 8 at 23:46
4
$\begingroup$

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.

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.