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Let $G \cong PSL(2, q)$ with $q \equiv 3 \pmod{8}$ or $q \equiv 5 \pmod{8}$. If $u \in G$ is an involution, then $C_G(u)$ is a dihedral subgroup of order $q \pm 1$.

I know that $PSL(2,q)$ could be identified with all mappings on $\mathbb F_q \cup \{\infty\}$ of the form $$ x \mapsto \frac{ax + b}{cx + d} $$ with $a,b,c,d \in \mathbb F_q$ and $ad - bc = 1$ and the definitions that $\infty$ mapsto $\infty$ for $c = 0$, and to $a / c$ otherwise, and $-d/c$ maps to $\infty$ for $c \ne 0$.

I tried to compute what form an involutory map should have, but I am stuck in the computations and do not know how to fix it. For specific involutory maps I can show that the normalizer must be dihedral, for example suppose $x^g = a^2x$ (as $a^2 = ax / a$ this fulfills the condition $ad - bc = 1$) with $a^4 = 1$. Then I can show that the normalizer of $\langle g \rangle$ is a dihedral group. And in general I cannot derive a characterisation of all involutory maps in the first place as this lead me to solving equations in finite fields which I cannot solve. But I guess the arguments are simple in the end.

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  • $\begingroup$ $b+b =0$ is not possible for odd $q$. And for odd $q$, all involutions in ${\rm PSL}(2,q)$ are conjugate. $\endgroup$ – Derek Holt Jan 16 '16 at 19:54
  • $\begingroup$ @DerekHolt Okay, yes $b + b = 0$ is stupid for $q$ odd. As they are all conjugate it would be enough to look at a specific involution. But as said for $x \mapsto a^2x$ with $a^4 = 1$ I can just show that the normalizer is dihedral, but not for the centralizer. Do you have any idea how to show the claim for one specific involution? $\endgroup$ – StefanH Jan 16 '16 at 20:05
  • $\begingroup$ What you say about the normalizer and the centralizer does not make much sense. Think about it. $\endgroup$ – Derek Holt Jan 16 '16 at 20:49
  • $\begingroup$ @DerekHolt If $x^g = a^2x$, then for all $h$ of the form $x^g = b^2x, b \ne 0$ we have $hg = gh$, further $x^s = -x^{-1}$ normalizes $\langle g \rangle$ with $s^{-1}gs = g^{-1}$, hence $\langle g, s \rangle$ lies in the normalizer and is dihedral, with a similar reasoning I "showed" that the normalizer is dihedral. So whats wrong with that??? And the centralizer is a subgroup of the normalizer, hence cyclic or dihedral??? Have I made a mistake somewhere??? $\endgroup$ – StefanH Jan 16 '16 at 21:08
  • $\begingroup$ No you haven't made a mistake but you are missing something obvious. $\endgroup$ – Derek Holt Jan 16 '16 at 22:11

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