# The Sylow $2$-subgroups of $PSL(2, p^n)$ for $p \ne 2$. Question on derivation.

Consider $G = PSL(2, q)$ with $q = p^n$, i.e. the projective special linear group over a finite field of order $q$. Then $G$ could be considered as the group of all mappings $$x \mapsto \frac{ax + b}{cx + d}$$ on $\mathbb F_q \cup \{\infty\}$ with $a,b,c,d \in \mathbb F_q$ such that $ad - bc$ is a square in $\mathbb F_q$. We have $$|G| = \frac{(q+1)q(q-1)}{k}$$ with $k = \operatorname{ggT}(2, q - 1)$. For the cyclic subgroup $U = \{ \alpha \in G : x^{\alpha} = a^2 x \mbox{ and } a \in \mathbb F_q^{\ast} \}$ we have $$|U| = \frac{q - 1}{k}$$ and for each $\alpha \in U$ we have that $N_G(\langle x \rangle)$ is a dihedral group of order $2|U|$. Further it is possible to construct another cyclic group $S$ of order $|S| = (q+1)/k$ such that for $\beta \in S$ we have that $N_G(\langle \beta \rangle)$ is a dihedral group of order $2|S|$ (this is done with the help of a so called Singer cycle in $GL(2, q)$ and using the definition of $G$ as a factor group of a subgroup of $GL(2,q)$).

Now I have a question on the proof, that for $p \ne 2$ the $2$-Sylow subgroups of $G$ are dihedral groups. The proof uses the above mentioned facts.

Let $D_1 = N_G(U), D_2 = N_G(S)$ with the notation as above. For $p \ne 2$, one of the numbers $p^n - 1$ or $p^n + 1$ is divisble by four. The corresponding group $D_i$ ($i = 1,2$) contains a Sylow $2$-subgroup then, which therefore is dihedral.

Suppose $p^n - 1$ is divisble by four, then $D_1$ has order $p^n - 1$, but $D_2$ has order $p^n + 1$, and this is an even number then. So $D_1$ could not contain a full Sylow $2$-subgroup, for otherwise $p^n + 1$ is not allowed to be divisble by two? Further the subgroups of a dihedral group are either themselve dihedral or cyclic, but I do not see why cyclic Sylow $2$-subgroups are excluded in the above argument?

• Please think about it a bit more! The statement that one of $D_1$ and $D_2$ contains a Sylow $2$-subgroup is correct, and it seems pretty obvious that a Sylow $2$-subgroup of a dihedral group (of order divisible by $4$) is itself dihedral. (They are regarding $C_2 \times C_2$ as a dihedral group here.) – Derek Holt Dec 17 '15 at 19:01
• Thank you, I wrote an answer myself. – StefanH Dec 17 '15 at 19:55

With Derek's hint it was not that difficult in the end. I somehow overlooked that for $p \ne 2$, we have $k = 2$ and so the number which is divisible by four contains all powers of $2$. For suppose $p^n - 1$ is divisible by four, then $p^n + 1$ is divisible by two as the highest power of $2$. Hence $\frac{p^n(p^n + 1)}{2}$ is not divisible by $2$ and the index of $D_1$ is odd, which gives the claim.
For the Sylow $2$-subgroups. The subgroups of a dihedral group of order $2n$ with rotation $r$ and reflection $s$ are of the form $\langle r^d \rangle$ with $d \mid n$ or $\langle r^d, r^i s\rangle$ with $d \mid n$ and $0 \le i \le d - 1$. As the subgroups of the cyclic rotation group have even index, they are not Sylow subgroups, hence the Sylow subgroups must be the remaining dihedral groups.