Sylow 2-subgroups of the group $\mathrm{PSL}(2,q)$

Solution 1:

When $q$ is a power of $2,$ we have ${\rm PSL}(2,q) = {\rm SL}(2,q)$ and a Sylow $2$-normalizer is a Borel subgroup of order $q(q-1).$ Hence there are $q+1$ Sylow $2$-subgroups as ${\rm SL}(2,q)$ has order $(q-1)q(q+1)$. When $q$ is odd, the order of ${\rm PSL}(2,q)$ is $\frac{q(q-1)(q+1)}{2}.$ A Sylow $2$-subgroup of ${\rm SL}(2,q)$ is (quaternion or) generalized quaternion and a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is either a Klein $4$-group or a dihedral $2$-group with $8$ or more elements. In all these cases, a Sylow $2$-subgroup of ${\rm SL}(2,q)$ contains its centralizer, and some elementary group theory allows us to conclude that the same is true in ${\rm PSL}(2,q).$ The outer automorphism group of a dihedral $2$-group with $8$ or more elements is a $2$-group. Hence a Sylow $2$-subgroup of ${\rm PSL}(2,q)$ is self-normalizing when $q \equiv \pm 1$ (mod 8), and in that case the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(q^{2}-1)_{2^{\prime}}$ where $n_{2^{\prime}}$ denotes the largest positive odd divisor of the positive integer $n.$ When $q \equiv \pm 3$ (mod 8), then a Sylow $2$-normalizer of ${\rm PSL}(2,q)$ must have order $12$ ( a Sylow $2$-subgroup is a self-centralizing Klein $4$-group, but there must be an element of order $3$ in its normalizer by Burnside's transfer theorem). In this case, the number of Sylow $2$-subgroups of ${\rm PSL}(2,q)$ is $q(\frac{q^{2}-1}{24})$