The order of the normalizer of Sylow subgroup in $PSL(2,q)$ Let $G = PSL(2, q)$ with $q = r^m$. If $q$ is odd, this group has order $q(q-1)(q+1)/2$, if $q$ is a power of $2$ then it has order $q(q-1)(q+1)$. If $S$ is a Sylow $r$-subgroup of $G$, what could be said about the order of its normaliser $N_G(S)$? Do we have $|N_G(S)| = (q-1)/2$?
 A: Let's start by doing the calcualtion in $X:= {\rm SL}(2,q)$ with $q=p^e$. Then the subgroup $P$ of upper unitriangular matrices is a Sylow $p$-subgroup, and $N_X(P)$ cosnsits of the upper triangular matrices in $X$. That is
$$N_X(P) = \left\{ \left( \begin{array}{cc}\lambda&\mu\\0&\lambda^{-1}\end{array}\right) \mid \lambda,\mu \in {\mathbb F}_q,\,\lambda \ne 0 \right\}.$$
So $|N_X(P)| = q(q-1)$.
The intersection of $N_X(P)$ with the subgroup of scalar matrices is trivial when $q$ is even and $\{\pm I_2\}$ when $q$ is odd. So, in $G={\rm PSL}(2,q)$, with $S \in {\rm Syl}_p(G)$, $|N_G(S)| = q(q-1)$ when $q$ is even and $q(q-1)/2$ when $q$ is odd.
A: You obviously forgot the factor of $q$ in the count of the normaliser (as $S$ is a subgroup in it). And I think you shouldn't divide by 2 [edit - see the note below]: consider the sequence 
$$ 1 \rightarrow \def\F {{\mathbb F_q}} \bar\F^* \rightarrow GL_2(\bar\F ) \rightarrow PGL_2(\bar\F) \rightarrow 1.$$
($PGL_2$ is the same as $PSL_2$.[edit -  again, see note]) Taking the long exact sequence of (Galois) cohomology, and using the fact that $H^1(\F,\bar\F^*)=1$, one sees that the following is also exact:
$$ 1 \rightarrow  \F^* \rightarrow GL_2(\F ) \rightarrow PGL_2(\F) \rightarrow 1.$$
In $GL_2(\F)$, a $q$-Sylow is the group of upper triangular matrices with $1$-on the diagonal, and its normalizer is the upper triangular matrices, which has a count of $(q-1)^2q$. Dividing out by $\F^*$, we get a count of $q(q-1)$. 
Note/Edit: for an argument that $PSL_n$ should be considered to coincide with $PGL_n$ (as I did, above), see http://www.jmilne.org/math/CourseNotes/AGS.pdf, Example 2.2 Chapter IX, 2.2. 
However! Remark 9.3.4 of http://math.stanford.edu/~conrad/249APage/handouts/alggroups.pdf points out that tradition  defines $PSL_2(k)$ as $SL_2(k)/ \{\pm 1\}$. (N.B. $-1=1$, when $q$ is even.) That is how Wikipedia defines it, and how it is being used in this question. 
So the (desired) cardinality of the normalizer is not as I gave, but rather, exactly as in Derek Holt's answer: $q(q-1)/2$ when $q$ is odd and $q(q-1)$ when $q$ is even. The factor of $2$ between my  and Derek's  answers arises because the two definitions do not coincide. Buyer beware!
