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 $|N_G(S)| = (q-1)/2$?

• The answer is $|N_G(S)| = q(q-1)/2$ whe $q$ is odd and $q(q-1)$ when $q$ is even. – Derek Holt Oct 23 '15 at 22:01
• @DerekHolt - could you make that answer - and maybe address what is wrong in my answer? thanks! please look at my comments to my answer. – peter a g Oct 23 '15 at 22:42
• @Stefan - please see my note/edit to my answer - I agree that Derek's answer is the one you should have accepted. – peter a g Oct 25 '15 at 3:33

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.

• Thanks Derek +1 - and see my note/edit to my (conflicting) answer. – peter a g Oct 25 '15 at 3:31

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!

• What is $\overline{\mathbb F}^{\ast}$, and what has exactness to do with the problem at hand (I must confess I am not familiar with (Galois)cohomology)? All I know is that a sequence of groups $1 \to A \to B \to C \to 1$ is exakt if there exists an injection $i : A \to B$ and a surjection $\pi : B \to C$ such that $\mbox{im}(i) = \mbox{ker}(\pi)$, but cannot connect that to what you are saying. – StefanH Oct 23 '15 at 18:23
• Then it's a bit difficult to explain... The idea is that these are algebraic groups, and algebraic morphisms, and one has an exact sequence over the algebraic closure of tjhe field $\mathbb F_q$, which I am denoting with a bar on top. If $k$ is a ring, $k^*$ denotes the invertible elements. In general a sequence over a bigger field is not exact over a smaller field, and the obstruction is measured in a cohomology group. For instance, (continued) – peter a g Oct 23 '15 at 18:32
• (continued) $$1 \rightarrow \pm 1 \rightarrow \mathbb C^* \rightarrow \mathbb C^* \rightarrow 1$$ is exact, where the map $\mathbb C^* \rightarrow \mathbb C^*$ is squaring. However, if one looks at the sequence with the reals in the place of the complex numbers, the sequence is not exact - the negatives are not squares. The obstruction is measured in Galois group $H^1({\mathbb R}. \mathbb \pm 1)$. But you obviously should not 'accept' this answer if you can't connect with what I am saying... sorry! – peter a g Oct 23 '15 at 18:37
• Edit of my first comment - more than 5 mins have gone by: "an EXACT sequence over the bigger field" – peter a g Oct 23 '15 at 18:38
• Also, rereading your question - I wonder whether we are even using the same definitions: using the argument in my answer (and my implied assumption that $PSL(2,q)$ is the group of $\mathbb F_q$-rational points of the corresponding algebraic group), there should have been no division by $2$ for the total count of elements in the group. I'll look into this over the weekend.... – peter a g Oct 23 '15 at 19:03