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Let q be an odd prime power. By a classic result, a Sylow 2−subgroup $P$ of $SL(2,q) $ is generalized quaternion. It is an irreducible subgroup of $GL(2,q)$ (since otherwise its natural representation would be the sum of two linear ones and so $G$ would be abelian, contradiction). By Schur's lemma, its centralizer in $SL(2,q)$ is the subgroup of diagonal matrices and so its centralizer in $SL(2,q)$ is $\{-I_2,I_2\}$. Therefore $C_{SL(2,q)}(P)\le P$. I would please like to show that the same is true for a Sylow $2-$subgroup of $PSL(2,q)$ (as per mentionned in the following post):

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

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Let $Z = \{\pm I_2\}$ be the centre of ${\rm SL}(2,q)$. So a Sylow $2$-subgroup of ${\rm PSL}(2,q) = {\rm SL}(2,q)/Z$ has the form $P/Z$.

If the centralizer of $P/Z$ was not contained in $P/Z$, then it would not be a $2$-group, so it would have a nontrivial element $gZ$ of odd order, and we can assume that $g$ has odd order in ${\rm SL}(2,q)$.

Then $gZ$ centralizes $P/Z$ but $g$ does not centralize $P$, so there exists $h \in P$ with $g^{-1}hg=hz$, where $z=-I_2$. Then, for any $k \ge 0$, $g^{-k}hg^k=hz^k$, which gives a contradiction if we take $k = o(g)$.

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  • $\begingroup$ Very nice. Thank you ! $\endgroup$ – Nicolas Jun 13 '16 at 14:55

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