# 2-sylow subgroups

Dear All, by the paper of Carter and Fong, we know the structure of 2-sylow subgroups of $GL(n,q)$. Let $W_{r-1}=Z_2\wr Z_2\wr...Z_2$ (r-1 times), and $W$ is a 2-sylow subgroup of $GL(2,q)$, then $W_r=W \wr W_{r-1}$ is a 2-sylow subgroup of $GL(2^r,q)$.If $n=2^{r_1}+2^{r_2}+...2^{r_k}$, then $P=W_{r_1} \times ...\times W_{r_k}$ is 2-sylow subgroup of $GL(n,q)$. I need to know the permutations which generate $W_{r-1}$. so by stather's paper I could have the structure of $W_r$ as a subgroup of $GL(n,q)$ (I could have the matrices which generate $W_r$). Thank you.

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You take a set of permutations that generate a Sylow $2$-subgroup of the symmetric group $S_{2^{r-1}}$.Then make them into permutation matrices of degree $2^r$ that permute $2 \times 2$ block matrices. These permutation matrices, together with a Sylow $2$−subgroup of ${\rm GL}_2(q)$ acting on one of the $2 \times 2$ blocks, generate a Sylow $2$−subgroup of ${\rm GL}_{2^r}(q)$. –  Derek Holt Apr 18 at 8:10
Is there any routine technique to find a set of permutations which generate a Sylow 2-subgroup of the symmetric group $S_{2^{r-1}}$? –  user33209 Apr 21 at 7:51
@unknown, yes there is. Suppose you've got a set of permutations generating the Sylow 2-subgroup of $S_{2^{r-2}}$. Write two copies of these, one as permutations on $[1, 2^{r-2}]$ and the other on $[2^{r-2}+1, 2^{r-1}]$. Add the permutation $(1,2^{r-2}+1)(2,2^{r-2}+2)\cdots$. This generates a wreath product on $[1,2^{r-1}]$ and you're done. –  Nick Gill Apr 22 at 9:06

It should be pretty clear how to turn a Sylow 2-subgroup of $S_{2^{r-1}}$ into a set of $2^r\times 2^r$-block matrices with blocks of size $2$. Which means that the only (potentially) tricky thing is to write down matrices for a Sylow $2$-subgroup of $GL_2(q)$. For this, there are two easy cases and a hard case:
• If $q$ is even, then you can take your Sylow $2$-subgroup to be upper-unitriangular matrices. Easy.
• If $q\equiv 1\pmod 4$, then a split torus of size $(q-1)^2$ will contain a Sylow $2$-subgroup of $GL_2(q)$, and you can just find a bunch of nice diagonal matrices to do the job.
• If $q\equiv 3 \pmod 4$, then a Sylow $2$-subgroup will lie in a non-split torus, isomorphic to a cyclic subgroup of order $q^2-1$. So you'll need to write down a bunch of matrices corresponding to this torus - you can do it using the method outlined in Carter's Finite groups of Lie type: you need to `twist' the torus $T$ in $GL_2(q^2)$ consisting of diagonal matrices by an element $g$ such that $g^{-1}F(g)$ corresponds to a non-trivial element of the Weyl group with respect to $T$ (here $F$ is the Frobenius endomorphism). A nice way of doing this is to let $x$ be an element of $\mathbb{F}_{q^2}$ satisfying $x^q=-x$, and then let $g=\left(\begin{array}{cc} a & b \newline c& d \\ \end{array}\right)$ such that $$\left(\begin{matrix} a^q & b^q \newline c^q & d^q \end{matrix}\right) = \left(\begin{matrix}a & b \newline c& d \end{matrix}\right)\left(\begin{matrix} & -x \newline x & \end{matrix}\right).$$ You can make your life easier by specifying that $\det g=1$, and you end up with $$g = \left(\begin{matrix} a & a^qx \newline c & c^qx \end{matrix}\right)$$ where $ac^qx - ca^qx=1$. Now the torus you are seeking is $gTg^{-1}$. The intersection of this torus with $GL_2(q)$ is a cyclic group of order $q^2-1$, and it will contain a Sylow $2$-subgroup of $GL_2(q)$ as required.