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Consider the group $\operatorname{GL}(2,3)$, the group of invertible $2 × 2$ matrices over the field of 3 elements. It is of order $48 = 2^4 \cdot 3$. A 3-Sylow subgroup is easily seen to be the Heisenberg group (unitary upper triangular matrices).

What about the 2-Sylow subgroups? Is there a nice way to identify one of them?

frafour
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  • What does $GL(2,3)$ denote? I understand that they are invertible matrices but over what? – C.S. Oct 27 '17 at 09:45
  • I just edited the question – frafour Oct 27 '17 at 09:47
  • @S.C. Matrices over the field of three elements. Finite group -people are known for writing $GL_n(\Bbb{F}_q)$ as $GL(n,q)$, or even $L(n,q)$. I guess its easier to typeset (particularly in the pre-TeX days), and the context helps (as always). – Jyrki Lahtonen Oct 27 '17 at 09:48
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    Don't know right away if there is a nice description for the matrices in one of the Sylow $2$-subgroups. But $GL(2,3)$ acts doubly transitively on the set of four lines thru the origin in the space $\Bbb{F}_3^2$. This gives us a homomorphism $f$ from $GL(2,3)$ to $S_4$ that is easily seen to be surjective: the Heisenberg group acts as a $3$-cycle, and the non-central diagonal matrices have two fixed points, and thus act as $2$-cycles. Clearly $Z(GL(2,3))$ is the kernel of $f$. We thus get a Sylow $2$-subgroup of $GL(2,3)$ as the inverse image of one of the three copies of $D_4\le S_4$. – Jyrki Lahtonen Oct 27 '17 at 09:55
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    Another description for a Sylow $2$-subgroup is the following. $GL(2,3)$ contains copies of the multiplicative group $G$ of the field of nine elements, i.e. $G\simeq C_8$. To such a copy $G$ there exists a matrix $A\in N(G)$ such that conjugation by $A$ gives the Frobenius automorphism $x\mapsto x^3$ on $G$ (Skolem-Noether theorem). This gives us a semidirect product $G\rtimes C_2$ with the latter group generated by $A$. – Jyrki Lahtonen Oct 27 '17 at 10:08

1 Answers1

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The Sylow $2$-subgroups of the finite classical groups were described by Carter and Fong (J. Algebra, 1964).

Your example is a special case of $G = \operatorname{GL}_2(\mathbb{F}_q)$ with $q \equiv 3 \mod{4}$. Let $2^s$ be the largest power of $2$ dividing $q+1$, so a Sylow $2$-subgroup of $G$ has order $2^{s+2}$.

The generators Carter and Fong give for a $2$-Sylow subgroup $P < G$ are $x = \begin{pmatrix} 0 & 1 \\ 1 & \varepsilon + \varepsilon^q \end{pmatrix}$ and $y = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$, where $\varepsilon$ is a primitive $2^{s+1}$th root of unity in $\mathbb{F}_{q^2}$. For $a = x$ and $b = yx$ we have the relations $a^{2^{s+1}} = b^2 = 1$ and $bab^{-1} = a^{2^s - 1}$. Hence it turns out that $P$ is a semidihedral group: $$P \cong \langle a, b | a^{2^{s+1}} = b^2 = 1, bab^{-1} = a^{2^s - 1} \rangle.$$

So that is fine, but you might wonder how one could come up with these matrices.

For this, we can consider $V = \mathbb{F}_{q^2}$ as a $2$-dimensional vector space over $\mathbb{F}_q$, so then we can identify $G$ with $\operatorname{GL}(V)$. We first want to find an element of order $2^{s+1}$. For this, take $f \in G$ to be the map defined by $v \mapsto \varepsilon v$, where $\varepsilon$ is a primitive $2^{s+1}$th root of unity. With respect to the basis $\{1, \varepsilon \}$ of $V$, you can verify that the matrix of $f$ becomes the matrix $x$ defined above. We also have the Frobenius map $g \in G$ defined by $v \mapsto v^q$. You can check that the matrix of $g$ with respect to the basis $\{1, \varepsilon \}$ is the matrix $yx$.

Mikko Korhonen
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