# 2-Sylow subgroup of $\operatorname{GL}(2,3)$

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?

• What does $GL(2,3)$ denote? I understand that they are invertible matrices but over what?
– C.S.
Oct 27, 2017 at 9:45
• I just edited the question Oct 27, 2017 at 9:47
• @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). Oct 27, 2017 at 9:48
• 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 $D_4\le S_4$. Oct 27, 2017 at 9:55
• 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 $A$. Oct 27, 2017 at 10:08

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: $$P \cong \langle a, b | a^{2^{s+1}} = b^2 = 1, bab^{-1} = a^{2^s - 1} \rangle.$$
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 $$yx$$.