Sylow 2-Groups of a Special Linear Group

Question

Let $SL_2(\mathbb{F}_3)$ be the special linear group over the finite field $\mathbb{F}_3$. Show that any Sylow 2-group of $SL_2(\mathbb{F}_3)$ is isomorphic to the quaternion group of order 8.

Answer 1

Hints:

$$|GL_2(\Bbb F_3)|=(3^2-1)(3^2-3)=8\cdot 6=48\implies |SL_2(\Bbb F_3)|=\frac{48}2=24$$

Define the following:

$$e=\begin{pmatrix}1&0\\0&1\end{pmatrix}\;,\;\;-e=\begin{pmatrix}2&0\\0&2\end{pmatrix}\;,\;\;i=\begin{pmatrix}0&1\\2&0\end{pmatrix}\;,\;\;j=\begin{pmatrix}1&2\\2&2\end{pmatrix}\;,\;\;k=\begin{pmatrix}2&2\\2&1\end{pmatrix}$$

And now check the following (hey, after all the field $\,\Bbb F_3\;$ is pretty small and easy to work with!):

$$i^2=j^2=k^2=-e\;,\;\;ij=k\;,\;\;jk=i\;,\;\;ki=j\;\ldots\ldots$$

Answer 2

$SL(2,\mathbb{F}_3)$ has unique element of order $2$ (an interesting exercise!!!); a $2$-group with unique element of order $2$ is either cyclic or quaternion. $|SL(2,\mathbb{F}_3)|=2^3.3$. If Sylow-$2$ subgroup is cyclic, then group will contain an element of order $8$. But a matrix with entries in $\mathbb{F}_3$ whose order is $8$ has minimal polynomial a divisor of $x^8-1/(x^4-1)=x^4+1=(x^2+x-1)(x^2-x-1)$; the last two polynomials are irrefucible over $\mathbb{F}_3$, but their companion matrices have determinant $-1$, so they are not in $SL(2,\mathbb{F}_3)$). Therefore, there is no element of order $8$ in $SL(2,\mathbb{F}_3)$.

(Thanks to Jack Schmidt for noticing me an error in the last statement, which is now modified.)