Special linear group of order $2$ over field of order $3$ Let $G = \text{SL}(2, F_3)$ (group of matrices of determinant $1$ over the field of order $3$).


*

*Find $|G|$.

*Show that $Z(G)$ is not $\{1_G\}$.

*Determine the number of Sylow $3$-subgroups of $G$.

*What is the isomorphism type of a Sylow $3$-subgroup of $G$?

*Determine the number of Sylow $2$-subgroups of $G$.

*What is the isomorphism type of a Sylow $2$-subgroup of $G$?


This is what I've got so far:


*

*$|G| = 24$

*I proved that the matrix $\bigl(\begin{smallmatrix}
2&0 \\ 0&2
\end{smallmatrix} \bigr)$ is in the center.

*Using Sylow's theorem I know the answer is $1$ or $4$ but I don't know which.

*Using Sylow's theorem I know the answer is $1$ or $3$ but I don't know which.


And I can't do parts $4$ or $6$. Any help is appreciated.
 A: Let $G=GL(2,3)$, then the order of $G$ is $(3^{2}-1)(3^{2}-3)=48$;
Since $N=SL(2,3)$ is the kernel of the determinant map into $F_{3}^{*}$ it's index is $2$ and so $|SL(2,3)|=24$.
In general the center $Z(SL(n,p))$ is the group of the scalar matrices;
In our case the matrix $\bigl(\begin{smallmatrix}
-1&0 \\ 0&-1
\end{smallmatrix} \bigr)$ is in the center.
We can prove that in $N$ there are: $1$ element of order $1$, $1$ element of order $2$, $6$ elements of order $4$ and no elements of order $8$ (and so we conclude that with this elements we can construct only one subgroup of order $8$, i.e. a $2$-Sylow):
in fact, the element $z= \bigl(\begin{smallmatrix}
-1&0 \\ 0&-1
\end{smallmatrix} \bigr)$ is the only element of order $2$ in $N$:
if $g\in N$ have order $2$ than $g^{2}-1=0$ and so the minimal polinomyal of $g$ divides $x^{2}-1$, so it can be: $x^{2}-1,x+1,x-1$; but in the third case $g=1$ and this is not the case; in the first case the minimal polinomyal of $g$ divides the characteristic polinomyal of $g$ but the last one have degree $2$, then the characteristic polinomyal of $g$ is $x^{2}-1$ then $det(g)=-1$ and so $g\notin N$ and this is not the case; then we must have $g=-1$.
If the order of $g$ is $4$ the minimal polinomyal of $g$ divides $x^{4}-1=(x^{2}+1)(X-1)(x+1)$ but so is divisible by $x^{2}+1$ otherwise the minimal polinomyal of $g$ divides $x^{2}-1$ and so the matrix $g$ have order $2$. So $x^{2}+1$ divides the characteristic polinomyal of $g$ and so coincide with him; we conclude that $g\in N$ ($det(g)=1$). So all the elements of order $4$ is in $N$. But if $g^{4}=1$ then $g^{2}=-1$ and we have only $6$ matrices in $N$ with this property (check yourself with the hand).
Finally there aren't elements of order $8$: indeed, the minimal polinomyal of $g$ divides $x^{8}-1=(x^{4}+1)(X^{4}-1)$ and in $F_{3}$ we have $(x^{4}+1)=(x^{2}+x-1)(x^{2}-x-1)$; since the minimal polinomyal of $g$ doesn't divide $x^{4}-1$ (otherwise $g$ have order $\leq 4$), the minimal polinomyal of $g$ is one of the above polynomial of degree $2$, but in both the case $det(g)=-1$ and so $g\notin N$. 
This $2-$ Sylow is not abelian, otherwise it have more element of order $2$, so is a $2^{3}-$ group not abelian, different from $D_{4}$ (by it's structure); but then is $Q_{4}$, from the classification of group of order $p^{3}$, with $p$ a prime;
Let $P=\bigl(\begin{smallmatrix}
1&1 \\ 0&1
\end{smallmatrix} \bigr) \cong C_{3}$, is a $3$-Sylow of $N$.
Finally we have only four $3-$Sylow in $N$:
in fact, $n_{3}(G)\equiv 1\pmod{3}$ and $n_{3}(G)| 16$;
So $n_{3}(G)=1,4,16$; but is different from $1$ because $P$ is not normal;
is different from $16$ because $n_{3}(G)=n_{3}(N)$ because $N$ is normal in $G$ and contains a $3$-Sylow of $G$ (so if I take a $3$-Sylow of $G$ it will be conjugate to a $3$-Sylow of $N$, e.g. to $P$, so by the normality of $N$ it is inside $N$). From this we get $n_{3}(G)| 8$ and it is different from $16$.
So it is equal to $4$.
A: For 3, if the answer were $1$, how many elements of order $3$ would be contained in $G$? Show that it is impossible (by finding "many" elements of order $3$ in $G$).
For 4, a sylow $3$-subgroup $S$ is by definition a group of order $3$, take a non-trivial element $s$ and look at the group generated by $s$ to show that $\langle s\rangle=S$ (try to compute the order of $s$ using Lagrange's theorem).
For 5, first realize that $Z(G)$ will always be contained in any $2$-Sylow (because it is a normal $2$-subgroup of $G$). Now take $g:=\begin{pmatrix}0&1\\2&0\end{pmatrix}$ and $h:=\begin{pmatrix}1&0\\0&2\end{pmatrix}$. Justify that $S_2:=\langle g,h\rangle$ is a $2$-Sylow of $G$. Show that it is not normal and conclude.
For 6, try to find the list of all groups of order $8$. There are $5$ of them, among them, only $2$ are not commutative. One easily see that $S_2$ is not commutative, so $S_2$ can only be isomorphic to $D_4$ or $Q_8$. Justify that $S_2$ contains two non-commutative elements of order $4$ and conclude that $S_2=Q_8$.
A: A Sylow $\;3 -$subgroup here is a group of order $\;3\;$ and there's only one group of order a prime up to isomorphism. 
Perhaps with what you did in (2) you already know that 
$$Z(G)=\left\{\;I,\,-I\;\right\}\implies|Z(G)|=2$$
and as a normal $\;2-$subgroup the center is contained in any Sylow $\;2-$subgroup of $\;G\;$ , which are of order $\;8\;$ .
We need to look for some elements of order a power of $\;2\;$, say:
$$i:=\begin{pmatrix}0&1\\\!\!-1&0\end{pmatrix}\implies i^2=-I\;,\;\;i^4=1$$
and thus there exists a cyclic subgroup of order $\;4\;$ in $\;G\;$ , yet
$$j:=\begin{pmatrix}1&1\\1&\!\!-1\end{pmatrix}\implies j^2=I\;,\;\;j^4=1$$
But
$$ij=\begin{pmatrix}1&\!\!-1\\\!\!-1&\!\!-1\end{pmatrix}\neq\begin{pmatrix}\!\!-1&1\\1&1\end{pmatrix}=ji$$
So we have that the Sylow $\;2-$subgroups are non-abelian, so each of them is either the dihedral $\;D_4\;$ or the quaternion group $\; Q\;$.
You are left now the reasonable easy task to show that it is the second option.
