Derived subgroups of $SL(2,2)$, $SL(2,3)$ and $GL(2,3)$

I would like to compute the commutator subgroups of $SL(2,2)$, $SL(2,3)$ and $GL(2,3)$.

For the group $G:=SL(2,2)$, we have $|G|=6$. We can easily show that $G$ is nonabelian. And hence either $|G'|=2$ or $|G'|=3$. Since the set of transvection generate $G$, $G'$ can not contain a transvection, which is conjugate to $\left(\begin{array}{cc}1&0\\1 &1 \end{array}\right)$. Hence $|G/G'|=2$, and so $|G'|=3$. Hence $G'=C_3$ is a cyclic group of order $3$.

Is the above argument correct? How can I do for other groups $SL(2,3$ and $GL(2,3)$?

• The argument is correct, given the fact that the transvections generate ${\rm SL}(2,2),$ (which is true), but that fact is probably as hard as determining the derived subgroup. Commented Jan 7, 2013 at 22:15

$$G=SL(2,2)=GL(2,2)\,\,\,\text{of order six and non abelian}\,\,\Longrightarrow G\cong S_3\Longrightarrow G'=S_3^{'}=A_3\cong C_3$$

To see that $\,S_3^{'}=A_3\,$ note that $\,S_3/A_3\cong C_2\,$ is abelian, and thus $\,S_3^{'}\leq A_3\,$ , and since $\,S_3^{'}\neq 1\,$ (why?) then $\,S_3^{'}=A_3\,$

$$|G:=GL(2,3)|=(3^2-1)(3^2-3)=48\,\,,\,\,|SL(2,3)|=\frac{48}{2}=24$$

Thus

$$G/SL(2,3)\cong C_2\,\,\,\text{abelian}\,\,\Longrightarrow G'\leq SL(2,3)$$

Now note that

$$\begin{pmatrix}1&a/2\\0&1\end{pmatrix}\begin{pmatrix}1&0\\0&\!\!\!\!-1\end{pmatrix}\begin{pmatrix}1&-a/2\\0&\;\;1\end{pmatrix}\begin{pmatrix}1&0\\0&\!\!\!-\!1\end{pmatrix}=\begin{pmatrix}1&a\\0&1\end{pmatrix}$$

and since transvections like the one above generate $\,SL(2,3)\,$, we get the equality $\,G'=SL(2,3)\,$

• thank you very much. How about the group $G=SL(2,3)$ then?
– 9999
Commented Jan 8, 2013 at 22:00

The derived subgroup of $G=SL(2,3)$ is the quaternion group $Q_8$. The possible minimal polynomials of an element of order $2$ in $G$ must divide $X^2-1$ and cannot be $X^2-1$ (since the determinant must be $1$). Therefore $X=-1$ (scalar multiplication) is the unique element of order $2$ (generating the center of $G$). In a similar way (determinant and trace of the minimal polynomial) one shows that there are no elements of order $8$ and $6$ elements of order $4$. Therefore there is a unique $2$-Sylow subgroup which must be normal. By multiplication of the elements of order $4$ you found, you show that the group is non-abelian. The dihedral group $D_4$ and the quaternion group $Q_8$ are up to isomorphism the only non-abelian groups of order $8$. Since our group contains a unique element of order $2$, it must be $Q_8$. Since $G/Q_8$ is abelian we have $G'$ is contained in $Q_8$. To see that it is $Q_8$ we note that $G/Z(G)$ is a non-abelian group of order $12$ whose 2-Sylow is isomorphic to $Q_8/Z(G)$ which is isomorphic to the Klein four-group and therefore must be the alternating group $A_4$ whose derive subgroup is of order $4$. Since $Q_8'$ of order $2$ and in $G'$, we obtain that $G'$ is of order $8$.

• can you explain why $X=-1$ is the only element of order $2$? the minimal polinomial should be $P(X).(X^2-1)$ and can we compuse $P(X)$? Commented Nov 9, 2019 at 10:05