I am trying to prove that the quotient group $SL_2(\mathbb Z_3)/Z(SL_2(\mathbb Z_3))$ is isomorphic to $\mathbb A_4$.

I could show that $SL_2(\mathbb Z_3)$ has $24$ elements (one can see this calculating all the elements by hand or simply by noticing that $GL_2(\mathbb Z_3)$ has $48$ elements and that $SL_2(\mathbb Z_3)$ is the kernel of the morphism $\det:GL_2(\mathbb Z_3) \to G_2$). The center of $SL_2(\mathbb Z_3)$ is $\{\begin{pmatrix}1& 0\\0& 1\end{pmatrix}, \begin{pmatrix}-1& 0\\0& -1\end{pmatrix}\}$.

I got stuck trying to show these two groups are isomorphic. By Sylow theory I can affirm that $n_3=4$, where $n_p$=number of $p-$Sylows. Then I can define the action by conjugation on $X$ the set of all $3-$Sylow groups.

So I have a morphism $\phi:SL_2(\mathbb Z_3) \to S(X)$. It is clear that $Z(SL_2(\mathbb Z_3)) \subset Ker \phi$. I don't know what to do from here,

I would appreciate suggestions or an alternative solution to this problem. Thanks in advance.

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    $\begingroup$ You cannot have $n_3=8$ because $8 \not\equiv 1 \bmod 3$. $\endgroup$ – Derek Holt Jun 14 '15 at 19:38
  • $\begingroup$ Thanks for the remark, I've edited my answer $\endgroup$ – user156441 Jun 14 '15 at 20:15
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    $\begingroup$ So the image of $\phi$ has order at least $24/2=12$. But this image also has $4$ Sylow $3$-subgroups and $S_4$ has only $4$ Sylow $3$-subgroups, which generate $A_4$, so the image of $\phi$ is $A_4$ and you are done. $\endgroup$ – Derek Holt Jun 14 '15 at 20:53
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    $\begingroup$ Sorry, I meant the image of $\phi$ has order at most $12$. $\endgroup$ – Derek Holt Jun 14 '15 at 21:08
  • $\begingroup$ From geometric POV the homomorphism to $A_4$ comes from the action of $SL_2(\mathbb Z_3)$ on 4-element set $P^1(\mathbb Z_3)$ (by fractional-linear transformations, if you will). Cf. $PSL_2(\mathbb F_4)\cong A_5$ etc. $\endgroup$ – Grigory M Jun 15 '15 at 10:07

Rather than the action on the $3$-Sylow, I think there is a more natural set on which $SL_2(\mathbb{Z}_3)/Z(SL_2(\mathbb{Z}_3))$ acts faithfully. Set $X:=\{\text{ vectorial lines in } \mathbb{F}_3\times \mathbb{F}_3\}$.

The cardinal of $X$ is easy to find, indeed, any vectorial line is given by a non-null vector and furthermore two colinear vectors give the same line hence :

$$X=\frac{\mathbb{F}_3\times \mathbb{F}_3\setminus \{0\}}{\mathbb{F}_3^*} $$

Hence $|X|=\frac{9-1}{2}=4$. It is clear that $SL_2(\mathbb{Z}_3)$ acts on $X$ furthermore a matrix $A$ will leave any line fixed if and only if it is in the center. This is easy to show (say $(e_1,e_2)$ is the canonical base of $\mathbb{F}_3\times \mathbb{F}_3$), take $A$ such a matrix then it leaves $\mathbb{F}_3e_1$, $\mathbb{F}_3e_2$ and $\mathbb{F}_3(e_1+e_2)$ invariant that is : $Ae_1=\lambda e_1$, $Ae_2=\mu e_2$ and $A(e_1+e_2)=\gamma (e_1+e_2)$, hence we have :

$$\gamma e_1+\gamma e_2=A(e_1+e_2)=Ae_1+Ae_2=\lambda e_1+\mu e_2 $$

Hence we have $\lambda=\gamma=\mu$, so that in the base $(e_1,e_2)$ the matrix of $A$ is :

$$\begin{pmatrix}\gamma& 0\\ 0& \gamma\end{pmatrix} $$

It is easy to show that this in the center of $SL_2(\mathbb{Z}_3)$. Conversly, any matrix in the center of $SL_2(\mathbb{Z}_3)$ is of this form and will fix any line. Hence we have shown that the action of $SL_2(\mathbb{Z}_3)$ on $X$ quotient through its center to give a faithfull action on $X$. In other words we have constructed an injective morphism :

$$\rho: SL_2(\mathbb{Z}_3)/Z(SL_2(\mathbb{Z}_3))\rightarrow \mathfrak{S}_X=\mathfrak{S}_4 $$

Now by cardinality calculus you have already done $SL_2(\mathbb{Z}_3)/Z(SL_2(\mathbb{Z}_3))=12$ and $\mathfrak{S}_4$ is of cardinal $24$. Hence the image of $\rho$ is a subgroup of index $2$ in $\mathfrak{S}_4$, it cannot but be the antisymmetric group $\mathfrak{A}_4$.


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