2
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $\mathfrak{A}_4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.