# Psl(2, 9)

• Jul 2nd 2008, 11:43 PM
dimuk
Psl(2, 9)
Can you explain me how to show that the PSL(2, 9) is isomorphic to Alt(6)?
• Jul 3rd 2008, 02:15 AM
NonCommAlg
Quote:

Originally Posted by dimuk
Can you explain me how to show that the PSL(2, 9) is isomorphic to Alt(6)?

clearly we have $\mathbb{F}_9=\{m+ni: \ m,n \in \mathbb{F}_3 \},$ where $i^2 = -1.$ now consider these two elements of $\text{SL}(2,9):$

$A=\begin{pmatrix} 1 & 1+i \\ 0 & 1 \end{pmatrix}, \ \ \ B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$ then a simple calculation shows that $A^3=I_2, \ B^2=(AB)^5=-I_2.$ let

$a=AZ(\text{SL}(2,9)), \ b=BZ(\text{SL}(2,9))$ be the elements of $G=\text{PSL}(2,9)$ corresponding to $A$ and $B.$ then:

$a^3=b^2=(ab)^5=1_G.$ let $H= \simeq A_5.$ (since $H$ and $A_5$ have the same presentation.) so $[G:H]=6.$

now let $X=\{a_jH: \ j=1, \ ... \ , 6 \}$ be the set of left cosets of $H.$ define $f:G \rightarrow S_X$ by $f(g)=\sigma_g, \ \forall g \in G,$

where $\sigma_g(a_jH)=ga_jH, \ j=1, \ ... \ , 6.$ clearly $\sigma_g \in S_X$ and $f$ is a non-trivial homomorphism. since G is simple,

we must have $\ker f = \{1 \},$ i.e. $f$ is an embedding. so G is isomorphic to a subgroup of $S_X \simeq S_6.$ since $|G|=360$

and $A_6$ is the only subgroup of $S_6$ which has order 360, we must have $G \simeq A_6. \ \ \ \square$
• Jul 3rd 2008, 02:33 AM
dimuk
Psl(2,9)
Thank you. I got the idea.
• Jul 3rd 2008, 09:16 AM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
obviously the number of Sylow 5-subgroups of G is 6.

What is the problem with having 36 Sylow 5-subgroups?
• Jul 4th 2008, 12:29 AM
NonCommAlg
Quote:

Originally Posted by ThePerfectHacker
What is the problem with having 36 Sylow 5-subgroups?

that proof wasn't complete. see the edited version!