1. ## Psl(2, 9)

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

2. 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 $G \simeq A_6. \ \ \ \square$

3. ## Psl(2,9)

Thank you. I got the idea.

4. 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?

5. Originally Posted by ThePerfectHacker
What is the problem with having 36 Sylow 5-subgroups?
that proof wasn't complete. see the edited version!