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

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- Jul 2nd 2008, 11:43 PM #1

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- Jul 3rd 2008, 02:15 AM #2

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

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

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

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

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

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

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

and $\displaystyle A_6$ is the only subgroup of $\displaystyle S_6$ which has order 360, we must have $\displaystyle G \simeq A_6. \ \ \ \square$

- Jul 3rd 2008, 02:33 AM #3

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- Jul 3rd 2008, 09:16 AM #4

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- Jul 4th 2008, 12:29 AM #5

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