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Here is my problem:

Let $X=\begin{pmatrix} 10 & 8 \\ 8 & 1 \end{pmatrix}$ and $Y=\begin{pmatrix} 5 & 7 \\ 5 & 5\end{pmatrix}$ be two elements of $SL_2(11)$. Find a subgroup of $PSL_2(11)$ isomorphic to $A_5$.

I know that $A_5$ has a presentation as $A_5=\langle x,y|x^2=y^3=(xy)^5=1\rangle$ and $PSL_2(11)=\displaystyle\frac{SL_2(11)}{\{\lambda I|\lambda^2=1, \lambda\in GF^*(11) \}}$. How can I use $X$ and $Y$? Any help will be appreciated. Thanks

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    $X$ has order $4,$ its image in ${\rm PSL}(2,11)$ has order $2$. $Y$ has order $3.$ That's suggestive.2012-06-29

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Note that $X^2 = -I$, and $Y^3 = I$, so in $PSL_2(11)$, the coset containing $X$ has order $2$, and the coset containing $Y$ has order $3$ (since neither $X$ nor $Y$ is in $SZ_2(11) = \{I,-I\}$).

Now $XY = \begin{bmatrix}2&0\\1&6 \end{bmatrix}$ and some short calculations show that:

$(XY)^2 = \begin{bmatrix}4&0\\8&3 \end{bmatrix}$, $(XY)^3 = \begin{bmatrix}8&0\\8&7 \end{bmatrix}$, $(XY)^4 = \begin{bmatrix}5&0\\1&9 \end{bmatrix}$, $(XY)^5 = \begin{bmatrix}10&0\\0&10 \end{bmatrix} = -I$

so the coset containing $XY$ has order $5$.

Letting $X',Y'$ be the images of $X,Y$ (respectively) in $PSL_2(11)$, we see that $\langle X',Y' \rangle$ is isomorphic to $A_5$.