# Finding a subgroup of $PSL_2(11)$

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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Have you looked at the powers of $X$ and $Y$? For instance, what is $X^2$ in $SL_2(11)$? What is it in $PSL_2(11)$? – Steven Stadnicki Jun 26 '12 at 15:20
I would just verify that the images of X and Y in PSL(2,11) satisfy the relations of A5 (and that the image of X is not the identity, so that you don't have a proper quotient of A5). It is slightly nicer numbers if you use [-1,-3;-3,1] and [5,-4;5,-6] for X and Y. – Jack Schmidt Jun 26 '12 at 15:31
@StevenStadnicki: I wanted to take $<X,Y>$ as possible subgroup here, but I hesitated. Because I don't know "if $X$ and $Y$ staisfy in relations of $A_5$ so we have an image of it" or not. – Babak S. Jun 26 '12 at 15:51
If $X$ and $Y$ satisfy the defining relations of $A_5$ then you can define a homomorphism $\phi : A_5 \ \to\ <X,Y>$, simply by setting $\phi(x)=X$ and $\phi(y)=Y$. Clearly $\phi$ is surjective, and since the kernel of a homomorphism is always a normal subgroup you just need one final observation... – newguy Jun 26 '12 at 15:52
$X$ has order $4,$ its image in ${\rm PSL}(2,11)$ has order $2$. $Y$ has order $3.$ That's suggestive. – Geoff Robinson Jun 29 '12 at 11:53

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 $A_5$.