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there is a problem in my textbook as follows: Why the finite group $SL(2,5)$ is isomorphic to a subgroup of $SL(2,11)$? Thanks for the answers

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  • $\begingroup$ Welcome to Math SE ! People here are willing to help you. So, could you tell what you already tried and where you are stuck ? $\endgroup$ – Claude Leibovici Sep 27 '14 at 12:12
  • $\begingroup$ What is SL(2,5) & SL(2,11)? $\endgroup$ – Ri-Li Sep 27 '14 at 12:51
  • $\begingroup$ @user152715 The special linear groups with coefficients in $\Bbb{F}_5$ and $\Bbb{F}_{11}$ respectively:en.wikipedia.org/wiki/Special_linear_group $\endgroup$ – Marc Bogaerts Sep 27 '14 at 14:38
  • $\begingroup$ That's right. Special linear group on 2*2 matrices over the field with 11 and 5 elements. $\endgroup$ – BHZ Sep 27 '14 at 15:40
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I would say because it can. The subgroup of $SL(2,11)$ generated by $\begin{pmatrix} 0&-1\\ 1 & 0 \\\end{pmatrix} $ and $\begin{pmatrix} 3 &1\\ 0 & 4 \\\end{pmatrix}$ is isomorphic to $SL(2,5)$

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Let's give a little more explanation here. First of all, $SL(2,5)$ can be presented as $<x,y,z|o(x)=5,o(y)=3,o(z)=2,xz=zx,yz=zy,(xy)^2=z>$. (See Passman, Permutation Groups, Prop 13.7). This can be realized over any field of characteristic $0$ or greater than $5$ that has square roots of -1 and of 5 via $Y=\begin{pmatrix} -1&1\\ -1 & 0 \\\end{pmatrix}$, $X=\begin{pmatrix} 0&\sqrt{-1}\\ \sqrt{-1}&\frac{\sqrt{5}-1}{2}\\\end{pmatrix}$.

For the remainder of this article please note that $p$ denotes a prime greater than $5$ and $q$ a power of a prime greater than $5$. I do not intend to discuss the modular case where the characteristic of the field divides $|SL(2,5)|$.

Thus, if $p>5$ then $SL(2,5)$ is a subgroup of any $SL(2,p^{2n})$ since $\Bbb{F}_{p^{2n}}$ contains the requisite square roots.

It is also sufficient to have a fifth root of unity (but not necessary....e.g., $\Bbb{F}_{49}$). In that case you can take $x=\begin{pmatrix} a&1\\ 0&a^{-1}\\\end{pmatrix}$ and $y=\begin{pmatrix} 0&a^{-1}\\ -a&-1 \\\end{pmatrix}$, with $a$ a primitive fifth root of unit. Since $5$ divides $10=11-1$, this is why $SL(2,11)$ has an $SL(2,5)$ subgroup. This exact same methods exhibits $SL(2,5)$ as a subgroup of $SL(2,31)$, $SL(2,41)$, $SL(2,61)$, $SL(2,71)$, etc., so it's a bit simplistic to say that $SL(2,5)$ is a subgroup of $SL(2,11)$ just "because it can".

It is necessary to have a square root of $5$, since that shows up in the character table of any $2$-dimensional complex representation of $SL(2,5)$. More simply, having a square root of $5$ in $\Bbb{F}_q$ also (by quadratic reciprocity and Galois theory) happens, if $q$ is an odd power of $p>5$, to be equivalent to $q\equiv\pm1\pmod5$, which is necessary for $5$ to divide $|SL(2,q)|=(q-1)q(q+1)$. It is sufficient to have a fifth root of unity OR to have square roots of $-1$ and of $5$. Neither of these last two is necessary. The harder problem is whether $SL(2,5)$ is a subgroup of $SL(2,p)$ for $p\equiv 19\pmod{20}$, the primes for which $\Bbb{F}_p$ has square roots of $5$ but neither square roots of $-1$ nor fifth roots of unity. This reference shows the answer is positive for $p\equiv19\pmod{20}$: https://www.staff.ncl.ac.uk/o.h.king/KingBCC05.pdf, since $A_5\equiv PSL(2,5)$ as a subgroup of $PSL(2,q)$ implies $SL(2,5)$ is a subgroup of $SL(2,q)$.

Thus, for $\Bbb{F}_q$ of characteristic greater than $5$, $SL(2,5)$ embeds in $SL(2,\Bbb{F}_q)$ if and only if $q\equiv\pm 1\pmod{10}$.

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