Is $SL(2, 3) $ a subgroup of $SL(2, p)$ for $ p>3$? As the title says, I was wondering whether $SL(2,3)$ is a subgroup of $SL(2,p)$ for $p>3$.  I know that it is for $p=5$ (it can be found explicitly using the quaternionic representation), and I have some evidence that it is for the other $p$'s, but I'm not sure how I would go about proving it, or indeed if it's true!
If anybody knows one way or another, please let me know.  If it's true, it would be helpful to give a hint about how I might think about proving it, but please don't give a proof as I'd like to figure it out myself.
 A: $SL_2(3)$ is the semidirect product of the quaternion group
$Q = \langle i,j\mid i^4, j^4, i^2=j^2, i^{-1}ji=j^{-1}\rangle$
with a cyclic group $Z_3 = \langle z\rangle$ of order $3$ via
$i^z = j$ and $j^z = ij$.
Hint: Find a subgroup of $SL_2(p)$ ($p$ any odd prime) isomorphic to
$Q$, and then extend it to full group $SL_2(3)$ by using this
little gem from an answer by Geoff Robinson: The units of the
Hurwitz quaternion
are the quaternions plus 16 elements that can be written as
$\frac{\pm 1\pm i\pm j\pm k}{2}$ (with $k = ij$).
As we are in odd characteristic, $\frac{1}{2}$ exists and we have our
candidate for $SL_2(3)$.
To simplify calculations these two easy observations might be useful:
An element of the form $\frac{\pm 1\pm i\pm j\pm k}{2}$ can be inverted
by flipping the signs of $i$, $j$ and $k$, but keeping $1$'s.
This can be best seen by observing that the mixed terms cancel out, and
only those contributing to the coefficient of $1$ add up.
The coefficient at $1$ of the square of an element of the form
$\frac{\pm 1\pm i\pm j\pm k}{2}$ is $-1$, so one expects elements of
order $3$ to contain the summand $-1$ and those of order $6$ the summand
$+1$.

Proof: Let $p$ be an odd prime.
The square of the matrix
$i = \left(\begin{array}{cc} 0 & 1\\-1& 0\end{array} \right)$ is
$\left(\begin{array}{cc} -1 & 0\\0& -1\end{array} \right)$.
Its order is $4$ and its determinant $1$.
For $j$ let's take an arbitrary
$\left(\begin{array}{cc} a & b\\c& d\end{array} \right) \in SL_2(p)$
with $ad-bc = 1$.
Using the equation $i^{-1}ji = j^{-1}$ we get
$\left(\begin{array}{cc} 0 & -1\\1& 0\end{array} \right)
\left(\begin{array}{cc} a & b\\c& d\end{array} \right)
\left(\begin{array}{cc} 0 & 1\\-1& 0\end{array} \right) =
\left(\begin{array}{cc} -c & -d\\a& b\end{array} \right)
\left(\begin{array}{cc} 0 & 1\\-1& 0\end{array} \right) =
\left(\begin{array}{cc} d & -c\\-b& a\end{array} \right) \stackrel{!}{=}
\left(\begin{array}{cc} d & -b\\-c& a\end{array} \right)$, hence $b=c$.
On the other hand $j^2 = i^2$ yields
$\left(\begin{array}{cc} a^2+bc & b(a+d)\\c(a+d) & d^2+bc\end{array}
\right) \stackrel{!}{=}
\left(\begin{array}{cc} -1 & 0\\0& -1\end{array} \right)$,
which holds in case $a = -d$.
So in case $a^2+b^2 = -1$ we get the candidate
$j = \left(\begin{array}{cc} a & b\\b& -a\end{array} \right)$
for $Q = \langle i, j\rangle$.
As there are $\frac{p+1}{2}$ squares in $\mathbb F_{p}$, by the
Cauchy-Davenport theorem
all $p$ elements of $\mathbb F_{p}$ are sums of two squares.
So pick solutions $a$ and $b$ of $a^2+b^2 = -1$, and we found $Q$.
[If $p = -1 \bmod 4$ there exist also solutions with $a=0$ or $b=0$
which we seemed to avoid when choosing $a = -d$ before.]
Defining $z = \frac{-1+i+j+k}{2} = -\frac{(1-j)\cdot(1-i)}{2}$ we get an
element of $SL_2(p)$ of order $3$, as
$det(z) = det(1-j)\cdot det(1-i)\cdot(-\frac{1}{2})^2
= \left|\begin{array}{cc} 1-a & -b\\-b& 1+a\end{array} \right| \cdot
\left|\begin{array}{cc} 1 & -1\\1& 1\end{array} \right| \cdot\frac{1}{4} =
(1-a^2-b^2)\cdot 2\cdot\frac{1}{4} = 1$ and
$z^2 = \frac{-1-i-j-k}{2} = z^{-1}$.
Now $i^z = z^{-1}\cdot i\cdot z =
\frac{-1-i-j-k}{2}\cdot i\cdot\frac{-1+i+j+k}{2} =
\frac{1-i-j+k}{2} \cdot\frac{-1+i+j+k}{2} =
j\cdot\frac{-1-i-j-k}{2}\cdot\frac{-1+i+j+k}{2} = j\cdot z^{-1}\cdot z = j$
and $j^z = z^{-1}\cdot j\cdot z =
\frac{-1-i-j-k}{2}\cdot j\cdot\frac{-1+i+j+k}{2} =
\frac{1+i-j-k}{2}\cdot\frac{-1+i+j+k}{2} =
k\cdot\frac{-1-i-j-k}{2}\cdot\frac{-1+i+j+k}{2} = k\cdot z^{-1}\cdot z = k$
finish the proof.

Final remark: For $p=3$ the proof gives us the semidirect product $Q\rtimes Z_3$ as subgroup of $SL_2(3)$, hence proving the statement in the first sentence of my answer, as both groups have the same order.
