Determine the Sylow $p$-subgroups of $SL(2, \mathbb{F}_5)$ I'm working on the following problem

Let $H := SL_2(\mathbb{F}_5)$.
  Determine, for every primefactor $p$ of $|H|$, a Sylow $p$-subgroup.

What I have so far:
Let $G := GL_2(\mathbb{F}_5)$ and $G' := \mathbb{F}_5^\times$, then is $f: G \to G'$ where $A \mapsto \det(A)$ a group homomorphism, since for $A, B \in G$
$$f(A) \cdot f(B) = \det(A) \cdot \det(B) = \det(A \cdot B) = f(A \cdot B)$$
As $f$ is a homomorphism we know that $\ker(f) \triangleleft G$, where we see that
$$\ker(f) = \{A \in G \mid f(A) = 1\} = H \triangleleft G$$
hence $G_{/H} \to G'$ is an isomorphism according to the first isomorphism theorem.
This means that $G_{/H} \cong G'$ and since $|G'| = 4$ follows $|G_{/H}| = [G:H] = 4$.
So since
$$|G| = (5^2-1)(5^2-5) = 24 \cdot 20 = 480$$
we have
$$|G| = [G:H] \cdot |H| \implies |H| = \frac{|G|}{[G:H]} = \frac{480}{4} = 120$$
where we see that $|H| = 120 = 2^3 \cdot 3 \cdot 5$, so according to the first Sylow theorem, there are subgroups $K < H$ or order $2^3$, $3$ and $5$.
So far so good, but now I struggle to determine the Sylow $p$-subgroups for $p = 2, 3, 5$.
I know that I can use the third Sylow theorem to determine how many subgroups there must be, but I don't see how this would help me find a concrete subgroup.
Is there any systematic approach to find a concrete Sylow $p$-subgroup or is it a brute-force exercise which involves detailed knowledge of the properties of $H$?
 A: As was already pointed out, the element of order $5$ is canonical. For the element of order $3$ I just tried some stuff out and noticed that
$ \left(
\begin{array} 
.4 & 4 \\
1 & 0 \\
\end{array}
\right)
$ works.
The most interesting is the group of order $8$. You should keep in mind that there are few possible isomorphism classes: a group of order $8$ is either abelian, or isomorphic to one of $D_8$ (the dihedral group of order $8$) or $Q_8$ (the quaternion group). To start, you know that $\mathbb{F}_5^\times$ is cyclic of order $4$, generated by $2$ or $3 = -2$. This gives you the following elements of order $4$:
$ I = \left(
\begin{array} 
.2 & 0 \\
0 & 3 \\
\end{array}
\right),
J = \left(
\begin{array} 
 .0 & 2 \\
2 & 0 \\
\end{array}
\right)
$ and there is always the usual element of order $4$ (using that $4 = -1$):
$K = \left(
\begin{array} 
.0 & 4 \\
1 & 0 \\
\end{array}
\right)
$. So now you have three elements of order $4$ all squaring to $-Id$. This should be reminiscent of the quaternion group. And indeed, you can check that the group generated by these three matrices is indeed $\langle I, J, K | I^2 = J^2 = K^2 = IJK = -1 \rangle = \{ \pm Id, \pm I, \pm J, \pm K \} \cong Q_8.$
