Let $G:={\rm SL}(2,5)$. I want to prove that if $P \in {\rm Syl}_2(G)$, then $P \cong Q_8$. Also i want to know $n_2(G)=5$.
MY TRY: I know $|G|=120=2^3 \times 3 \times 5$. Thus $|P|=8$. If $P$ is cyclic, then $G$ is solvable, a contradiction. So $P$ is non-cyclic. I need to show $P$ is non-abelian and all maximal subgroups of $P$ are cyclic. I know $Z(G) \cong \Bbb{Z}_2$ is only non-trivial normal subgroup of $G$ and $G/Z(G) \cong A_5$.
Let $P$ be a $2$-Sylow group in $G=SL(2,\Bbb F_5)$. The general linear group over this field has $(5^2-1)(5^2-5)=480$ elements. The determinant $\det :GL(2,\Bbb F_5)\to \Bbb F_5^\times$ is surjective, so the kernel has $480/4=120=2^3\cdot 3\cdot 5$ elements. The group $P$ has thus $8$ elements. To see which can be the its structure, let us first find all $g\in G$ which have order among $1,2,4,8$. There is only one element in $G$ of order $1$, the unit (matrix) $1\in G$, $$1=\begin{bmatrix}1&\\&1\end{bmatrix}\ .$$ Let $$g=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ be an element of order $2$. Then its minimal polynomial divides $X^2-1=(X-1)(X+1)$. It cannot be $X^2-1$, since the determinant is $1$. So it is $X\pm 1$. So there exists exactly one element of order two in $G$, this is $-1$, $$-1=\begin{bmatrix}-1&\\&-1\end{bmatrix}\ .$$ Now let us find all matrices $g\in G$ with order $4$ or $8$. The minimal polynomial of $g$ divides $X^4-1$, respectively $X^8-1$. We have the decompositions in $R=\Bbb F_5[X]$: $$ \begin{aligned} X^4-1 &= (X-1)(X-2)(X-3)(X-4)\ ,\\ X^8-1 &= (X^4-1)(X^4+1)\\ &=(X-1)(X-2)(X-3)(X-4)\cdot (X^2-2)(X^2-3)\ . \end{aligned} $$ The only chance to use two eigenvalues in the list of the roots of the above polynomials, so that their product is (the determinant) $1$, is to use $2,3$. (Or double eigenvalues, $+1$, or $-1$, already considered.)
This implies that we have no element $g\in G$ of order $8$.
An element in $G$ has order $4$, iff its characteristic polynomial is $(X-2)(X-3)=X^2+1$, so iff it has trace zero, i.e. $$g=\begin{bmatrix}a&b\\c&-a\end{bmatrix} \ ,\qquad -a^2-bc=1\ .$$
We may choose $a\in 2,3$, then $bc=0$, and we have $5+5-1$ possibilities for the pair $(b,c)$. This counting gives us $18$ elements of order $4$.
We may choose $a\in 1,4$, then $bc=3$, and there are $4$ possibilities for $b\ne 0$, which determine $c$. We obtain $8$ more elements of order $4$.
The final choice is $a=0$, getting $4$ more elements of order $4$.
So there are totally $$ 18+ 8 +4 = 30$$ elements of order $4$ in $G$. (And no elements of order $8$, so $P$ is not cyclic, $P\not\cong \Bbb Z/8$.)
From here, $P$ cannot be abelian. Since else it would have the structure $\Bbb Z/4\times \Bbb Z/2$, (or $\Bbb Z/2\times \Bbb Z/2\times \Bbb Z/2$...) contradiction with the existence of exactly one element of order two in $G$.
So $P$ is a dihedral group of order $8$.
Now we need further information to get $n_2(G)$. So let us list all matrices of order four:
We can explicitly construct now a group of order $8$ by suitably choosing two different elements $s,t$ in the above list. The group $P(s,t)=\langle s, t\rangle$, if it is a group, has then the elements $$1,t,-1=t^2=s^2,-t=t^3;\ s,-s;\ st, ts\ .$$ So $sts$ is an other element, it must be in the list. The only chance is $$sts=t\ .$$ So each $P$ is of the form: $$ P(s,t)=\langle s, t\rangle =\{\ 1,-1,\ t,-t,\ s,-s,\ st, -st=ts\ \}\ ,\qquad sts=t\ ,\ tst=s\ . $$
One possible realization is with: $$ P_0= \left\langle \ \begin{bmatrix} 0 & 2\\ 2& 0\end{bmatrix} \ ,\ \begin{bmatrix}2 & 0\\ 0& 3\end{bmatrix} \ \right\rangle $$ We can draw a graph by joining two elements $(s,t)$ among the $30$ iff they satisfy $sts=t$. (Then $tst=t$.) Even better, use in the graph only one vertex for a pair $\{s,-s\}$ generating the same group with four elements.
Here is the restricted list of $15$ representatives $\hat s$ for the pair $\{s,s^{-1}\}$:
The corresponding graph is
it splits in five triangle connected components, coresponding to edges of the shape $s\to t\to \pm st\to s$ for suitable element. With this analysis, we see there are exactly $$n_2(G)=5$$ $2$-Sylow subgroups, explicitly realized.
An other way to get this number is as follows. The element $$u= \begin{bmatrix} 1 &1\\2&3 \end{bmatrix} $$ has order $3$ and normalizes $P_0$, since $$ u \begin{bmatrix} 2 &\\&3 \end{bmatrix} u^{-1} = \begin{bmatrix} &2\\2& \end{bmatrix} \ . $$ So $u$ is an element in the normalizer $N$ of $P_0$. A Sylow theorem tells us that $n_2(G)=|G:N|$. How many element can have $N$? It contains the group $P$ and an element of order $3$, so $|N|$ is a multiple of $24$. Only $24$ and $120$ can be possible. In the first case there are $120:24=5$ Sylow groups, which is the case. Since the second case with exactly $120:120=1$ Sylow group can be excluded, one can construct two.