Sylow subgroups of $2\times 2$ invertible matrices with elements in $\mathbb Z_5$. Let $G$ be the group of all $2 \times 2$ invertible matrices with elements in $\mathbb Z_5$.  This group has order of $480=2^5\cdot 3 \cdot 5$.
I want to know if there is a quick and efficient strategy to find an example of each type of Sylow subgroup.  For example, I know that there could be $1, 3, 5,$ or $15$ $2$-Sylow subgroups with $32$ elements.

You want matrices such that their order is $1\le2^n\le32$.  So try finding "roots" of $I$:
$$\begin{align} \begin{pmatrix}a&b\\c&d\end{pmatrix}^2&\equiv I\pmod{5}\\ \begin{pmatrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{pmatrix}&\equiv I\pmod{5}\\
a^2+bc\equiv d^2+bc&\equiv 1\pmod 5\\
b(a+d)\equiv c(a+d)&\equiv 0\pmod 5 
\end{align}$$
Since $a^2-d^2=(a+d)(a-d)\equiv0\pmod 5$, $5|(a+d)$ or $5|(a-d)$.
Case $1$: $5|(a+d)$
If $a\equiv d\equiv0\pmod 5$, then $bc\equiv 1$, so we have solutions $\begin{pmatrix}0&1\\1&0\end{pmatrix},\begin{pmatrix}0&2\\3&0\end{pmatrix},\begin{pmatrix}0&3\\2&0\end{pmatrix},\begin{pmatrix}0&4\\4&0\end{pmatrix}$
Otherwise, if $(a,d)=(1,4)\lor(4,1)\Longrightarrow bc\equiv0$ and if $(a,b)=(2,3)\lor(3,2)\Longrightarrow bc\equiv 2$.  In the former case, we get solutions $(b,c)=(0,0),(0,1),(0,2),(0,3),(0,4),(1,0),(2,0),(3,0),(4,0)$ while the latter gets $(b,c)=(1,2),(2,1),(3,4),(4,3)$.
Case 2: $5|(a-d)$
We already covered $a\equiv d\equiv 0$, so we can move on
Since $a\equiv d\not\equiv0\pmod 5$, $a+d\equiv 2a\not\equiv 0$, so $b\equiv c\equiv 0\pmod 5$.  We then have $a^2\equiv d^2\equiv 1$, so the only solutions in this case are $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}4&0\\0&4\end{pmatrix}$
So that's all $32$ roots of $I$, and these will generate more than one $2$-Sylow (you can check that the roots aren't closed).  In general $n^{th}$ roots of $e$ for groups is a good start, but it depends on the group's structure.
