An example of a non-abelian group of order $p^3$ Let $p$ be prime and consider the set of all matrices $$\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}$$ where $a,b,c \in \mathbb{Z_p}$. This set forms a non-abelian group of order $p^3$.
Any ideas how to prove using group actions? 
 A: $\begin{bmatrix}1&a&b\\0&1&c\\0&0&1\end{bmatrix}^{-1}=\begin{bmatrix}1&-a&ac-b\\0&1&-c\\0&0&1\end{bmatrix}$  
Use the two-step group test to show that it is a group.  
For the non-abelian part pick two matrices such as
$\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$ and $\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}$
and show that they does not commute.  
The order part should be obvious.
A: $${{H}_{p}}=\left\{ \left. \left( \begin{matrix}
   1 & a & b  \\
   1 & 1 & c  \\
   1 & 1 & 1  \\
\end{matrix} \right)\,\, \right|\,\,\,a,b,c\in {{Z}_{{{p}^{3}}}} \right\}
$$
$$\left( \begin{matrix}
   1 & a & b  \\
   0 & 1 & c  \\
   0 & 0 & 1  \\
\end{matrix} \right)\left( \begin{matrix}
   1 & a' & b'  \\
   0 & 1 & c  \\
   0 & 0 & 1  \\
\end{matrix}' \right)=\left( \begin{matrix}
   1 & a+a' & b+b'+ac'  \\
   0 & 1 & c+c'  \\
   0 & 0 & 1  \\
\end{matrix} \right)\ne \left( \begin{matrix}
   1 & a+a' & b+b'+a'c  \\
   0 & 1 & c+c'  \\
   0 & 0 & 1  \\
\end{matrix} \right)
$$
$$\left( \begin{matrix}
   1 & a & b  \\
   0 & 1 & c  \\
   0 & 0 & 1  \\
\end{matrix} \right)\left( \begin{matrix}
   1 & 0 & 0  \\
   0 & 1 & 0  \\
   0 & 0 & 1  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 & 0  \\
   0 & 1 & 0  \\
   0 & 0 & 1  \\
\end{matrix} \right)\left( \begin{matrix}
   1 & a & b  \\
   0 & 1 & c  \\
   0 & 0 & 1  \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & a & b  \\
   0 & 1 & c  \\
   0 & 0 & 1  \\
\end{matrix} \right)
$$
$${{\left( \begin{matrix}
   1 & a & b  \\
   0 & 1 & c  \\
   0 & 0 & 1  \\
\end{matrix} \right)}^{-1}}=\left( \begin{matrix}
   1 & -a & ac-b  \\
   0 & 1 & -c  \\
   0 & 0 & 1  \\
\end{matrix} \right)
$$

let 
$$G=\left\{ \left. \left( \begin{matrix}
   x & y  \\
   0 & 1  \\
\end{matrix} \right)\, \right|\,\,x,y\in {{Z}_{{{p}^{2}}}}\,,\,\,x=1\,\,\bmod \,p\, \right\}
$$Indeed we can say $H_p$ and $G$ are isomorphic!
