Find normal subgroup of the following group Is the group $G$ given by 
$$\left\{\begin{bmatrix} 1 & \alpha &\beta \\0& 1 &\gamma\\0 &0 &1\end{bmatrix}:\alpha,\beta,\gamma \in \Bbb R\right\}$$
simple?
My try:
Obviously $G$ is a subgroup of $\text{SL}_n(\Bbb R)$
I tried with matrices like 
$$\left\{\begin{bmatrix} 1 & \alpha &\alpha \\0& 1 &\alpha\\0 &0 &1\end{bmatrix}:$\alpha,\beta,\gamma \in \Bbb R\right\}$$
but they did not help. 
How should I do it?
 A: Inverse of \begin{bmatrix} 
  1 & x & y \\ 
   0 &1 &z \\ 
   0 & 0 & 1 
\end{bmatrix}
is given by 
$\begin{bmatrix} 
   1 & -x & xz-y \\ 
      0 & 1 & -z \\ 
    0 & 0 & 1 
 \end{bmatrix}\tag{1}$
Denote the center of this group by $Z\big(G\big)=\left\{\, \begin{bmatrix} 
  1 & 0 & y \\ 
   0 &1 &0 \\ 
   0 & 0 & 1 
\end{bmatrix} \quad \middle| \quad \text{ for any } y\in \mathbb{R} \, \right \}.$
Now show that $Z(G)$ is normal in $G$. 
Let $A\in G, B\in Z(G)$ be any arbitrary element. Now show that $ABA^{-1}\in Z(G)$
$G$ is also known as Heisenberg Group(in our case over $\mathbb{R}$

Question by OP in comments: How to find $Z(G)$
Let $M=\begin{bmatrix} 
  1 & x & y \\ 
   0 &1 &z \\ 
   0 & 0 & 1 
\end{bmatrix}\in Z(G)$ and $A=\begin{bmatrix} 
  1 & a & b \\ 
   0 &1 &c \\ 
   0 & 0 & 1 
\end{bmatrix}$ be an arbitrary element in $G$. By definition we have $AM=MA$
$$\begin{bmatrix} 
  1 & a & b \\ 
   0 &1 &c \\ 
   0 & 0 & 1 
\end{bmatrix} 
\begin{bmatrix} 
  1 & x & y \\ 
   0 &1 &z \\ 
   0 & 0 & 1 
\end{bmatrix} =\begin{bmatrix} 
  1 & x & y \\ 
   0 &1 &z \\ 
   0 & 0 & 1 
\end{bmatrix}\begin{bmatrix} 
  1 & a & b \\ 
   0 &1 &c \\ 
   0 & 0 & 1 
\end{bmatrix}.$$
Take the product.
$$\begin{bmatrix} 
  1 & x+a & y+az+b \\ 
   0 &1 &z+c \\ 
   0 & 0 & 1 
\end{bmatrix}=\begin{bmatrix} 
  1 & a+x & b+cx+y \\ 
   0 &1 &c+z \\ 
   0 & 0 & 1 
\end{bmatrix}.$$
Since $A$ was arbitrarily chosen. As a particular case,
Take $a=0 , c=1$ you get $x=0$ 
Take $a=1,c=0$ you get $z=0$
A: The map below is a homomorphism $G \to \text{SL}_2(\Bbb R)$:
$$
\begin{bmatrix} 1 & \alpha &\beta \\0& 1 &\gamma\\0 &0 &1\end{bmatrix}
\mapsto
\begin{bmatrix} 1 & \alpha \\0& 1 \end{bmatrix}
$$
Therefore, its kernel is a normal subgroup of $G$.
The kernel is the set of matrices in $G$ with $\alpha=0$, and so is a nontrivial proper subgroup of $G$.
A: No, it is not simple. The matrices of the form$$\begin{bmatrix}1&0&\alpha\\0&1&0\\0&0&1\end{bmatrix}$$($\alpha\in\mathbb R$) form a normal subgroup.
