Find normal subgroup of the following group

Question

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?

Answer 1

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.

Answer 2

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 $\mathbb{R}$

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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 $z=0$

Answer 3

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 $G$.