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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?

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3 Answers 3

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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 $z=0$

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  • $\begingroup$ Explain the down votes please. $\endgroup$ Oct 24, 2018 at 10:19
  • $\begingroup$ No idea. Perhaps because it is more or less identical to the earlier answer? I've upvoted it anyway. $\endgroup$
    – Derek Holt
    Oct 24, 2018 at 10:21
  • $\begingroup$ I was writing the answer when other user posted his answer. It took me some time to write matrices in latex. @DerekHolt $\endgroup$ Oct 24, 2018 at 10:22
  • $\begingroup$ But is not $Z(G)=\{I\}$?Did You check? $\endgroup$
    – Learnmore
    Oct 24, 2018 at 10:25
  • $\begingroup$ @Learnmore What I have written is correct. $\endgroup$ Oct 24, 2018 at 10:26
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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 $G$.

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  • $\begingroup$ Nice approach (+1) $\endgroup$ Oct 24, 2018 at 12:20
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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.

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  • $\begingroup$ This represents the center $Z(G)$ of the given group $G$ and centers are normal. $\endgroup$
    – Learning
    Sep 27, 2019 at 9:20

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