The Heisenberg Group

The Heisenberg group is the set of $3\times 3$ matrices of the form

(1)
\begin{pmatrix} 1 & a & b \\0 & 1 & c \\0 & 0 & 1 \end{pmatrix}

where $a,b,c$ are elements of $\mathbb{R}$

Suppose $a,b,c ,d,e,f,g,h,i$ are elemnts of $\mathbb{R}$
then

(2)
\begin{pmatrix} 1 & a & b \\0 & 1 & c \\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & d & e \\0 & 1 & f \\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & g & h \\0 & 1 & i \\0 & 0 & 1 \end{pmatrix}

Do the first multiplication,then do then second one

(3)
\begin{pmatrix} 1 & a+d & e+af+b \\0 & 1 & c+f \\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & g & h \\0 & 1 & i \\0 & 0 & 1 \end{pmatrix}

then we get

(4)
\begin{pmatrix} 1 & a+d+g & h+ai+ad+e+af+b \\0 & 1 & c+f+i \\0 & 0 & 1 \end{pmatrix}

Do the second multiplication first

(5)
\begin{pmatrix} 1 & a & b \\0 & 1 & c \\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & d+g & e+di+h \\0 & 1 & f+i \\0 & 0 & 1 \end{pmatrix}

Then we get

(6)
\begin{pmatrix} 1 & a+d+g & h+ai+ad+e+af+b \\0 & 1 & c+f+i \\0 & 0 & 1 \end{pmatrix}

So Associative Axiom holds
For identity Axiom
It not hard to find that a=0,b=0,c=0, so it is

(7)
\begin{pmatrix} 1 & 0 & 0\\0 & 1 & 0 \\0 & 0 & 1 \end{pmatrix}

For Inverse Axiom
for any $a,b,c\in R$

(8)
\begin{align} \left|\begin{array}{ccc}1 &a & b \\0 & 1 & c \\0 & 0 & 1\end{array}\right|=1>0 \end{align}

So the inverse of this form of a matrix always exsits
then we just need to find the inverse is in the group
so find inverse of

(9)
\begin{pmatrix} 1 & a & b \\0 & 1 & c \\0 & 0 & 1 \end{pmatrix}


$a,b,c\in R$

(10)
\begin{pmatrix} 1 & a & b & 1 & 0 & 0\\0 & 1 & c & 0 & 1 &0 \\0 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}


Do row and column operations

(11)
\begin{pmatrix} 1 & 0 & 0 & 1 & -a & ac-b \\0 & 1 & 0 & 0 & 1 &-c \\0 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}


so the inverse is

(12)
\begin{pmatrix} 1 & -a & ac-b \\0 & 1 & -c \\0 & 0 & 1 \end{pmatrix}

since -a, ac-b,-c are elements of R
So it is in the set of The Heisenberg Group
So the Inverse Axiom holds

Good job…, but you showed that the inverse of the matrix exists, but it is it also of the same form? This means, if

$\begin{pmatrix}1&a&b\\0&1&c\\ 0&0&1\end{pmatrix}$ is a matrix, can you determine the inverse matrix formula in terms of $a,b, c$?

A subgroup for The Heisenberg Group
My guess will be the matrix in the following form with mutiplication

(13)
\begin{pmatrix} 1 & 0 & b \\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}

where $b$ are elements of $\mathbb{R}$

Proof
closure

(14)
\begin{pmatrix} 1 & 0 & a \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & b \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 & 0 & a+b \\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}

so it is in the form

(15)
\begin{pmatrix} 1 & 0 & b \\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}

Identity
when b=0,it turns out to be identity

(16)
\begin{pmatrix} 1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}

inverse

(17)
\begin{align} \left|\begin{array}{ccc}1 &0 & b \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right|=1>0 \end{align}

so it is invertible

(18)
\begin{pmatrix} 1 & a & b & 1 & 0 & 0\\0 & 1 & c & 0 & 1 &0 \\0 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}


Do row and column operations

(19)
\begin{pmatrix} 1 & 0 & 0 & 1 & 0 & -b\\0 & 1 & 0 & 0 & 1 &0 \\0 & 0 & 1 & 0 & 0 & 1 \end{pmatrix}

so the inverse is

(20)
\begin{pmatrix} 1 & 0 & -b \\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}

fiind some cosets of this subgroup

(21)
\begin{pmatrix} 1 & 1 & 0 \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & b \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 & 1 & b \\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}
(22)
\begin{pmatrix} 1 & 1 & 1 \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & b \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 & 1 & b+1\\0 & 1 & 0\\0 & 0 & 1 \end{pmatrix}
(23)
\begin{pmatrix} 1 & 1 & 1 \\0 & 1 & 1\\0 & 0 & 1\end{pmatrix}\begin{pmatrix}1 & 0 & b \\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 & 1 & b+1\\0 & 1 & 1\\0 & 0 & 1 \end{pmatrix}
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