3
$\begingroup$

Dears,

Let $H$ be Heisenberg group, a group of $3\times 3$ matrices with $1$ on the main diagonal, zeros below, and elements of $\Bbb R$ above the main diagonal. Its center is the subgroup of all matrices with $0$ on the first diagonal above the main diagonal.

My question is - is it also a commutator subgroup of that group? The quotient group $H/Z(H)$ is abelian (this group is nilpotent of class two), so commutator subgroup must be inside the center. I can't imagine other subgroups being CS, but I want someone smarter to let me know.

Have a nice day.

$\endgroup$
7
$\begingroup$

$$\begin{pmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{pmatrix}^{-1}=\begin{pmatrix}1&-x&xz-y\\ 0&1&-z\\ 0&0&1\end{pmatrix}$$

Therefore \begin{align}ABA^{-1}B^{-1}&=\begin{pmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&s&t\\ 0&1&u\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&-x&xz-y\\ 0&1&-z\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&-s&su-t\\ 0&1&-u\\ 0&0&1\end{pmatrix}=\\&=\begin{pmatrix}1&s+x&t+ux+y\\ 0&1&u+z\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&-x-s&ux+xz-y+su-t\\ 0&1&-z-u\\ 0&0&1\end{pmatrix}=\\&=\begin{pmatrix}1&0&t+ux+y+ux+xz-y+su-t-(z+u)(s+x)\\ 0&1&0\\ 0&0&1\end{pmatrix}=\\&=\begin{pmatrix}1&0&ux-zs\\ 0&1&0\\ 0&0&1\end{pmatrix}\end{align}

It is therefore apparent that commutators are exactly the elements in the form $\begin{pmatrix}1&0&\alpha\\0&1&0\\ 0&0&1\end{pmatrix}$, which incidentally form a subgroup.

$\endgroup$
6
$\begingroup$

Since$$\begin{bmatrix}0&a&c\\0&1&b\\0&0&1\end{bmatrix}.\begin{bmatrix}1&d&f\\0&1&e\\0&0&1\end{bmatrix}\begin{bmatrix}0&a&c\\0&1&b\\0&0&1\end{bmatrix}^{-1}\begin{bmatrix}1&d&f\\0&1&e\\0&0&1\end{bmatrix}^{-1}=\begin{bmatrix}0 & 0 & a e-b d \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},$$yes, the commutator group is the center.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.