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.


2 Answers 2


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

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.


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.


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