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