For the Heisenberg group
$ H = \{ \begin{bmatrix} 1 & x & y\\ 0 & 1 & z\\ 0 & 0 & 1 \end{bmatrix} | \:x,y,z \in \mathbb{R}\} $
compute the one-parameter subgroups and the exponential map.
I have so far found a basis $B = \{ \frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, x \cdot \frac{\partial}{\partial{y}} + \frac{\partial}{\partial{z}}\}$ for the left-invariant vector-fields on H. But I can't manage to find the corresponding integral-curves.
Thanks in advance for any help !
Wouldn't a basis like this be easier: $$X = \begin{pmatrix} 0&1&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}, Y = \begin{pmatrix} 0&0&0\\ 0&0&1\\ 0&0&0 \end{pmatrix}, Z = \begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&0 \end{pmatrix} $$
General elements of the Heisenberg Lie algebra are of the form: $$\begin{pmatrix} 0&a&c\\ 0&0&b\\ 0&0&0 \end{pmatrix} $$ This has exponential: $$\begin{pmatrix} 1&a&\frac{1}{2}ab+c\\ 0&1&b\\ 0&0&1 \end{pmatrix} $$ So a generic one parameter subgroup is of the form: $$\begin{pmatrix} 1&at&\frac{1}{2} abt^2+ct\\ 0&1&bt\\ 0&0&1 \end{pmatrix}$$