The Heisenberg group $G$ over the field $k$ is the subgroup of $GL_{3}(k)$ defined by the matrices of the form
$$ \begin{pmatrix}
1 & x & z\\
0 & 1 & y \\
0 & 0 & 1
\end{pmatrix},\qquad (x, y, z \in k)$$
Find $\operatorname{Tor}(G)$.
How can I find $\operatorname{Tor}(G)$? could anyone help me please?
You can verify (by induction) that $$\begin{pmatrix}1&x&z\\0&1&y\\0&0&1\end{pmatrix}^n=\begin{pmatrix}1&nx&\frac{1}{2}n^2xy-\frac{1}{2}nxy+nz\\0&1&ny\\0&0&1\end{pmatrix}.$$
I assume that by $\text{Tor}(G)$ you mean $\text{Tor}(G)=\{A\in G\mid A^m=I_3 \text{ for some }m\geq 1\}$? If so, the above depends on the characteristic of your field (and I probably am using bad notation if said characteristic were $2$!).
EDIT: Let's do the induction. For $n=1$, there is nothing to prove, so assume that equality holds for $n\geq 1$. Then $$\begin{pmatrix}1&x&z\\0&1&y\\0&0&1\end{pmatrix}^{n+1}=\begin{pmatrix}1&x&z\\0&1&y\\0&0&1\end{pmatrix}\begin{pmatrix}1&nx&\frac{1}{2}n^2xy-\frac{1}{2}nxy+nz\\0&1&ny\\0&0&1\end{pmatrix}=\begin{pmatrix}1&(n+1)x&(\frac{1}{2}n^2xy-\frac{1}{2}nxy+nz)+nxy+z\\0&1&(n+1)y\\0&0&1\end{pmatrix}.$$ Note that $(\frac{1}{2}n^2xy-\frac{1}{2}nxy+nz)+nxy+z=(\frac{1}{2}(n+1)^2xy-\frac{1}{2}(n+1)xy+(n+1)z)$. This shows the induction argument.