We define the Heisenberg group $H^{n}$ for $n\geq1$ as follows. As an analytic manifold $H^{n}=\mathbb{R}^{2n+1}$. We denote elements in $H^{n}$ by $(t_{i},q_{i},p_{i})$ with $t_{i}\in\mathbb{R}$ and $q_{i},p_{i}\in\mathbb{R}^{n}$, and we denote the standard inner product on $\mathbb{R}^{n}$ by $\cdot $. Then we define multiplication in $H^{n}$ by: $$(t_{1},q_{1}.p_{1})(t_{2},q_{2},p_{2})=\left(t_{1}+t_{2}+\frac{1}{2}(p_{2}\cdot q_{1}-p_{1}\cdot q_{2} ),q_{1}+q_{2},p_{1}+p_{2}\right).$$
Show that $H^{n}$ is a Lie group.
My attempt: Clearly, by definition $H^{n}$ is a smooth manifold. It is not hard to see that $H^{n}$ is a group with the above operation.
The problem that motivates me to write this post is to show that the map $\mu:H^{n}\times H^{n} \longrightarrow H^{n}$ where $$\mu \left((t_{1},q_{1}.p_{1}),(t_{2},q_{2},p_{2})\right)=\left(t_{1}+t_{2}+\frac{1}{2}(p_{2}\cdot q_{1}-p_{1}\cdot q_{2} ),q_{1}+q_{2},p_{1}+p_{2}\right)$$ is $C^{\infty}$.