 Solution

Use the canonical Poisson bracket convention $\{H,G\}={\sum }_k(H_{q_k}{G}_{p_k}-H_{p_k}{G}_{q_k})$. The Hamiltonian flow generated by $G$ has ${\dot{q}}_k=G_{p_k}$ and ${\dot{p}}_k=-G_{q_k}$. The chain rule therefore givesThus $H$ is invariant under every such trajectory exactly when $\boxed{\{H,G\}=0}$ identically. Antisymmetry then gives $dG/dt=\{G,H\}=0$ along the flow of $H$, so $G$ is a constant of motion.

For $G=\alpha {\sum }_k{p}_k$, the generated equations are ${\dot{q}}_k=\alpha$, ${\dot{p}}_k=0$, hence $q_k(t)=q_k(0)+\alpha t$, $p_k(t)=p_k(0)$. Invariance under this simultaneous translation implies conservation of $\boxed{\sum _k{p}_k}$ when $\alpha \ne 0$; the zero generator alone conveys no additional conservation law.