The infinitesimal generator of a symplectic action is symplectic

For fixed $p\in M$ put $\theta (t)={\exp }_G(-t\xi )\cdot p$. The action law and [F1] give, for every $t$, ${\theta }^{\mathrm{\prime }}(t)={\frac{d}{ds}\mid }_0{\exp }_G(-(t+s)\xi )\cdot p={\frac{d}{ds}\mid }_0{\exp }_G(-s\xi )\cdot ({\exp }_G(-t\xi )\cdot p)={\xi }_M(\theta (t)),$ so $\theta$ is an integral curve of ${\xi }_M$ with $\theta (0)=p$. Its domain is all of $\mathbb{R}$, because the action and the exponential are defined for all real parameters and all group elements.

By [F4] the maximal integral curve of ${\xi }_M$ through $p$ is unique, so $\theta$ is it; hence the flow of ${\xi }_M$ is the global map ${\mathrm{\Phi }}_t(p)={\exp }_G(-t\xi )\cdot p$ on $\mathbb{R}\times M$.

For each real $t$ the map ${\mathrm{\Phi }}_t=a_{{\exp }_G(-t\xi )}$ is the diffeomorphism induced by the group element ${\exp }_G(-t\xi )$, so [F2] gives ${\mathrm{\Phi }}_t^{\ast }\omega =\omega$; therefore the curve $t\mapsto {\mathrm{\Phi }}_t^{\ast }\omega$ is the constant two-form $\omega$ on its flow domain.

Differentiating this constant curve at $t=0$ and applying the flow characterization [F3] with $X={\xi }_M$ and $T=\omega$ yields ${\mathcal{L}}_{{\xi }_M}\omega =0$.