Since I could use the reputation... letting $\psi: G \to Diff(M)$ \operatorname{Diff}(M)$be the action then$X' = \psi_* X$, where we identify$Lie(Diff(M))$\operatorname{Lie}(\operatorname{Diff}(M))$ with vector fields on $M$. This is because they both correspond to the one-parameter subgroup $t\mapsto \psi(\exp tX)$ of $Diff(M)$. \operatorname{Diff}(M)$. Then$[X', Y'] = [X,Y]'$is immediate. 2 added 107 characters in body Since I could use the reputation... letting$\psi: G \to Diff(M)$be the action then$X' = \psi_* X$, where we identify$Lie(Diff(M))$with vector fields on$M$. This is because they both correspond to the one-parameter subgroup$t\mapsto \psi(\exp tX)$of$Diff(M)$. Then$[X', Y'] = [X,Y]'$is immediate. 1 Since I could use the reputation... letting$\psi: G \to Diff(M)$be the action then$X' = \psi_* X$, where we identify$Lie(Diff(M))$with vector fields on$M$. Then$[X', Y'] = [X,Y]'\$ is immediate.