Notes for AKT-140228/0:41:45

'Showing '$(A^{g})^{h}=A^{(gh)}$

${\begin{aligned}A^{(gh)}&=(gh)^{-1}A(gh)+(gh)^{-1}\mathrm {d} (gh)\\&=(gh)^{-1}A(gh)+(gh)^{-1}{\Big (}(\mathrm {d} g)h+g(\mathrm {d} h){\Big )}\\&=(gh)^{-1}A(gh)+(gh)^{-1}(\mathrm {d} g)h+(gh)^{-1}g(\mathrm {d} h)\\&=h^{-1}(g^{-1}Ag)h+h^{-1}(g^{-1}\mathrm {d} g)h+h^{-1}\mathrm {d} h\\&=h^{-1}{\Big (}g^{-1}Ag+g^{-1}\mathrm {d} g{\Big )}h+h^{-1}\mathrm {d} h\\&=(A^{g})^{h}.\end{aligned}}$

The equality shows that the action is a group action.