Groups and FieldsChallengeGroup Actions ShareGroup Actions · Challenge problemFor a non-abelian group GGG, does the rule g⋅x=g−1xg\cdot x=g^{-1}xg⋅x=g−1x (with x∈Gx\in Gx∈G) define an action of GGG on itself?ANo: g⋅(h⋅x)=g−1h−1x=(hg)−1xg\cdot(h\cdot x)=g^{-1}h^{-1}x=(hg)^{-1}xg⋅(h⋅x)=g−1h−1x=(hg)−1x, which is (hg)⋅x(hg)\cdot x(hg)⋅x, not (gh)⋅x(gh)\cdot x(gh)⋅xBYes, it is always an actionCNo: the identity element does not fix every xxxDYes, but only when GGG is finiteCheck Show answer Walk through it step by step Practice more like this Review this concept