# Prove that $\phi: G / F \rightarrow \operatorname{Sym}(X)$ is a monomorphism

I'm doing this exercise 11 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.

If $$G$$ acts on $$X$$, and $$F$$ consists of those $$g \in G$$ fixing every $$x \in X$$, prove that $$F \trianglelefteq G$$. If $$p: G \rightarrow G / F$$ is the projection, prove that there is a unique action of $$G / F$$ on $$X$$ with $$(p g) x=g x$$. If $$\phi$$ maps $$p g$$ to the permutation $$x \mapsto g x$$ on $$X$$, prove that $$\phi: G / F \rightarrow \operatorname{Sym}(X)$$ is a monomorphism.

Because the authors mentioned "the permutation $$x \mapsto g x$$ on $$X$$", I tried to prove that $$x \mapsto g x$$ is bijective, but to no avail. Could you please elaborate on the correctness of this exercise?

• You are given $G$ acts on $X$, so $x\mapsto gx$ is, by definition, bijective. – user10354138 Jul 19 '20 at 17:08

If $$gx=gy$$, then multiplying both sides by $$g^{-1}$$, we have $$x=y$$. And given an arbitrary $$y\in X$$, take $$x=g^{-1}y$$.