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

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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?

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    $\begingroup$ You are given $G$ acts on $X$, so $x\mapsto gx$ is, by definition, bijective. $\endgroup$ – user10354138 Jul 19 '20 at 17:08
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If $gx=gy$, then multiplying both sides by $g^{-1}$, we have $x=y$. And given an arbitrary $y\in X$, take $x=g^{-1}y$.

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