# injectivity of a group action

The action of a group $G$ on $X$ is always "injective" in the following sense: if $x\not = y$ then $\forall g\in G$, $gx\not = gy$ indeed if $gx=gy$ then $g^{-1}(gx)=(g^{-1})gx=x=g^{-1}(gy)=(g^{-1}g)y=y$. Is this why the second axiom of group action is set: $g(hx)=(gh)x$? and what is the importance of this "injectivity"?

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If $g \in G$, the function $g. : x \mapsto gx$ is not only injective (the property you mention), it's a bijection. The reason is that its inverse is given by $g^{-1}. : x \mapsto g^{-1}x$. – Joel Cohen Jul 4 '11 at 11:47
Of course I don't begrudge upvotes for Joel, but it seems his comment is saying the same as my (earlier) answer? – wildildildlife Jul 4 '11 at 13:59

## 1 Answer

A $G$-action on $X$ is the same as a group morphism $G\to S_X$ into the symmetric group on $X$. Thus each element $g\in G$ induces a permutation $x\mapsto gx$ of $X$, and group multiplication corresponds to permutation composition. In particular, $x\mapsto gx$ being a bijection (=permutation), with inverse $y\mapsto g^{-1}y$, means it is both surjective and injective.

I guess my 'point' is that you shouldn't only concentrate on the injectivity, but rather think of permutations of the set.

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