Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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"?

share|improve this question
6  
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
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.