# question about definition of group actions

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I'm sorry this is a stupid question.

Obviously if ag = bg then ag(g inverse) = bg(g inverse) => a = b

I shouldn't post questions late at night.

In the definitionof a group action say of a group G on a set S,

there are two conditions:

1 for a in S and e in G where e is the identity, ae = a.

2 for a in S and g and h in G, (ag)h = a(gh)

Where does it say that if a and b in S and g in G, that ag is not equal to bg?

They always say that for ag = z then obiously there is an inverse g- that zg- = a.

One could add another condition but nobody ever does.

Thanks

Edited by jerryb
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Where does it say that if a and b in S and g in G, that ag is not equal to bg?

I don't quite understand what you mean?

Why should ag = bg if a and b are different points of S?

What does the condition (group axiom) that every member of G has an inverse imply about a and b and the equality of ag and bg?

I find it helps to discuss such statements in relation to a particular example, did you have one in mind?

Edited by studiot
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See the revised question.

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• 2 weeks later...

Where does it say that if a and b in S and g in G, that ag is not equal to bg?

They always say that for ag = z then obiously there is an inverse g- that zg- = a.

One could add another condition but nobody ever does.

The statement $a\ne b\Rightarrow ag\ne bg$ is the contrapositive of $ag=bg\Rightarrow a=b$ (which is what you proved in the opening statement of your post). There is no need to state it as an extra condition.

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