Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is very simple, but as far as I can tell it has not been asked yet.

Let the group $G$ act on the set $S$ and define an equivalence relation by $x \sim x'$ if there exists a $g \in G$ for which $gx=x'$.

Proving reflexivity and transitivity is easy, so let's look at the symmetric property: Say $x \sim x'$ with $gx=x'$. Then we have $x=x'g^{-1}$. So $x$ is equal to $x'$ multiplied by an element of $G$, but does this work since we are now using right multiplication? Can we do something 'clever' like $ex=x'g^{-1} \Rightarrow xe=g^{-1}x' \Rightarrow x=g^{-1}x' \Rightarrow x' \sim x$? Something about that last bit seems foul to me.

The group theory tag isn't really appropriate here. I would create a 'group actions' tag if I were able.

share|cite|improve this question
Why do you have $x = x'g^{-1}$? If the action is on the left, you should just have $x = g^{-1}gx = g^{-1}x'$ – user29743 Apr 28 '12 at 2:31
There need not be any such thing as "right multiplication". An action tells you what $gs$ is for any $g\in G$ and $s\in S$, i.e. it is a function $f:G\times S\to S$ (satisfying a few axioms). – Zev Chonoles Apr 28 '12 at 2:35
up vote 6 down vote accepted

You're idea is right, but all your actions should be on the same side. Here's the correct version of the argument you were trying to construct:

If $gx=x^\prime$, then $x=ex=(g^{-1}g)x=g^{-1}(gx)=g^{-1}x^\prime$

share|cite|improve this answer
Makes sense. I should really stop trying to do math when I'm tired! Thanks. – Alex Petzke Apr 28 '12 at 3:08

Your Answer


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.