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

Theorem: Actions of a finite group G on finite sets $X$ and $Y$ are equivalent if and only if $|X^H|=|Y^H|$ for each subgroup $H$ of $G$

  1. I've proved the finite case inductively but I'm wondering if it generalizes to the infinite case. In my proof, it doesn't hold for the infinite case.

  2. Also, my proof didn't utilize the Cantor–Bernstein–Schroeder Theorem, although since $f:X\rightarrow Y$ is a stable bijection I think it can be used to find a simpler proof (at least that's what I've been told.)

  3. Finally, if anyone can help point to a proof of this statement (or a more general statement) in a book, that would be great. My instructor told me that a proof exists in Isaac's book Finite Group Theory, but I couldn't find it.

Def: $X^H$ is the set of $x\in X$ fixed by each $h\in H$

Def: Actions of G on sets $X$ and $Y$ are equivalent if there is a stable bijection $f:X\rightarrow Y$

Thm: The notion of equivalence is an equivalence relation on the set of all actions of a given group G.

share|cite|improve this question
How do you define "equivalent (group) actions"? – DonAntonio Jan 31 '13 at 3:00
As an equivalence relation – zzzzzzzzzzz Jan 31 '13 at 6:05
Uh? That doesn't seem to make much sense when talking about equivalent actions, does it? Equivalence relation is a subset of a cartesian product of some set with itself which fulfills certian conditions. How is that appliable here? – DonAntonio Jan 31 '13 at 6:22
"Equivalent group actions" is standard terminology, and means that there is a bijection between $X$ and $Y$ such that each group elements acts on $X$ in the same way as it acts on $Y$, using the bijection. You have to distinguish this from isomorphic group actions, which means you can apply an automorphism of $G$ to get from one action to the other. But I am not sure what $X^H$ means. I guess it must be the set of orbits of the action. – Derek Holt Jan 31 '13 at 8:58
Well, that already makes sense, though I'm not sure about "standard": it doesn't appear in several of the well-known book in the subject and after googling for a while I found a definition only in one of Keith Conrad's papers. About the notation $\,X^H\,$ : for me, that means the set of fixed points in $\,X\,$ by the subgroup $\,H\,$ , but I can't be sure the OP meant this. – DonAntonio Jan 31 '13 at 10:21
up vote 2 down vote accepted

The proof is found in I. Martin Isaac's Character Theory of Finite Groups on page 230 under the unattributed Lemma 13.23.

share|cite|improve this answer

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.