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 $G$
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.
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.)
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.