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Suppose we have a finite group $G$ acting on a finite set $X=\{ x_1, ..., x_n \}$. Then we can take the free Abelian group generated by elements of $X$ (which is of course isomorphic to $\mathbb{Z}^n$) and we get an induced action of $G$.
My question is if it is possible to have two non-isomorphic $G$ actions on $X$ that induce isomorphic actions on $\mathbb{Z}^n$.
Thanks!
There are fairly general examples due to Conlon in the paper:
Conlon, S.B., Monomial representations under integral similarity, J. Algebra 13, 496-508 (1969). ZBL0185.06702.
There is even a transitive example, due to Scott:
Scott, Leonard L., Integral equivalence of permutation representations, Sehgal, Surinder (ed.) et al., Group theory. Proceedings of the 21st biennial Ohio State-Denison mathematical conference, Granville, OH (USA), 14-16 May, 1992. Singapore: World Scientific. 262-274 (1993). ZBL0828.20004.
In this example, $G$ is $\text{PSL}(2,29)$ and the permutation actions are on the cosets of two non-conjugate subgroups both isomorphic to the alternating group $A_5$.
It might be a partial answer.
Claim: If $G$ has a nonisomorphic two actions having nonequal number of orbits then the induced actions on $\mathbb Z^n$ are also nonisomorphic.
Let $W=\{v\in\mathbb Z^n\mid gv=v \}$. Clearly $W$ is an $G$-submodule of $V$.
Subclaim: $dim(W)$ is equal to the number of the orbits of $G$ on $X$. (We can talk about dimenssion of $W$ as $\mathbb Z$ is an PID. I do not write the proof of this claim as I believe it will be routine as in the case $\mathbb C^n$ )
Since dimension of $W$ is invariant, two induced actions on $\mathbb Z^n$ are also nonisomorphic.
Thus, if such example exist, the number of the orbits must be same.
