9
$\begingroup$

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!

$\endgroup$
5
$\begingroup$

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 $A_5$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.