# Number of free actions from group to set

I'm currently stuck on a problem where we need to determine the number of free actions from $Z_{10}$ to the set $\lbrace 1,2,....,100\rbrace$. As I understand it, an action being 'free' implies that every stabilizer is just the set $\lbrace \epsilon \rbrace$. By the Orbit-Stabilizer theorem, this implies that the size of every orbit must be 10. As such, there will be ten different orbits for this action.

However, there's an immense number of different actions that will produce ten different orbits in the numbers 1-100... which always makes me doubt the validity of the answer. Is my reasoning sound? If so, might it be that only 'natural' actions of $Z_{10}$ are to be taken into consideration, as in, the ones that send an element of the set to its successor? In this case, the answer would be 10, I think.

I hope someone is able to shed some light on this problem, thank you in advance!

-

Everything you say makes sense; the question really comes down to what we mean by "number of ... actions."

Basically, you are looking for embeddings of $\mathbf{Z}_{10}$ into $S_{100}$ in which no element other than the identity fixes any point. Any such action will be completely determined by the action of $1$; and the property of not fixing any point means that the image of $1$ must be a product of 10 disjoint cycles, each of length 10.

Up to isomorphism (i.e., up to an automorphism of $S_{100}$), there is only one such element; that is, any two embeddings of $\mathbf{Z}_{10}$ into $S_{100}$ are related by conjugation (renaming the elements of ${1,2,\ldots,100}$). So in fact, we would have just one and only one action, "essentially", namely the action will, up to relabeling the points, correspond to the action determined by $$1\longmapsto (1,2,\ldots,10)(11,12,\ldots,20)\cdots(91,92,\ldots,100).$$

On the other hand, we may not want to consider the actions "the same" unless they are completely identical. In that case, even as simple a change as mapping $1$ to $$1 \longmapsto (2,1,3,4,\ldots,10)(11,12,\ldots,20)\cdots(91,92,\ldots,100)$$ instead will give us a "new" action. In this situation, the number of distinct actions will be equal to the index of the centralizer of $$(1,2,\ldots,10)(11,12,\ldots,20)(91,92,\ldots,100)$$ in $S_{100}$. This centralizer has order $100$, so the total number of actions would be $100!/100 = 99!$. Oops. As Derek Holt points out, this particular computation is incorrect and silly (I was centralizing each $10$-cycle, and then compounding the error by miscomputing $10\times 10$ instead of $10^{10}$). The centralizer actually has larger order: you can first permute the $10$ cycles amongst themselves ($10!$ ways), and then permute each cycle in $10$ different ways to obtain the same permutation. So you get $10!(10^{10})$ elements in the centralizer, giving you a total of $$\frac{100!}{10!10^{10}}$$ actions.

Or you may want actions to be "the same" if they agree up to an automorphism of $\mathbf{Z}_{10}$, but otherwise be identical in $S_{100}$; in that case, you would want to divide by $|\mathrm{Aut}(\mathbf{Z}_{10})|$ as well.

Or maybe somewhere in between.

If you want to consider two actions as being "the same" if the orbits are the same (which is, I think, what you describe under "the one(s) that send an element of the set to its successor", i.e., the next smallest element in the orbit, or in the case of the largest element in the orbit, to the smallest), then you want to count the number of partitions of $\{1,2,\ldots,100\}$ into subsets of size $10$ each. You get $$\frac{1}{10!}\binom{100}{10}\binom{90}{10}\cdots\binom{10}{10}$$ possible partitions.

-
Arturo: I think you are significantly underestimating the size of the centralizer in $S_{100}$ of a product of 10 disjoint 10-cycles. The centralizer is the wreath product of a cyclic group of order 10 with $S_{10}$, so it has order $10^{10} \cdot 10!$. – Derek Holt Nov 3 '11 at 17:04
@Derek: Oh, dear. You are right. I was centralizing each cycle by itself, which is quite silly of me. Thank you! – Arturo Magidin Nov 3 '11 at 17:10