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Let $G$ and $H$ be permutation groups on the natural numbers such that the orbits of $G$ and $H$ are all finite. Suppose that for all $\pi \in Sym(\mathbb{N})$, there is some $N$ (depending on $\pi$) such that for all $n \ge N$, the ordered tuple $(\pi(1),\pi(2),\dots,\pi(n))$ has a larger orbit (by a fixed ratio) under $G$ than it has under $H$.

Can $G$ and $H$ be conjugate in $Sym(\mathbb{N})$?

Edit: Answer is 'yes' (see Jim Belk's comment below); indeed $G$ can be conjugate to proper subgroups of itself of finite index, which makes the size of tuple orbit property automatic.

But what if $G$ only has finitely many orbits of size $n$ for each $n \in \mathbb{N}$? This would at least ensure that $G$ cannot be conjugate to one of its own subgroups.

Edit 2: An example would need to have the following property:

There is a tuple $t$, such that for any tuple $u$ for which $G_u$ is contained in $G_t$, then the $G$-orbit of $u$ is larger than the $H$-orbit of $u$.

So for instance if we pick a tuple $u$ by saying 'choose a large number $K$, then choose from among the $K$-tuples with no repeats one with smallest possible $G$-orbit', then $G_u$ would not be contained in $G_t$ no matter how large $K$ is. I think this rules out examples where the tuple stabilisers of $G$ are totally ordered, for instance if $G$ is cyclic and all orbits have length a power of a fixed prime.

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2 Answers 2

up vote 6 down vote accepted

Here's a case where $G$ and $H$ can be conjugate. First some notation: given a sequence $\{k_n\}$ of positive integers, let $[k_1,k_2,\ldots]$ denote the permutation

$$(1,\ldots,k_1)(k_1+1,\ldots,k_1+k_2)(k_1+k_2+1,\ldots,k_1+k_2+k_3)\cdots$$

with cycles of size $k_1,k_2,k_3\ldots$. For example, $[1,1,1,1,\ldots]$ denotes the identity, $[2,2,2,2,\ldots]$ denotes $(1,2)(3,4)(5,6)(7,8)\cdots$, and $[2,3,2,3\ldots]$ denotes $(1,2)(3,4,5)(6,7)(8,9,10)\cdots$.

Let $$g = [1,2,\;\;1,2,4,\;\;1,2,4,8,\;\;\ldots],$$ let $$h = [1,1,1,\;\;1,1,1,2,2,\;\;1,1,1,2,2,4,4,\;\;\ldots],$$ and let $G$ and $H$ be the cyclic subgroups generated by these elements. Since $g$ and $h$ have the same cycle structure, they are conjuagte in $Sym(\mathbb{N})$, so $G$ and $H$ are conjugate subgroups. However, for sufficiently large $n$, the orbit of $(\pi(1),\pi(2),\ldots,\pi(n))$ under $G$ will be precisely twice the size of the orbit under $H$.

Of course, in this example $G$ and $H$ both have infinitely many orbits of size $2^k$ for every $k$, so this does not answer the more restrictive version of the question.

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In fact, it looks like $g$ is conjugate to $g^2$ in this case. I can see now there is a lot of potential for examples of this kind. As you say though, this trick appears to require infinitely many orbits of a given size. –  Colin Reid Sep 4 '10 at 8:40

Having thought about it some more, there ought to be an abundance of examples here. This is not quite an answer as I haven't thought of an explicit example yet, but hopefully I am getting close to understanding what I wanted.

Let $G$ be a group, with subgroups $H$ and $K$ of different finite indices, such that there is an isomorphism $\phi$ from $H$ to $K$. Form the corresponding HNN extension $L = G *_\phi$ and let $T$ be the Bass-Serre covering tree associated to this extension. Then $T$ is locally finite and $G$ is a point stabiliser of the action of $L$ on $T$, so we have a countable set with a $G$-action such that all $G$-orbits are finite. As for there being only finitely many orbits of given size, it's enough to choose a $G$ that has finitely many subgroups of given index, and make sure that the stabiliser in $G$ of a ray in $T$ (by which I mean a non-self-intersecting path with one endpoint) has infinite index. We also want the action to be faithful and for some finite subset of $T$ to have stabiliser contained in $G \cap G^t$, but neither of these should be too difficult to arrange. Then $G$ and $G^t$ (where $t$ is the new generator in the HNN extension) would have the property I described that $G$ looks 'bigger' than $G^t$ (or bigger than $G^{t^{-1}}$, depending on which of $H$ and $K$ has the larger index in $G$).

Note here that an ascending HNN extension doesn't work, because in this situation there would be a ray in $T$ fixed by $G$. Conversely though, I imagine if a finite index subgroup of $G$ did fix a ray, then $L$ would have to be 'close' to being ascending, which is far from the typical case for HNN extensions.

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