MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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


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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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