6
$\begingroup$

There is a proposition about the number of elements not lying in any conjugate of a subgroup when the group is finite. It states:

If $G$ is a finite group and $H$ is a proper subgroup, then the number of elements not in any conjugate of $H$ is at least $|H|$.

(See, for example, Isaac's Finite Group Theory Page 7.

However, I wonder whether the statement still holds when $G$ is an infinite group. I suppose it's not so hard to prove the statement for the infinite case, but I failed to find a way. Though it's clear that the statement holds when $G$ has an element with infinite order, I couldn’t find a way if I already get $G$ has a finite subgroup contains $H$ and the existence of a such subgroup is something I can't work out as the bounded Burnside problem shows there exists finitely generated infinite group with every element having a bounded order.

In fact, if there's any estimation about this number of elements not in conjugate of a subgroup, it'll be helpful!

Hope for an answer, thanks in advance!

$\endgroup$
2
  • $\begingroup$ A proof for the finite case can also be found in Proof for a subgroup $H$ of the finite subgroup $G$ there are at least $|H|$ elements that are not in conjugates of $H$. $\endgroup$
    – FredH
    Commented Mar 16, 2019 at 20:54
  • 2
    $\begingroup$ You are incorrect in claiming that the statement holds when $G$ has an element of infinite order; as noted in the answer, you can start with a torsionfree infinite group (lots of elements of infinite order), and end up with a torsionfree group containing your original group, in which any two non-identity elements are conjugate. In particular, all elements lie in a conjugate of any nontrivial subgroup $\langle x\rangle$. $\endgroup$ Commented Mar 16, 2019 at 22:17

2 Answers 2

7
$\begingroup$

For an infinite group, it is in fact possible that every element is contained in some conjugate of a proper subgroup containing just two elements. That is, it is possible for an infinite group to consist of just two conjugacy classes: the identity and everything else.

The construction of such a group is not elementary. A very nice outline of the process has been given by in Arturo Magidin's answer to Infinite group with only two conjugacy classes.

$\endgroup$
1
  • $\begingroup$ Note that this example is countable. Then by a pullback argument, one can get an example of $G$ and $H$ in which $G$ is free of countable rank. Since Osin produced a (considerably harder) example for which $G$ is infinite finitely generated (with 2 conjugacy classes), one even has an example of $(G,H)$ with $G$ free of finite rank. $\endgroup$
    – YCor
    Commented Mar 17, 2019 at 0:36
6
$\begingroup$

Here's an explicit example:

let $G$ be the group of increasing self-homeomorphisms of $\mathbf{R}$ with bounded support (i.e., identity outside a compact subset), and $H$ the subgroup of those increasing self-homeomorphisms with support in $[-1,1]$.

Indeed, if $g\in G$, there exists $u\in G$ such that $u(\mathrm{Supp}(g))\subset [-1,1]$, and hence $ugu^{-1}\in H$.

Also, if we restrict to those elements in $G$ (and in $H$) that are piecewise affine, for which slopes are integral powers of $2$ and breakpoints are dyadic numbers, we get a countable example for $G$.

$\endgroup$
2
  • $\begingroup$ Thanks!By the way, may I ask how to construct a counterexample like that? (I know that question is a little bit too broad, but I'm just too curiuos about how to get such a counterexample, for instance, the specific one presented above.) $\endgroup$ Commented Mar 17, 2019 at 15:47
  • $\begingroup$ I don't know, I don't have a particular recipe...! $\endgroup$
    – YCor
    Commented Mar 18, 2019 at 19:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .