6
$\begingroup$

I would like to find a way to pick a set of generators in a group $G$ so that one can always find an element of $G$ of arbitrary length. I'm not sure whether or not this is always possible, and if not, what conditions on $G$ make it possible?

$\endgroup$
  • 4
    $\begingroup$ Are you talking about finitely generated groups? If so, and your group is infinite, then yes you can, since there are only finitely many elements less than or equal to some length, when you consider a finite generating set. $\endgroup$ – Paul Plummer Jul 21 '15 at 17:07
  • 2
    $\begingroup$ No, I am concerned with G uncountable. Sorry, I should have stated that. $\endgroup$ – Sam Eastridge Jul 21 '15 at 19:46
  • 1
    $\begingroup$ Do you know if it is true for countable groups, or are you just interested in the uncountable case? If you do know for countable groups, do you have a short proof or reference? $\endgroup$ – Paul Plummer Jul 21 '15 at 21:03
  • $\begingroup$ No I do not have any case other than finitely-generated, and although the question is particularly useful to me if G is uncountable, the countable case would certainly be interesting. I would even be interested if this was true just for non-locally finite groups, i.e. if we have an infinite, finitely-generated subgroup. Is there a way to pick generators of G/H without losing words of arbitrary length in H? $\endgroup$ – Sam Eastridge Jul 21 '15 at 23:37
9
$\begingroup$

In general there are uncountable groups where for all generating sets, every element has length bounded by some $n$.

  • In Generating infinite symmetric groups by George M. Bergman, it is shown that $S_\Omega$, where $\Omega$ is an infinite set then if you look at the Cayley graph generated by a generating set $U$, that Cayley graph is bounded. He also shows that for $S_\Omega$ there is not a uniform bound for all generating sets.
  • A paper by Yves de Cornulier, Strongly bounded groups and infinite powers of finite groups, shows that if $G$ is a finite perfect group and $I$ is a set, then $G^I$ has a bounded Cayley graph for every generating set.
  • Shelah constructed such a group in On a problem of Kurosh, Jónsson groups, and applications too. In fact there is a uniform bound that can be choosen (unlike $S_\Omega$).

It seems like it may be open if there are countable groups with that property. (This is question 8 in Bergman's paper)

Prop 2.4 in Yves's paper says that these properties for groups are equivalent:

  1. (not strongly bounded) There is a metric space it acts on with an unbounded orbit
  2. (not cayley bounded or has cofinality=$\omega$) There is an unbounded Cayley graph, or $G$ is the countable union of a strictly increasing chain of subgroups

That condition could be useful, if you know that your group has cofinality$\neq \omega$, and it acts on a metric space with unbounded orbits (so there must be an unbounded Cayley graph). There is another condition you could use, Prop 2.7, from the same paper.

Maybe something else that could be useful, say your group can be a topological group which is not compact, but it is compactly generated, then you have that the compact set gives an unbounded Cayley graph. Infinite finitely generated groups can be given the discrete topology and then the finite generating set will be the compact generating set, $(\mathbb{R}^n,+)$, with standard topology is generated by $[0,1]^n$.

$\endgroup$
  • $\begingroup$ I will continue to look around for general conditions, I would be surprised if there were "really general" conditions, so you (@Sam) might benefit from being more specific about the sort of groups you are looking at, maybe ask a different question about that group. I suspect something like "being close to free" would be an area to look into (maybe freely acting on some spaces). Prop 2.4 in Yves's paper above could be useful to you, it gives some equivalences between some properties involving Cayley bounded, so maybe the negation of those statements could be useful to you. $\endgroup$ – Paul Plummer Jul 22 '15 at 2:46

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.