Supposing that $\Gamma$ is an infinite, discrete group and that $\beta\Gamma$ is the Stone-Cech compactification of $\Gamma$, the group structure of $\Gamma$ can be extended to a semigroup structure on $\beta\Gamma$ by means of its universal property, for which the right multiplication maps over $\beta\Gamma$ are all continuous. It is well-known that any minimal left ideal (i.e., subsets of the form $(\beta\Gamma)x$ for some $x\in\beta\Gamma$) contains an idempotent.

However, the following question eludes me:

Does $\beta\Gamma\setminus\Gamma$ contain a non-idempotent element?

It seems to be so, as this question indicates, but is there any easy way, say, to just pick one out? I hoped originally that one could just take an idempotent $e\in\beta\Gamma\setminus\Gamma$ and consider $te$ for $t\in\Gamma$, but this also seems to be an idempotent, so maybe the answer is more convoluted than I once thought...

  • $\begingroup$ Your idea of picking $te$ is a good idea. $\endgroup$ – André Henriques Dec 24 '14 at 9:54

Let $\Gamma=\mathbb Z$, and consider the sequence of all odd numbers, viewed as a filter. Pick an ultrafilter containing this filter. Then this ultrafilter is not idempotent.

The reason is simple: odd + odd = even.

  • $\begingroup$ I agree completely. :) But is this in any way possible to generalize to any infinite group? We would have to consider some net indexed by $\Gamma$, right? $\endgroup$ – Bryder Dec 24 '14 at 12:01
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    $\begingroup$ You just need to find some infinite subset $S\subset \Gamma$ (like the odd numbers) such that $SS$ is disjoint from $S$. That shouldn't bee too hard. $\endgroup$ – André Henriques Dec 24 '14 at 12:20
  • $\begingroup$ It is easier to find an infinite subset $S\subseteq\Gamma$ such that $\{st|s,t\in S,s\neq t\}\cap S=\emptyset$ rather than satisfying the slightly stronger condition $SS\cap S=\emptyset$ and one can still construct a non-idempotent non-principal ultrafilter in this case on any infinite group. $\endgroup$ – Joseph Van Name Dec 24 '14 at 14:51
  • $\begingroup$ It turns out that any infinite group $G$ has an infinite subset $A\subseteq G$ such that $a^{2}\neq b$ whenever $a,b\in A$. If $A\subseteq G$ is an infinite set and $S$ is a maximal subset of $A$ such that $\{st|s,t\in S,s\neq t\}\cap S=\emptyset$, then $S$ is infinite. In particular, if $G$ is a group and $A$ is an infinite subset with $a^{2}\neq b$ for $a,b\in A$ and $S$ is a maximal subset of $A$ such that $SS\cap S=\emptyset$, then $S$ is an infinite set. $\endgroup$ – Joseph Van Name Dec 24 '14 at 15:33
  • $\begingroup$ Thanks for the ideas, both of you! May I just ask whether Zorn's lemma is required to show the existence of such a subset (just to know if I'm on the right track)? $\endgroup$ – Bryder Dec 25 '14 at 10:58

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