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](https://mathoverflow.net/questions/128756/how-much-do-idempotent-ultrafilters-generate-in-terms-of-semigroups) 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...



Non-idempotent ultrafilters in the Stone-Cech compactification
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.


