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I have to prove that the fundamental group of the union of three open convex subsets of $\mathbb{R}^n$ is trivial or $\mathbb{Z}$. I can show that it has only one generator, but I can't prove that if the generator is not null-homotopic it is not nilpotent.

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It looks like you posted the question while still typing it... Where's the rest of the last sentence? – Henry T. Horton Jan 29 at 1:47
There it is. Sorry, I didn´t notice. I´ve edited it. – Frank Jan 29 at 2:05
You can pick a point in each of your given open sets and argue that any path is homotopic to a path which, whenever it intersects one of these sets, includes the distinguished point in that set. Now you've essentially reduced the problem to looking at the fundamental group of a graph on three vertices. If you've already proven that the group is cyclic, this should get you the rest of the way. – Brett Frankel Jan 29 at 2:33
That's the problem. I've proved that the fundamental group of a graph on three vertices is trivial or $\mathbb{Z}$. Now if I suppose it is not trivial (the graph is a triangle homeomorphic to a circle), I only know that the fundamental group of the original space has one generator. I have to check that the homotopy class of the generator in the original space is not nilpotent. How can I check that? – Frank Jan 29 at 2:45

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