# The fundamental group of an open set in $\mathbb{R} ^n$ does not have nilpotent elements.

I am studying a little of basic algebraic topology and I thought that this statement could be true. If you have an open connected set $U \subset \mathbb{R}^m$ and a loop $\gamma$ that is not null-homotopic (homotopic to a constant) then $\gamma \ast \gamma ... \ast \gamma$ (n-times) is never null-homotopic for $n \in \mathbb{N}$.

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See my answer here: math.stackexchange.com/questions/287009/… – Henry T. Horton Jan 29 '13 at 0:45
Is this true for $m \le 3$? – Tunococ Jan 29 '13 at 1:08
Henry, thanks a lot! :) – Frank Jan 29 '13 at 1:09
Usually when talking about a group one would say that it has no elements of finite order. Or in a more fancy way that the group is torsion-free. – JSchlather Jan 29 '13 at 1:37