Existence of primitive element in a non abelian group Let $G$ be a group. an element $g \in G$ is called primitive if, whenever
$h \in G$ satisfies $h^k = g$ for some $k > 0$ it follows that $k = 1$ and $h = g$.
Are there any example of infinite non-abelian torsion-free groups with no primitive elements?
Is it true that any infinite torsion-free non abelian group has a primitive element?
 A: The way to prevent primitive elements is to pick a group where everything has a "root." Lie groups might be a nice place to try. If the exponential map is surjective, then everything lies in the image of a one-parameter subgroup, so it's easy to take roots. For instance, $n$th roots of a rotation in a rotation group can be calculated by simply dividing the angle of rotation by $n$. But rotation groups, for instance, are full of torsion elements.
One must go to at least the $2\times 2$ case to see nonabelian matrix groups. The full general linear group $\mathrm{GL}_2(\mathbb{R})$ (almost) equals central scalar matrices times $\mathrm{SL}_2(\mathbb{R})$, that latter of which has an Iwasawa decomposition $KAN$ where $K=\mathrm{SO}(2)$ is the rotation group (containing torsion elements no matter the dimension), $A$ is comprised of diagonal matrices with determinant $1$ (abelian in any dimension), and $N$ is the group of upper triangular matrices. 
While in the $2\times 2$ case the upper triangular group is abelian (it's just $\cong\mathbb{R}$), if we go to the $3\times 3$ case it's nonabelian and has no torsion elements. Over any field of characteristic zero, writing an element of this Heisenberg group as $I+X$ where $X$ is a nilpotent matrix, the binomial theorem can be used to show that taking it to powers scales the one-above-diagonal entries, which implies lack of torsion elements. It also has $n$th roots: to evaluate $\sqrt[n]{I+X}$ just plug it into the Newton-binomial series, which will terminate since $X$ is nilpotent.
(This is just a long-winded lead up to Derek's example in the comments.)
Another idea is to start with a free group $\langle a,b,c,\cdots\rangle$ and adjoin new radical expressions that look like $\sqrt[3]{a\sqrt{b^2c^3}}$ for instance. That should work.
