Does there exist a non-abelian group with roots?

Question

I just learned the definition of a division group:

An abelian group $G$ is called divisible if for every $x\in G$ and every $k\in\mathbb{Z}^+$, there exists $y\in G$ such that $y^k=x$ (or $ky=x$, in additive notation).

So I was wondering if the property of "existence of roots" implies that a group is abelian. I believe this is not the case, but I couldn't find a counterexample. I had some ideas of using free groups and/or semidirect products (involving $\mathbb{Q}$, which I believe is the "simplest" division group), but I didn't get anywhere.

Answer 1

Consider the complex affine group $\mathrm{Aff}(\mathbb{C})$, which we can regard as the group of affine maps $$\mathbb{C} \to \mathbb{C},~~ f(z) = a z + b,~~ a \neq 0$$ under composition. Induction gives that

$$f^k(z) = a^k z + (a^{k - 1} + \cdots + a^2 + a) b.$$

So, for any $g \in \mathrm{Aff}(\mathbb{C})$, say, $g(z) = cz + d$, we can find $f$ such that $f^k = g$, that is, we can find $a,b$ such that $a^k z + (a^{k - 1} + \cdots + a^2 + a) b = c z + d$: There are $k$ distinct solutions $a$ to $a^k = c$; since $a^{k - 1} + \cdots + a^2 + a = 0$ for at most $k - 1$ values of $a$, there is some solution $a$ for which $a^{k - 1} + \cdots + a^2 + a \neq 0$. For such solutions $a$, we can solve $(a^{k - 1} + \cdots + a^2 + a)b = d$.

In fact, this should work just as well for the affine group over any algebraically closed field, perhaps requiring zero characteristic.

Remark 1 There's actually a simpler related example: Consider the group $\mathrm{Aff}_+(\mathbb{R}) = \mathbb{R} \rtimes \mathbb{R}_+$ of invertible, orientation-preserving affine transformations of $\mathbb{R}$, namely the maps

$$\mathbb{R} \to \mathbb{R},~~f(x) = a x + b,~~a > 0$$

under composition. In fact, using the above formula, which applies formally here, we can show that each $f \in \textrm{Aff}_+(\mathbb{R})$ has precisely one $k$th root for all $k \in \mathbb{Z}$.

Remark 2 In fact, both of these examples are subordinate to Micah's example, as we can show that the exponential maps for these groups are both onto:

If we embed $\textrm{Aff}(\mathbb{C})$ into $GL(2, \mathbb{C})$ in the usual way, namely via $$(z \mapsto a z + b) \mapsto \left(\begin{array}{c}1&0\\b&a\end{array}\right),$$ then we can identify the Lie algebra $\mathfrak{aff}(\mathbb{C})$ of $\textrm{Aff}(\mathbb{C})$ with $$\left\{\left(\begin{array}{c}0&0\\ \beta&\alpha\end{array}\right) : \alpha, \beta \in \mathbb{C} \right\} \subset \mathfrak{gl}(2, \mathbb{C}).$$ Computing gives that $$\exp \left(\begin{array}{c}0&0\\ \beta&\alpha\end{array}\right) = \left(\begin{array}{c}1&0\\ \frac{\exp \alpha - 1}{\alpha}\beta&\exp \alpha\end{array}\right) \leftrightarrow \left(z \mapsto (\exp \alpha) z + \left(\frac{\exp \alpha - 1}{\alpha}\right)\beta\right),$$ and it's not hard to show that this is surjective. The same argument works for $\textrm{Aff}_+(\mathbb{R})$.

Answer 2

The multiplicative group of (nonzero) quaternions is non-abelian. However, any single element is contained in a subgroup isomorphic to the multiplicative group of complex numbers, and thus has $n$th roots for all $n$.

In general, any Lie group whose exponential map is surjective (in particular, any compact connected Lie group) will be divisible in this sense: if $g \in G$ with $g=\exp(v)$ for some $v \in \mathfrak{g}$, then $\exp\left(\frac{1}{n}v\right)$ is an $n$th root of $g$.

Answer 3

You can also consider the set of the isometries of $\Bbb R^2$ that also preserve the standard orientation. Here noncommutatity is witnessed by translation to a certain point and rotation by a certain angle.

Arriving at the $n$th root of a pure rotation or translation is fairly easy, getting an $n$th root of a translation composed with a rotation is fairly tricky. You can think of taking a piecewise linear approximation to a circular arc starting at the origin going to the final point you're rotating around.