# Does there exist a non-abelian group with roots?

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.

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 $\textrm{Aff}_+(\mathbb{R})$.

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 $g$.

• The argument for the group of nonzero quaternions is especially nice; one can formulate the corresponding problem for power-associative magmas, and your (first) argument shows equally well that the magma (in fact loop) of nonzero octonions is divisible in this sense. – Travis Sep 7 '14 at 7:55
• It is also interesting to note that your first example ($\mathbb{H} - \{0\}$) and Bryan's (positive-determinant matrices in $M(2, \mathbb{R})$) can be derived in similar ways: The Cayley-Dickson construction identifies the quaternions with $\mathbb{C} \times \mathbb{C}$ endowed with the multiplication rule $(a, b)(c, d) := (ac - \bar{d}b, da + b\bar{c})$. If one replaces the $-$ with $+$, the resulting algebra is isomorphic to $M(2, \mathbb{R})$. Then, Bryan's example is exactly the connected component of the identity in the group $GL(2, \mathbb{R})$ of invertible elements. – Travis Sep 7 '14 at 8:03
• Alas, Bryan's original example of $GL_+(2, \mathbb{R}$ turns out not to work, and he has since replaced it with the working example $SO(2, \mathbb{R})$, so unfortunately this connection does not exist as such. – Travis Sep 8 '14 at 3:11
• Theorem A in the reference jlms.oxfordjournals.org/content/s2-27/3/427.full.pdf+html given in an answer to math.stackexchange.com/questions/924557/… shows that for connected Lie groups $G$, surjectivity of $\exp: \mathfrak{g} \to G$ is not only sufficient but also necessary for divisibility of $G$. – Travis Sep 10 '14 at 3:25

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.

• This group doesn't appear to be divisible. For example, suppose $A := \left(\begin{array}{c}a&b\\c&d\end{array}\right)$ is a square root of $C := \left(\begin{array}{c}-1&0\\0&-2\end{array}\right)$. The off-diagonal entries of $A^2$ are $b(a + d)$ and $c(a + d)$, so either $b = c = 0$ or $a + d = 0$. In the former case, $A^2 = \left(\begin{array}{c}a^2&0\\0&d^2\end{array}\right)$, but there is no real $a$ such that $a^2 = -1$. The latter case implies that the diagonal entries $a^2 + bc$ and $d^2 + bc$ coincide, which again is a contradiction, and hence $A$ has no square root in the set. – Travis Sep 7 '14 at 14:47
• @Travis thank you for the correction. I had isometries in mind but wrote the wrong thing. – Robert Wolfe Sep 7 '14 at 19:33
• You're welcome. I was fooled myself and some nagging thought led to me looking for a counterexample in an old course text. In fact, $SL(2, \mathbb{R})$ is isomorphic to to $Spin_+(1, 2)$, which acts by isometries on Lorentzian $3$-space $\mathbb{R}^{1, 2}$, so we can see that the definite signature is in a way essential in your example. (That said I don't know whether $SO_+(1, 2) = PSL(2, \mathbb{R}) = SL(2, \mathbb{R}) / \{\pm I\}$ is divisible---here, the counterexample $C$ projects to $[C]$, which does have a square root, namely $[\textrm{diag}(1, \sqrt{2})]$.) – Travis Sep 8 '14 at 3:07