## Uniquely divisible non-abelian groups

Posted: March 9, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Problem. Give an example of a non-abelian group $G$ that satisfies the following condition: for every $n \in \mathbb{N}$ and every $g$ in $G,$ there exists a unique element $x \in G$ such that $x^n=g.$

Solution.  Consider $\mathbb{Q}$ and $\mathbb{R}$ as additive groups and define the map $\theta : \mathbb{Q} \longrightarrow \text{Aut}(\mathbb{R})$ by $\theta(q)(r)=2^qr,$ for all  $q \in \mathbb{Q}$ and $r \in \mathbb{R}.$ See that $\theta$ is a group homomorphism. Let $G=\mathbb{R} \rtimes_{\mathbb{\theta}} \mathbb{Q}.$ Then $G$ is non-abelian; in fact $Z(G)=\{(0,0)\}$. To prove that $G$ has the required property, see that for any $x=(r,q) \in G$ and integer $n \geq 1$ we have $x^n=((1+2^q+ \cdots + 2^{(n-1)q})r, nq).$ So if $g=(r',q') \in G$ and $n \geq 1$ are given, then the only $x \in G$ satisfying $x^n=g$ is $x=(r,q),$ where $\displaystyle q=\frac{q'}{n}$ and  $\displaystyle r=\frac{r'}{1+2^q + \cdots + 2^{(n-1)q}}. \ \Box$