Uniquely divisible non-abelian groups

Posted: March 9, 2010 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

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

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s