GAPlus(1,R) is rationally powered

This article gives the statement, and possibly proof, of a particular group or type of group (namely, GAPlus(1,R) (?)) satisfying a particular group property (namely, Rationally powered group (?)).

Statement

The group $GA^{+}(1,\mathbb {R} )$, denoted GAPlus(1,R), defined explicitly as the group under composition of maps from $\mathbb {R}$ to $\mathbb {R}$ of the form:

$x\mapsto ax+b,\qquad a,b\in \mathbb {R} ,a>0$

is a rationally powered group. In other words, for any positive integer $n$ and any element $u\in GA^{+}(1,\mathbb {R} )$, there is a unique $v\in GA^{+}(1,\mathbb {R} )$ such that $v^{n}=u$.

Proof

Suppose $u$ is the map:

$x\mapsto a_{1}x+b_{1}$

We want to find all elements $v$ that are maps of the form $x\mapsto a_{2}x+b_{2}$ such that $v^{n}=u$. By composing $v$ with itself $n$ times, we get

$v^{n}=x\mapsto a_{2}^{n}x+b_{2}(1+a_{2}+\dots +a_{2}^{n-1})$

For this to equal $u$, we know that the coefficient of $x$ and the constant term should match up separately, so we get:

$a_{1}=a_{2}^{n}$

and:

$b_{1}=b_{2}(1+a_{2}+\dots +a_{2}^{n-1})$

Solving, we get that the unique solution is the element $v$ with:

$a_{2}=a_{1}^{1/n},\qquad b_{2}={\frac {b_{1}}{1+a_{1}^{1/n}+\dots +a_{1}^{(n-1)/n}}}$

In other words, the solution is:

$x\mapsto a_{1}^{1/n}x+{\frac {b_{1}}{1+a_{1}^{1/n}+\dots +a_{1}^{(n-1)/n}}}$