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

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

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

Proof

Suppose $u$ is the map:

$x \mapsto a_1x + b_1$

We want to find all elements $v$ that are maps of the form $x \mapsto a_2x +b_2$ such that $v^n = u$. By composing $v$ with itself $n$ times, we get

$v^n = x \mapsto a_2^nx + b_2(1 + a_2 + \dots + a_2^{n-1})$

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:

$x \mapsto a_1^{1/n}x + \frac{b_1}{1 + a_1^{1/n} + \dots + a_1^{(n-1)/n}}$