Show a set of maps of the complex plane $x\alpha = ax + b$ is of finite order for given variables $a, b$ where $a \ne 1$ and $a$ is root of unity

Question

I'm trying some exercises from the Introduction to Group Theory book by Ledermann and Weir and I'm already stuck on an exercise from Chapter 1:

Let $\alpha$ be the map of the complex plane defined by $x\alpha = ax + b$, where $a$ and $b$ are given complex numbers and $a \ne 1$. Obtain a formula for $\alpha^n$ where $n$ is a positive integer, and show that $\alpha$ is of finite order if and only if $a$ is a root of unity.

My complex numbers knowledge is a bit basic so I'm unsure how to start here. I guess if a complex number is not a root of unity, repeatedly applying this $\alpha$ to $x$ would carry on yielding more and more different numbers? But how could I put that into a formula that demonstrates it algebraically?

Answer 1

The group here is going to be a set of maps acting on the complex plane. The maps in question are of the form $x\mapsto ax+b$. To form a group, the maps all have to have inverses, so we can't allow maps with $a=0$, but otherwise $a$ and $b$ can be any complex numbers.

Important things to check are that the identity map is $x\mapsto x=1x+0$ and that two maps, $x\mapsto a_1x+b_1$, $x\mapsto a_2x+b_2$ are the same map if and only if $a_1=a_2$ and $b_1=b_2$.

To answer your question, you need to determine under what circumstances a group element $\alpha$ given by $x\alpha=ax+b$ has the property that $\alpha^n$ is the identity map, $x\mapsto x=1x+0$. If $a=1$ and $b\ne0$, you should be able to see that this will never happen. So we are left with the case $a\notin\{0,1\}$.

Try iterating the map $\alpha$ a few times, $$ \begin{aligned} x\alpha&=ax+b\\ x\alpha^2&=a^2x+ab+b\\ x\alpha^3&=a^3+a^2b+ab+b, \end{aligned} $$ and you should be able to find the general formula for $x\alpha^n$. Since we are assuming $a\ne1$, you can simplify this using the formula for geometric sums, $1+a+x^2+\ldots+a^{n-1}=\frac{a^n-1}{a-1}$. You should then be able to determine the circumstances under which $x\mapsto x\alpha^n$ is the same map as $x\mapsto x$.

Answer 2

It is not hard to prove by induction that$$\alpha^n(x)=a^nx+a^{n-1}b+a^{n-2}b+\cdots+b.$$If $a^n=1$ and $a\ne1$, then, since$$0=a^n-1=\overbrace{(a-1)}^{\phantom0\ne0}\left(a^{n-1}+a^{n-2}+\cdots+1\right),$$we have\begin{align}a^nx+a^{n-1}b+a^{n-2}b+\cdots+b&=x+\left(a^{n-1}+a^{n-2}+\cdots+1\right)b\\&=x.\end{align}

Answer 3

As Jose Carlos Santos mentioned $\alpha^n(x) $ will be of that form replacing $b^i$'s by $b$ .

For the converse part consider the equality, $\alpha ^n(x) =x\ \forall x \in \mathbb C$ so, it's an identity hence $a^n=1$.