Determining whether this is a group action

Question

I'm having trouble with an exercise we were given. I have to determine for which values $a,b\in\mathbb{R}$ $$n\cdot t=\phi_n(t)=2^nt+a^n+b$$ defines a group action of the group $(\mathbb{Z},+)$ on $(-1,\infty)$.
I am not really sure what exactly I am supposed to show. (I've read the definition of group actions over and over, but it just doesn't seem to be clear in my head. I was unable to participate in the course of Group Theory, so I have not had a proper introduction to all of that. (This course is introducing us to topology.)) I just don't know where to start, so I'm hoping someone can clarify what exactly it is that I need to show and perhaps give me a little push in the right direction.

I believe I need to determine for which $a,b$ the function $\phi_n:(-1,\infty)\to(-1,\infty),t\mapsto 2^nt+a^n+b$ is a homeomorphism for all $n\in\mathbb{Z}$.

This is the book we're using: http://www.staff.science.uu.nl/~crain101/topologie2013/aaa-main-2013-2014.pdf
Group actions are introduced on page 52.

Thanks a lot in advance!

P.S.: I hope I'm using this forum correctly. I'm new here, so please correct me if I did something wrong.

Answer 1

For $\phi$ to be an action, you have to answer the following questions:

(1) For which $a,b$ is $\phi_n\colon (-1,\infty)\to (-1,\infty)$?

As $\phi_n$ is increasing and continuous for all $n$, this happens exactly if $\lim_{t\to -1} \phi_n(t) \in [-1,\infty)$, we have $$ \lim\limits_{x\to -1}\phi_n(t) = -2^n+a^n +b $$ So we must have $a^n-2^n \ge b-1$ for all $n$, that is $a \ge 2$.

(2) Do we have $\phi_n \circ \phi_m = \phi_{n+m}$?

Just compute. For $t > -1$, we have \begin{align*} \phi_n\phi_m(t) &= 2^n\phi_m(t)+a^n + b\\ &= 2^{n+m}t + 2^na^m + 2^nb + a^n + b\\ &= 2^{n+m}t + a^{n+m} + b + 2^na^m + 2^nb + a^n - a^{n+m}\\ &= \phi_{n+m}(t) + 2^n(a^m + b) + a^{n}(1 - a^m) \end{align*} So $\phi$ is an action iff $2^n(a^m + b) = a^{n}(a^m-1)$ holds for all $n,m$. For $n=m=0$ we get $1+b = 0$, that is $b=-1$. That is $$ 2^n(a^m - 1) = a^n(a^m - 1)$$ As $a \ge 2$ by (1), we have $a^m -1 \ne 0$ for $m=1$, hence $a=2$.

Hence $\phi$ is only an action for $a=2$, $b=-1$, giving $$ \phi_n(t) = 2^nt + 2^n - 1 = 2^n(t+1) - 1. $$