$G=\{f_n(x):n\in \mathbb{Z}\}$ is cyclic

Question

Define $f_n(x)=x+n \;\;\forall n \in \mathbb{Z}$. Let $G=\{f_n:n\in \mathbb{Z}\}$.

I proved that $G$ is a subgroup of $S_\mathbb{R}$ ($f_n$ is a permutation of $\mathbb{R}$), and now I am trying to show that $G$ is cyclic.

I figured that I could do this by finding a generator of $G$. Then I realized I don't understand too much about generators, so I want to see if this approach is correct:

$G=\langle x+m \rangle : m\in \mathbb{Z}$ because $\forall f_n\in G$ it is clear that $f_n(x+m)=f_{n+m}(x)\in G$.

Is this sufficient to prove $x+m$ is a generator? (Is it even a generator?)

If that isn't correct, I was thinking of showing that $x+1$ is a generator (the other integers could be unnecessary). Please steer me in the right direction here. Just hints please, no full solutions.

Answer 1

You need to define a composition law, not just a set, for this to be a group.

For this to be a group (and a subgroup of $S_{\mathbb{R}})$ you probably wanted to define composition as $f_n,f_m \mapsto f_n \circ f_m$.

Then $f_n, f_m \mapsto (x + m) + n = x + (m+n) = f_{m+n}$. This shows that $G \cong \mathbb{Z}^+$ so it must be cyclic.

Alternatively, you can show that $f_1 = x+1$ is a generator. Any $f_n$ is obtained by composing $f_1$ with itself $n$ times, and any $f_{-n}$ is obtained by composing the inverse of $f_1$ with itself $n$ times.

Answer 2

$x+m$ is not an element of $G$. Perhaps what you meant to say is that $f_m$ is a generator for any $m$.

But that is not true either. For example, how can you get $f_1$ by composing $f_2$ with itself?