Proving that a cyclic group is generated by a single element.

Question

I am currently reading "The Theory of Finite Groups" by Kurzweil and Stellmacher.

I am already stuck on page 4.

On page 3, a cyclic group is defined as:

The group G is cyclic if every element of G is a power of a fixed element g.

Then on page 4 a proof is given for:

1.1.2 Let $G = \langle g \rangle$ be a cyclic group of order $n$. Then $G = \{1, g, \cdots, g^{n-1}\}$

I don't understand the given proof. But I think this is partially due to the fact that I do not know why the author is proving 1.1.2 in the first place. For me, the definition and 1.1.2 look pretty much the same.

Could someone please explain to in which way they differ and why someone has to prove 1.1.2 given the definition.

Answer 1

By definition, if $G$ is cyclic with generator $g$, then every element of $G$ is a power of $g$. What 1.1.2 is saying is that if $G$ has order $n$, then more specifically every element of $G$ is equal to $g^m$ for some $m$ such that $0\leq m<n$. This is stronger than the definition, because of the restriction that $0\leq m<n$.

Answer 2

It does seem rather straightforward, I guess that the thing that we can extract is that the order of the group coincides with the order of the generator. I think you can keep reading without too much concern for this.