Why are $-1$ and $1$ generators for the Set of integers under addition? I'm reading my textbook and I'm confused why $-1$ and $1$ are generators for the group of integers under addition.
For example for 1:
we have $1, 1+1=2, 1+1+1=3, 1+1+1+1=4,$ etc.
So shouldn't $1$ be a generator for only the group of positive integers under addition?
For $-1$: we have
$-1, -1+-1=-2, -1+-1+-1=-3$.  So shouldn't $-1$ be a generator for only the group of negative integers under addition?
Or is it that both $1$ and $-1$ are generators for the set of integers under addition because in the definition of cyclic subgroup $\{a^n; n\in \Bbb Z \}$, $n$ can take on negative powers? Or am I just confused? Thanks.

In group theory, the word "generates" has the following meaning: a set $S$ generates a group $G$ if every element of $G$ can be written as a string over $S\cup S^{-1}$ (here, if $S=\{a, b, \ldots\}$ then $S^{-1}=\{a^{-1}, b^{-1}, \ldots\}$). Equivalently, $G$ is the minimal subgroup of $G$ containing all the elements of $S$. In your example, $S=\{1\}$ so $S\cup S^{-1}=\{1, -1\}$ (there is a notation issue here...but I am sure that you can cope).
If $S=S^{-1}$ then we say that $S$ is symmetrised.
In finite groups we do not need to consider the set $S\cup S^{-1}$ as if $a\in S^{-1}$ then there exists $b\in S$ and $n\in\mathbb{N}$ such that $b^n=a$ (indeed, $b=a^{-1}$).
