A group $G$ is called cyclic if there exists a $g \in G$ such that
$G=\{g^k : k\in \mathbb{Z}\}$
So $k$ does not have to be positive,
When it is said that the generators of $\mathbb{Z}$ are $\pm1$ does it mean either $\mathbb{Z}$ $=\langle -1 \rangle$ or $\mathbb{Z}$ $=\langle 1\rangle$?
I was always confused because I thought that $k$ had to be positive just like for the cyclic group $C_n$ and so thought it was meant $-1$ generated the negative side and $1$ generated the positive side or that it meant that any member of $\mathbb{Z}$ could be attained by $p(-1)+q(+1)$ i.e the set $\{-1,+1\}$ is a generating set which for some reason I thought was an ok exception to the fact that a cyclic group is generated by only one element.
So to confirm- it's true that: $\{-1,+1\}$ is a generating set for $\mathbb{Z}$
but this isnt what makes it a cyclic group because stronger still-
$\{-1\}$ is a generating set which contains one element so is cyclic and also $\{1\}$.
Correct?
Best Answer
Yes, $\{-1, 1\}$ is a generating set for $\mathbb{Z}$ so are $\{-1, 1, 754\}$, $\{-23, 2\}$, and plenty of other things.
But the reason that $\mathbb{Z}$ is considered as cyclic is that it has a generating set of cardinality $1$, for example $\{1\}$ or also $\{-1\}$ as you said.
The reason why we want $k$ to be an integer there is that one wants the structure generated to be a subgroup and using the positive integers we would not get a group.
Thus, it is not really the case that in the finite case the $k$ there has to be positive. The definition rather also will say it is an integer, it just so happens that it yields the same in the finite case.
If one lets $k$ to be a positive integer then one gets the sub-semi-group generated by $g$. [One could debate if one prefer non-negative rather than positive to have a subgroup with identity.]