How is $0$ generated from $1$ or $-1$? The set of integers $\Bbb Z$ under ordinary addition is cyclic. Both $1$ and $-1$ are generators. But I am a bit confused how can $1$ generate $0$ and how $-1$ generates $0$? What is the order of $1$ and $-1$ on this group of integers?
 A: In $(\mathbb Z,+)$ there is no positive integer $m$ such that $m \times 1 =0$. We say that the order of $1$ (and of every other integer apart from $0$) is infinite. The order of $0$ is $1$.
A: The subgroup generated by a set of elements of a group is the smallest subgroup that contains all the elements. A group by definition always includes the identity element.
The order of an element is the size of the subgroup generated by it. Both $1$ and $-1$ generate all of $(\mathbb Z,+)$ which has infinitely many elements, thus the order of them is infinite (as there are infinitely elements in the group).
Note that for any element $n$ of $\mathbb Z$, the generated subgroup consists of all multiples of $n$, which for non-zero $n$ consists of infinitely many elements as well, thus every non-zero integer has infinite order.
On the other hand, one can easily check that $\{0\}$ is a subgroup all be itself, called the trivial subgroup, and therefore this is the group generated by $0$ (note that again, it consists of all multiples of $0$, since $0n=0$ for all $n\in\mathbb Z$). Since the trivial subgroup only contains one element (namely $0$), the order of $0$ is $1$.
Indeed, it is easy to see that for any group $G$, its identity element generates the trivial subgroup and therefore is of order $1$. Also, the group of any non-identity element $g$ contains at least two elements (namely the identity and  $g$ itself), and therefore is of order $\ge 2$.
As a side note, even if we ignore that $0$ is by definition in the subgroup of $(\mathbb Z,+)$ generated by $1$, we can easily get it through the supported operations: Since $1$ is in the set, and the set is closed under negation, $-1$ is also in the set. And since the set is also closed under addition, $1+(-1)=0$ is also in the set. However while this works for groups, the same idea fails for other algebraic structures like monoids where you also have an identity, but generally no inverse; an example of this would be $(\mathbb Z_{>0},\times)$.
Therefore the right way to think of it is that the neutral element is already there by definition, as this works for all algebraic structures.
A: gandalf61's answer is technically correct. But let me try to offer a more intuitive explanation.
You can think of the group $G =(\mathbb{Z}, +)$ in terms of functions which permutes integers, i.e, functions of the type $\mathbb{Z} \to \mathbb{Z}$. For example, think of $7_G$ as the function that adds $7$ to every integer, i.e, $7_G(x) = 7 + x$. And when you are composing two elements of this group, what you are really doing is function composition. For example, $7_G + 2_G = 9_G$ is justified since for any $x \in \mathbb{Z}$, $7_G(2_G(x)) = 7 + (2 + x) = 9 + x = 9_G(x)$. Similarly, negatives correspond to function inverses, and so on. (This perspective is called group actions)
Now, there is an elelement $0_G$, which represents the identity function of this type. This function fixes every element of $\mathbb{Z}$, in other words, it does nothing.
We could notice that if we had the function $1_G$ (and it's inverse), we could repeatedly compose them (sarting from $0_G$) to get any desired element of this group of functions. To get $0_G$ out of the generators is trivial, since you do not have to apply it any number of times at all.
This is the idea that if you apply a function $0$ number of times, you have not changed the argument at all and hence you get the identity function. (Put more complicatedly, the $\text{id}$ morphism is the identity in the monoid of morphisms.) This is the same spirit as you get $1$ if you multiply something $0$ number of times in a ring, i.e, $a^0 = 1$.
