# Infinite Cyclic Group of Integers

I am new to group theory. While reading about cyclic groups, according to my understanding, A Cyclic group has a generator that generates all other elements by several copies of it. Now coming to set of integers $$Z$$ with addition as a binary operation, i read that $$Z$$ is an infinite cyclic group with generators $$1$$ and $$-1$$.

But $$1$$ cannot generate negative integers no matter how many copies are added and analogously for $$-1$$ which cannot generate positive integers. So does it mean $$1$$ generates positive integers and $$-1$$ generates negative integers? How about generating identity element $$0$$?

• Note that "generates" is context-dependent in mathematics. $1$ generates the group $\mathbb{Z}$ but $1$ does not generate the monoid $\mathbb{Z}$ (as in the latter case, the set of non-negative integers is a strictly smaller monoid containing $1$) Commented Apr 26, 2020 at 6:09

The definition of the group generated by an element or a set of elements allows inverses to be used when composing elements. The generated group always includes the identity, regarded as the composition of nothing.

$$-1$$ is the inverse of $$+1$$, so $$+1$$ alone generates all of $$\mathbb Z$$. Likewise, $$-1$$ alone generates all of $$\mathbb Z$$.

• so you mean to say $g^n=h$ is equivalent to $(g^{-1})^{-n}=h$ where $g$ and $g^{-1}$ are generators and $h$ is some element in the Group. Commented Apr 26, 2020 at 6:11
• @Umeshshankar Yes. The laws of exponents apply to groups as well. Commented Apr 26, 2020 at 6:12
• ok when we say $g^n=h$ it means $h$ is generated by $n$ copies of $g$ which also means $h$ is generated by $-n$ copies of $g^{-1}$. Does this makes sense? Commented Apr 26, 2020 at 6:15
• @Umeshshankar Yes. Commented Apr 26, 2020 at 6:15

This is one thing that may be a little easier to understand in multiplicative notation. The cyclic group generated by an element $$x$$ is $$\langle x\rangle=\{x^n:n\in\Bbb Z\}$$.

In additive it's: $$\langle x\rangle=\{n\cdot x:n\in\Bbb Z\}$$.

Thus $$\langle-1\rangle=\langle1\rangle=\Bbb Z$$.