[Math] How to Cyclic groups be infinite

Question

I am a little confused about how a cyclic group can be infinite.

To provide an example, look at $\langle 1\rangle$ under the binary operation of addition. You can never make any negative numbers with just $1$ and the addition opperation.

When we declare a cyclic group $\langle a\rangle $, does it go without saying that even if $a^n \neq a^{-1}, \forall n \in \mathbb{N}$ that $a^{-1} \in \langle a\rangle $?

If the inverse element can not be made with the generator and the operator, how can it be in the group? Do all groups come with an inverse operation such that $a \in S$ and $b \in S$, $a \circ^{-1}b \in S$?

Answer 1

When we talk about a "cyclic group", meaning a group generated by "single element", we really mean "that one element and its inverse."

It's really all the integer powers of the element, not just the natural number powers.

However, in a finite cyclic group, you can think of it as being generated by all natural number powers of the generating element, because if $g^n = e$, then $g^{n - 1} = g^{-1}$.

Edit:

Perhaps I shouldn't have said "we really mean." I think it's fair to say that you can think of a cyclic group as being created by products of the element and its inverse.

Here's an official definition, from Lang's Algebra, page 9:

Let $G$ be a group and $S$ a subset of $G$. We shall say that $S$ generates $G$, or that $S$ is a set of generators for $G$, if every element of $G$ can be expressed as a product of elements of $S$ or inverses of elements of $S$, i.e. as a product $x_1 \cdots x_n$ where each $x_i$ or $x_i^{-1}$ is in $S$.

That $T$, the set of all products of elements of $S$ or inverses of elements of $S$, is a subgroup of $G$, and in fact the smallest subgroup of $G$ containing $S$, follows quickly after this.

A cyclic group one for which $S$ is a singleton.