How can Cyclic groups be infinite 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 $b \in S$, $a \circ^{-1}b \in S$?


Do all groups come with an inverse operation such that $a \in S$ and $b \in S$, $a \circ^{-1}b \in S$?

Do you mean "$a^{-1} b \in S$"? If so, then yes, the existence of inverses is literally one of the group axioms!

I think a precise meaning of the word "generated" will help answer this question.
Let $G$ be a group, and let $S$ be any subset of $G$. The subgroup of $G$ generated by $S$, sometimes denoted $\langle S \rangle_G$, is defined to be the intersection of all subgroups of $G$ which contain $S$ as a subset.
From this definition, we see that $\langle S \rangle_G$ is the unique smallest subgroup of $G$ which contains $S$ as a subset, and from this it's not too hard to prove that $\langle S \rangle_G$ is also the set of elements of $G$ of the form $x_1 x_2 \cdots x_n$ where each $x_1, \dots, x_n$ is either an element of $S$ or the inverse of an element of $S$. So the inverses aren't coming out of nowhere: they arise naturally from this construction of the subgroup generated by a subset!
When we say that $G$ is generated by a subset $S$, we mean that $\langle S \rangle_G = G$; i.e. every element of $G$ can be written as a finite product of elements of $S$ and their inverses. "$G$ is cyclic" means that $G$ is generated by some singleton subset, i.e. there is some $a \in G$ such that every element of $G$ is a finite product of the terms $a$ and $a^{-1}$. In other words, "$G$ is cyclic" is equivalent to saying "there exists some $a \in G$ such that every element of $G$ is equal to $a^n$ for some integer $n$".
