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 $b \in S$, $a \circ^{-1}b \in S$?

All groups come with an inverse operation so if $a \in S, a^{-1}\in S$.  Closure of the operation then guarantees that $a^{-1} \circ b \in S$ so $a^{-n} \in S$  That still doesn't mean that the integers wrap around like the word cyclic implies in English, but if you look at the definition wrapping around is not required.  The definition of cyclic just says that $a^n$ is the whole group, where $n$ is an integer, not necessarily a natural.
When we extend definition from the finite to the infinite we often have to decide what is important to keep and what isn't.  For finite groups, cyclic implies that there is an element $a$ and a natural $n$ such that $a, a^2, a^3 \ldots a^n, e=a^{n+1}$ is the whole group.  As $n$ gets larger the cycle gets longer.  If we insisted on the wraparound, there would be no infinite cyclic groups.  We can give up the wraparound and just ask that $a$ generate the whole group.  That allows infinite cyclic groups like the integers under addition.  It was decided that was the proper extension.
