How can 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.

Answer 2

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.

Answer 3

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!

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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 $n$".

Answer 4

The cyclic group generated by an element $a\in G$ is by definition $G\ge\langle a\rangle:=\{a^n:n\in\Bbb Z\}$. From this it follows easily that $\Bbb Z\ge\langle 1\rangle=\Bbb Z$.

Interestingly enough, this is the only infinite cyclic group.

Also interestingly, for finite groups we have the simplification that $\langle a\rangle=\{a^n:n\in\Bbb Z^+\}$, since for some $n\gt0$, $a^n=e$.

Answer 5

Others have given some answers. My observation would be that to have something generated by $1$ using the operation $+$ you already need a context in which $1$ and $+$ make sense. This you have not given us. What kind of object is $1$ and what properties does it have? What kind of operation is $+$ and what properties does it have (associative? commutative?).

To illustrate the problem, it is not clear whether what you generate includes something like $0$ which is an additive identity. The non-negative integers (see I've defined a set within which my operation makes sense) under the usual definition of addition form a monoid, but not a group. The difference is that a group has inverses for every element by definition.