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

• Probably a bad choice of word. It just means it can be generated by a single element. You can make negative numbers, if addition is the group operation then $-n = 1^{-n}$. Don't confuse the symbols with the usual usage. Oct 4 '20 at 2:40
• If $\langle a \rangle$ didn't contain $a^{-1}$, then it wouldn't be a group. Every element of a group must have an inverse. Oct 4 '20 at 2:43
• In my undergraduate textbook (in French, but words correspond to English: cyclic/cyclique, monogenic/monogène) a group generated by a single element was called monogenic, and a finite monogenic group was called cyclic (because of the cycle shape). Eventually [and before my textbook was edited!] it became practical to call all "cyclic". Similarly a permutation of a finite set splits into "cycles", but for permutations of infinite sets one also allows (bi-)infinite orbits, which are also called "cycles".
– YCor
Dec 14 '20 at 11:49

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.

• I take issue with "we really mean". A cyclic group is truly generated by one and only one element, OP has just misunderstood the meaning of "generated". A group G is said to be generated by a subset S if and only if there are no proper subgroups of G which contain S. This happens to be equivalent to saying that every element of G is some finite product of elements of S and their inverses. Oct 4 '20 at 2:57
• I am wondering how we get from 1 to -1. The two possibilities I have thought of are that either $<a>$ really means $<a,a^{-1}>$, or that for any binary operation $\circ$, we are also allowed to use $\circ^{-1}$. For example, if it is a group under addition, we are also allowed to use subtraction. If it is a group under multiplication, we are allowed to use division. If I understand this answer correctly, any group $<a>$ really means $<a,a^{-1}>$, even if $a^{-1}$ cant be created by $a$ and the binary operation? Oct 4 '20 at 14:38
• @RolandKillian I think you're on the wrong track thinking about inverse operations. In my (admittedly not super extensive) experience with abstract math, we consider sets with one or more operations on them (group, ring, field), and elements of the set may or may not have inverse elements with respect to a given operation. (In groups, every element has an inverse element with respect to the one group operation.) Oct 4 '20 at 19:07
• I have not seen any discussion of inverse operations. I know you can think of subtraction as the inverse of addition and division as the inverse of multiplication, but as far as I know, in advanced math, subtraction and division are just shorthand for adding or multiplying by an inverse element, respectively. Oct 4 '20 at 19:08
• Let's look at the example of the integers that you brought up. In this case we have a group $(\mathbb Z, +)$ and we let $S = \{ 1 \}$. We say that $S$ generates $\mathbb Z$ because every element of $\mathbb Z$ can expressed as a sum of $1$s or $-1$. The set $S$ itself is not necessarily a group (it's clearly not in this case), but the set of elements you can create using $S$ and inverses of the elements of $S$ is a group. (This is always true.) Oct 4 '20 at 19:08

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 $$n$$".

• No, I literally meant an inverse operation, $\circ^{-1}$. For example, the inverse of + would be -. The inverse of * would be /. I am confused about how we get from 1 to -1 with just the addition operation. Is this because we are allowed to use the inverse opperation -, or becasue <a> is really <a,a^{-1}>. Oct 4 '20 at 14:33
• I hope my post answered your question (so please read it): $\langle a \rangle$ really is just $\langle a \rangle$, you just had a (very understandable) misunderstanding of what "generates" means. Anyway, $\circ^{-1}$ is not the right way to communicate this idea. $\circ$ itself is a function $G \times G \to G$, so if $G$ is a finite group with more than one element, this map cannot be invertible. The map you're talking about is actually $(a,b) \mapsto a b^{-1}$ (when the group operation is addition, this is subtraction; when the group operation is multiplication, this is division; etc.) Oct 4 '20 at 18:37
• Anyway another way to communicate what I said in my post is: $\langle a \rangle$ is supposed to be the smallest group we can produce using only the element $a$. By the group axioms, this must contain $a^{-1}$ (or else $\langle a \rangle$ would not be a group at all!). That does not mean that $\langle a \rangle$ is really supposed to be written $\langle a, a^{-1} \rangle$; rather the thing to understand is that $\langle a \rangle$ does not mean "the set of finite products of $a$ with itself". Oct 4 '20 at 18:40

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.

• "we really mean." Who are the "we" ? Oct 4 '20 at 5:39
• @JCAA: the mathematics community Oct 4 '20 at 13:54
• What makes you think that you can speak for that community or that you even belong to it? Oct 4 '20 at 14:25
• This was unnecessarily hostile and rude, imo. Instead of attacking Ross personally, it would be more productive to explain why you think this is not what "the mathematics community really means". Also, I don't see the words "we really mean" in their answer, but perhaps it was edited. Oct 4 '20 at 18:46

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 $$n\gt0$$, $$a^n=e$$.

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.