I don't understand why the set of natural numbers constitutes a commutative monoid with addition, but is not considered an Abelian group.

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    $\begingroup$ What is the additive inverse of $n\neq 0$? $\endgroup$ – Andrew Dec 7 '12 at 21:22
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    $\begingroup$ It's not a nonabelian group. It's an abelian nongroup. $\endgroup$ – Qiaochu Yuan Dec 7 '12 at 21:23
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    $\begingroup$ Note that $\mathbb N$ is not considered a group (even less an abelian group), but your question title suggests that it is considered a non-abelian group - no it isn't even a non-abelian group. $\endgroup$ – Hagen von Eitzen Dec 7 '12 at 21:24
  • $\begingroup$ Such a "great" question deserves more than 3 upvotes! $\endgroup$ – user26857 Dec 9 '12 at 18:42

Addition on the natural numbers IS commutative ...

...but the natural numbers under addition do not form a group.

Why not a group?

  • if you define $\mathbb{N} = \{ n\in \mathbb{Z} \mid n\ge 1\}$, as we typically do, then it fails to be a group because it does not contain $0$, the additive identity.

    Indeed, $\mathbb{N} = \{ n\in \mathbb{Z} \mid n\ge 1\}$ fails to be a monoid, if the additive identity $0\notin \mathbb{N}$.

    So I will assume you are defining $\mathbb{N} = \mathbb{Z}^{+} = \{x \in \mathbb{Z} \mid x \ge 0\}$, which is indeed a commutative monoid: addition is both associative and commutative on $\mathbb{N}$, and $0 \in \mathbb{N}$ if $\mathbb{N} = \mathbb{Z}^{+} = \{x \in \mathbb{Z} \mid x \ge 0\}$.

  • There is no additive inverse for any $n\in \mathbb{N}, n \ge 1$. For example, there is no element $x \in \mathbb{N}$ such that $3 + x = 0.$ Hence the natural numbers cannot be a GROUP.

A monoid is a group if it also (in addition to being associative, has the additive identity for addition) satisfies the ADDED requirement that the inverse element of each element in the monoid is also in the monoid. (In such case, the monoid is said to be a group.)

  • $\begingroup$ n_ary_crazy: Are you familiar with the properties of a "group" - as that term is used in math? $\endgroup$ – Namaste Dec 7 '12 at 22:17
  • $\begingroup$ If OP states that $\Bbb N$ is a commutative monoid, then you may infer that she considers $0\in\Bbb N$. $\endgroup$ – Marc van Leeuwen Dec 8 '12 at 17:28

The natural numbers do not have inverses, hence they are not a group.


What is a monoid?

A monoid is an algebraic structure with a single associative binary operation and an identity element.

Formally, a monoid is a set, $S$, with a binary operation, $\cdot$, that satisfies the following three axioms: $$a\cdot b \in S \quad \forall a,b \in S.$$ $$(a \cdot b)\cdot c=a\cdot (b\cdot c) \quad \forall a,b,c \in S.$$ $$\exists e \in S: e\cdot a=a\cdot e=a \quad \forall a \in S.$$ These axioms are, in other words, closure, associativity, and the existence of an identity. (Some would formally say the existence of a left and right identity which are equivalent.)

Depending on your definition of $\mathbb{N}$, we may or may not have a monoid. If $\mathbb{N}=\{0,1,2,\cdots\}$, then we have a monoid $(\mathbb{N},+)$ since all three aforementioned axioms are satisfied. (In particular, we have $e=0$.)

Now, here's where I think your confusion lies: You don't know particularly what a group is. In fact, a group is a special kind of monoid! It is a monoid with the additional axiom below: $$\exists a^{-1}: a\cdot a^{-1}=a^{-1}\cdot a=e \quad \forall a\in S.$$ Thus, we see a group is a monoid with the addition of inverses. Since there are not inverses in $(\mathbb{N}, +)$, $(\mathbb{N},+)$ is not a group. There are not inverses for a clear reason: What do we add to $1$ to return $0$, for example? $-1$. That's not in $\mathbb{N}$, hence we have at least one example where there is not an inverse and that's enough to state $(\mathbb{N},+)$ is not a group.

What should be noted is the following awesomeness: $(\mathbb{Z},+)$ is a monoid and a group, and not just any group: An abelian group. ;-D

What you'll find throughout mathematics is the following awesomeness: There are generalizations between generalizations between generalizations between . . . ad infinitum (?). We're seeing this here. The algebraic structure of a group is a generalization of what we see in $(\mathbb{Z},+)$. Similarly, a monoid is an even weaker 'version' of a group. (This is informal.) On top of this, we have the notion of a semigroup: A generalization of a monoid by removing the necessity for an identity element. So, if we define $\mathbb{N}=\{1,2,3,\dots\}$, we have that $(\mathbb{N},+)$ is a semigroup, and not a monoid nor group.

What's awesome is the rabbit hole goes further: There is a structure known as a magma (some call it a groupoid, but this often causes confusion with a category theory groupoid). This is a semigroup without the axiom of associativity. That means that the only axiom it imposes on the operation and set is closure. Cool stuff, right?

In summary,

$$\dots \subset \text{Magma} \subset \text{Semigroup} \subset \text{Monoid} \subset \text{Group} \subset \text{Abelian Group} \subset \cdots$$


Assuming the axiom of choice one can actually endow any set with a group structure.

To get a group structure for $\mathbb N$ define for example $f: \mathbb Z \to \mathbb N$, $k \mapsto 2|k|$ if $k \leq 0$ and $k \mapsto 2k + 1$ otherwise. Now define addition on $\mathbb N$ as follows: let $n + m := f^{-1}(n) + f^{-1}(m)$.

Verify that $f$ is a (group) isomorphism, that $0$ is the additive identity and that the additive inverse of $n$ is $f(-f^{-1}(n))$.


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