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

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.)
