Natural Numbers under Multiplication do not form Group

Theorem

The algebraic structure $\left({\N, \times}\right)$ consisting of the set of natural numbers $\N$ under multiplication $\times$ is not a group.

Proof

Aiming for a contradiction, suppose that $\left({\N, \times}\right)$ is a group.

By the definition of the number $0 \in \N$:

$\forall n \in \N: n \times 0 = 0 = 0 \times n$

Thus $0$ is a zero in the abstract algebraic sense.

From Group with Zero Element is Trivial, $\left({\N, \times}\right)$ is the trivial group.

But $\N$ contains other elements besides $0$.

From this contradiction it follows that $\left({\N, \times}\right)$ is not a group.

$\blacksquare$

Sources

• 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): $\S 3$: Examples of Infinite Groups: $\text{(ii)}$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 7$: Example $7.2$