Natural Numbers under Multiplication do not form Group/Proof 1

Theorem

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

Proof

Aiming for a contradiction, suppose that $\struct {\N, \times}$ 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, $\struct {\N, \times}$ is the trivial group.

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

From this contradiction it follows that $\struct {\N, \times}$ is not a group.

$\blacksquare$