Natural Numbers under Multiplication do not form Group/Proof 2

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.

We have that $1 \times 1 = 1$ and so is idempotent.

From Identity is only Idempotent Element in Group it follows that $1$ is the identity of $\struct {\N, \times}$.

Let $x \in \N$ such that $x \ne 0$ and $x \ne 1$.

There exists no $y \in \N$ such that $x \times y = 1$

Hence $\struct {\N, \times}$ does not fulfil Group Axiom $\text G 3$: Existence of Inverse Element.

Hence by Proof by Contradiction $\struct {\N, \times}$ is not a group.

Sources

• 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 3$: Examples of Infinite Groups: $\text{(i)}$