Natural Numbers under Multiplication do not form Group
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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 $7.2$