# Integers under Multiplication do not form Group

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## Theorem

The set of integers under multiplication $\struct {\Z, \times}$ does not form a group.

## Proof

In order to be classified as a group, the algebraic structure $\struct {\Z, \times}$ needs to fulfil the group axioms.

From Integers under Multiplication form Monoid, $\struct {\Z, \times}$ forms a monoid.

Therefore group axioms $\text G 0$, $\text G 1$ and $\text G 2$ are satisfied.

However, from Invertible Integers under Multiplication, the only integers with inverses under multiplication are $1$ and $-1$.

As not all integers have inverses, it follows that $\struct {\Z, \times}$ is not a group.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $1$: Definitions and Examples: Example $1.5$