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 Axiom $\text G 0$: Closure, Group Axiom $\text G 1$: Associativity and Group Axiom $\text G 2$: Existence of Identity Element 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 $1.5$