One and Minus One form Subgroup of Multiplicative Group of Rational Numbers
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Theorem
Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers.
Let $S \subseteq \Q$ where $S = \set {1, -1}$.
Then $\struct {S, \times}$ is a subgroup of $\struct {\Q_{\ne 0}, \times}$.
Proof
By hypothesis, $S$ is not empty.
As $0 \notin S$, it follows that $S \subseteq \Q_{\ne 0}$.
Recall that $-1 \times -1 = 1$ and also $1 \times 1 = 1$.
Thus:
- $\forall x \in S: x \times y^{-1} \in S$
The result follows from the One-Step Subgroup Test.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$