One and Minus One form Subgroup of Multiplicative Group of Rational Numbers

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Theorem

Let $\left({\Q_{\ne 0}, \times}\right)$ be the multiplicative group of rational numbers.

Let $S \subseteq \Q$ where $S = \left\{{1, -1}\right\}$.


Then $\left({S, \times}\right)$ is a subgroup of $\left({\Q_{\ne 0}, \times}\right)$.


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.

$\blacksquare$


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