Proof
(related to Proposition: Algebraic Structure of NonZero Rational Numbers Together with Multiplication)
The set of rational numbers with the number zero \(0\) excluded, denoted by \(\mathbb Q^*\), together with the specific multiplication operation
"\(\cdot\)" is a commutative group. This is because:
 The multiplication operation is associative, i.e. \((x\cdot y)\cdot z=x\cdot (y\cdot z)\) is valid for all \(x,y,z\in\mathbb Q^*\).
 We have shown the existence of a neutral element of multiplication  the number \(1\in\mathbb Q^*\), i.e. such that \(1\cdot x=x\) for all \(x\in\mathbb Q^*\).
 For every \(x\in\mathbb Q^*\), there there exists an inverse rational number \(x^{1}\in\mathbb Q^*\), such that \(x\cdot x^{1}=1\).
 For every \(x\in\mathbb Q^*\), there there exists an inverse rational number \(x^{1}\in\mathbb Q^*\), such that \(x\cdot x^{1}=1\).
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References
Bibliography
 Kramer Jürg, von Pippich, AnnaMaria: "Von den natürlichen Zahlen zu den Quaternionen", SpringerSpektrum, 2013