ThmDex – An index of mathematical definitions, results, and conjectures.
Result R893 on D23: Abelian group
Cyclic group is Abelian
Formulation 0
Let $G$ be a D1303: Cyclic group.
Then $G$ is an D23: Abelian group.
Proofs
Proof 0
Let $G$ be a D1303: Cyclic group.
Let $g$ be a generator element of $G$ and let $x, y \in G$. Since $G$ is cyclic, result R627: Explicit expression for elements of cyclic group states that there are $n, m \in \mathbb{Z}$ such that $x = g^n$ and $y = g^m$. Now \begin{equation} x y = g^n g^m = g^{n + m} = g^{m + n} = g^m g^n = y x \end{equation} $\square$