Proof

Let $(G,\ast)$ be a cyclic group generated by the element $g$ , i.e. $G=\langle g \rangle$ .
Then any two elements $x,y\in G$ can be represented as $x=g^k$ and $y=g^l$ for some integers $k,l\in\mathbb Z$ .
It follows by the exponentiation in a group and commutativity of integer addition that $\begin{array}{rcl}x\ast y&=&g^k\ast g^l\\&=&g^{k+l}\\&=&g^{l+k}\\&=&g^l\ast g^k\\&=&y\ast x,\end{array}$ which means that $G$ is also Abelian.
∎

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