Cyclic Group is Abelian

Theorem

Let $G$ be a cyclic group.


Then $G$ is abelian.


Proof 1

Let $G$ be a cyclic group.

All elements of $G$ are of the form $a^n$, where $n \in \Z$.

Let $x, y \in G: x = a^p, y = a^q$.

From Powers of Group Elements: Sum of Indices:

$x y = a^p a^q = a^{p + q} = a^{q + p} = a^q a^p = y x$

Thus:

$\forall x, y \in G: x y = y x$

and $G$ is by definition abelian.

$\blacksquare$


Proof 2

We have that Integers under Addition form Abelian Group.

The result then follows from combining:

Epimorphism from Integers to Cyclic Group
Epimorphism Preserves Commutativity.

$\blacksquare$


Sources

  • 1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: $\S 9$: Cyclic Groups
  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 43$
  • 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts
  • 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Abelian group
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