Cyclic implies abelian automorphism group

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This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., cyclic group) must also satisfy the second group property (i.e., group whose automorphism group is abelian)
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Statement

Any cyclic group is a group whose automorphism group is abelian: the automorphism group of a cyclic group is an abelian group.

Related facts

Proof

Given: A cyclic group G with generator g. Automorphisms \alpha, \beta of G.

To prove: \alpha \circ \beta = \beta \circ \alpha.

Proof: Suppose \alpha, \beta are two automorphisms of G. Since G is cyclic on g, there exist integers a,b such that \alpha(g) = g^a, \beta(g) = g^b. Thus, we have:

\alpha \circ \beta (g) = \beta \circ \alpha (g) = g^{ab}.

In particular, \alpha \circ \beta and \beta \circ \alpha are equal on the generator g. Since g generates G, they must be equal as automorphisms.