Cyclic implies abelian automorphism group

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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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.