Automorphism Group of Cyclic Group is Abelian
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Theorem
Let $G$ be a cyclic group.
Let $\Aut G$ denote the automorphism group of $G$.
Then $\Aut G$ is abelian.
Proof
Let $G = \gen g$
Let $\phi, \psi \in \Aut G$.
As $G$ is cyclic:
\(\ds \exists a \in \Z: \, \) | \(\ds \map \phi g\) | \(=\) | \(\ds g^a\) | |||||||||||
\(\ds \exists b \in \Z: \, \) | \(\ds \map \psi g\) | \(=\) | \(\ds g^b\) |
Thus:
\(\ds \map {\phi \circ \psi} g\) | \(=\) | \(\ds \paren {g^a}^b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g^{a b}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds g^{b a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {g^b}^a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\psi \circ \phi} g\) |
Thus in particular, $\phi \circ \psi$ and $\psi \circ \phi$ are equal on the generator $g$.
Since $g$ generates $G$, they must be equal as automorphisms.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $27$