Automorphism Group of Cyclic Group is Abelian

Theorem

Let $G$ be a cyclic group.

Let $\Aut G$ denote the automorphism group of $G$.

Then $\Aut G$ is abelian.

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.

$\blacksquare$

1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $27$