Hi guys,
Is the automorphism group of a cyclic group always abelian?
I believe so but I need to prove it. I don't really know where to start...any ideas?
Hi
Let $\displaystyle (G,.)$ be a cyclic group, $\displaystyle a$ one of its generators. Two automorphisms $\displaystyle \phi$ and $\displaystyle \psi$ are determined by $\displaystyle \phi(a)=a^k$ and $\displaystyle \psi(a)=a^l$ for some $\displaystyle k,l \in\mathbb{Z}.$
Since $\displaystyle \psi\circ\phi$ and $\displaystyle \phi\circ\psi$ are automorphisms, they're also determined by $\displaystyle \psi\circ\phi(a)$ and $\displaystyle \phi\circ\psi(a),$ therefore if $\displaystyle \psi\circ\phi(a)=\phi\circ\psi(a)$ then $\displaystyle \psi\circ\phi=\phi\circ\psi$ and the automorphism group is abelian.
So is $\displaystyle \psi\circ\phi(a)=\phi\circ\psi(a)$ true?
You can do better that what clic-clac said if you want to try and additional exercise.
If $\displaystyle |G|=n$ is a cyclic group then $\displaystyle \text{Aut}(G) \simeq \mathbb{Z}_n^{\times}$.
If $\displaystyle |G|=\infty$ then $\displaystyle \text{Aut}(G) \simeq \mathbb{Z}_2$.