Prove or disprove: ${\rm Aut}(\Bbb Z_8)$ is abelian and cyclic.

Question

$\DeclareMathOperator{\Aut}{Aut}$ So for this class I was introduced to Automorphisms through the homework. We had to prove that Automorphisms under composition is a group, and the next question was asking whether $\Aut(\Bbb Z_8)$ is abelian and/or cyclic.

I have no idea how to approach this problem. To disprove it I'd need an example but I don't know how to find the elements of $\Aut(\Bbb Z_8)$. And if it is true, I'd need to write a solid proof which I also don't know how to approach.

Ideas I have so far:

I have a hunch that $\Aut(\Bbb Z_8)$ is not abelian. If it were, I'd basically need to show that $f(g(x)) = g(f(x))$, which I think is only true if the functions are inverses of each other. So I'd need to prove that every function in $\Aut(\Bbb Z_8)$ is an inverse of every other function in $\Aut(\Bbb Z_8)$, which I don't think is possible.

I am unsure about whether $\Aut(Z8)$ is cyclic or not. $\Bbb Z_8$ is cyclic and generated by $1$, but I don't know how/if I could use that information to prove that $\Aut(\Bbb Z_8)$ is cyclic.

Any help/hints/resources would be appreciated. Thanks!

Answer 1

Hint:

An automorphism $f$ of $\mathbf Z_8$ maps the generator $\bar 1$ onto another generator, and this image characterises $f$.

Now the generators of $\mathbf Z_8$ are $\;\{\bar 1,\bar 3,\bar 5,\bar 7\}$, hence $\operatorname{Aut}(\mathbf Z_8)$ has order $4$. Check that any automorphism $f$ satisfies $f^2=\text{id}$, and deduce from this relation that $\operatorname{Aut}(\mathbf Z_8)$ is commutative.

Note: if you know that there are, up to an isomorphism, only two groups of order $4$, and that they both are commutative, it's still shorter.

Answer 2

As @Bernard showed it's abelian.

But for cyclic notice again to what @Bernard said that

"Any automorphism $f$ satisfies $f^2=id$."

So it isn't cyclic, in fact: $${\rm Aut}(\mathbb{Z}_8) \equiv \mathbb{Z}_2 \times \mathbb{Z}_2$$