# Prove that any automorphism $\phi \in Aut(\mathbb{Z}_n)$ is determined by $\phi([1])$ and that $\phi([1])$ must be a generator for $\mathbb{Z}_n$

Relevant Proposition

Suppose $$G$$ and $$H$$ are groups and that $$e$$ denotes the identity in each. If $$\phi: G \to H$$ is a group homomorphism then $$\phi(e) = e$$. Furthermore for all $$g \in G$$ and $$n \in \mathbb{Z}$$ we have $$\phi(g)^n = \phi(g^n)$$

By the proposition we know that $$\phi([1])^k = \phi([1]^k)$$ where $$[1]^k = [1] + [1] + \cdots + [1] = [(k)( 1)] = [k]$$. This means that the mapping of any element $$[r] \in \mathbb{Z}_n$$ by the automorphism $$\phi$$, $$\phi([r])$$, is determined by iterating the map of $$[1]$$ under $$\phi$$ $$r$$-times. Hence any automorphism $$\phi \in Aut(\mathbb{Z}_n)$$ is determined by $$\phi([1])$$.

For the next part, showing that $$\phi([1])$$ is a generator of $$\mathbb{Z}_n$$ I always seem to struggle with showing that an element is a generator of a cyclic group. I saw a proof of a similar problem and have interpreted it as follows:

By the proposition we know that $$\phi([x]^k) = \phi([x])^k$$ and $$\phi([0]) = [0]$$ for all $$[x] \in \mathbb{Z}_n$$ and $$k \in \mathbb{Z}$$. Therefore, if $$\phi([1])^k = [0]$$ for $$k \lt n$$ then by applying the inverse map $$\phi^{-1}$$ (whose existence is guaranteed by the fact that $$\phi$$ is an automorphism). We find: $$[1]^k = [0]$$ Now, first, is this the correct proof that $$\phi([1])$$ is a generator for $$\mathbb{Z}_n$$ and second, I'm having trouble seeing why this proves $$\phi([1])$$ is a generator. Is it because we see that given the proposition and the assumptions our algebra leads us to the fact that $$[1]^k = [0]$$ where we know that $$[1]$$ is a generator for the (additive) group $$\mathbb{Z}_n$$? Thus $$\phi([1])$$ must also be a generator?

• Suppose we had a homomorphism $\psi : \Bbb Z_n \to \Bbb Z_n$ such that $\psi([1])$ didn't generate $\Bbb Z_n$: Then it would would generate some proper subgroup $H \subset \Bbb Z_n$, and the $\psi$ maps the arbitrary element $[k]$ to $\psi([k]) = \psi([1^k]) = \psi([1])^k \in H$ for all $k$, in which case $\psi$ wouldn't even be a bijection. Commented Jul 6, 2022 at 0:07

Since every $$k\in\Bbb Z_n$$ is $$1^k$$ (multiplicative notation), a homomorphism of a cyclic group is completely determined by what it does to any generator.

And the homomorphism $$\phi$$ is bijective precisely when $$\phi(1)$$, say, is a generator.

A nice application is that, since the generators of $$\Bbb Z_n$$ are the $$m$$ such that $$(m,n)=1$$, we easily get that $$\rm{Aut}(\Bbb Z_n)=(\Bbb Z_n)^×$$, the group of units.

But, this is a little more advanced, because it uses the idea that $$\Bbb Z_n$$ is a ring.

You just need to prove that $$o(\phi(1)) =n$$. Which is trivial since $$\phi(1)^n=\phi(1^n) =\phi(n) =\phi(0) =0$$

• This only proves one direction?? Commented Jul 5, 2022 at 23:33
• Also, this shows $\lvert\phi(1)\rvert\mid n$. Commented Jul 6, 2022 at 10:53