An ${\rm Aut}(\mathbb{Z}/n\mathbb{Z}) \rightarrow(\mathbb{Z}/n\mathbb{Z})^\times$ construction

Question

i would like to understand how to construct a morphism between 2 groups $${\rm Aut}(\mathbb{Z}/n\mathbb{Z}) \rightarrow (\mathbb{Z}/n\mathbb{Z})^\times$$ and prove that it is an isomorphism.

The only thing that I need is how to construct it.

I'm still not good in manipulating Automorphisms so I have a difficulty. Then, to prove that it is an isomorphism I'm sure it's pretty easy to prove it's surjection and injection.

I would probably consider a generator of $(\mathbb{Z}/n\mathbb{Z})^\times$ and construct an image of $\phi([r]_n)$, where $\phi \in{\rm Aut}(\mathbb{Z}/n\mathbb{Z})$ but I'm not sure how to manipulate all of this.

Thanks in advance

Answer 1

Hint: Consider the map $\Phi : (\Bbb Z/n \Bbb Z)^\times \to \mathrm{Aut}(\Bbb Z/n\Bbb Z)$ defined by $$ [\Phi([r]_n)]([s]_n) = [rs]_n. $$ I think that the trickiest part from there should be proving that $\Phi$ is surjective.

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If you insist on constructing a map in the opposite direction, then you could consider the map $\phi \mapsto \phi(1)$, as is discussed in the comments to this answer.

Answer 2

It turns out that for every $a \in \mathbb Z/n$, there is a unique map $\mathbb Z/n \longrightarrow \mathbb Z/n$ sending $1 \mapsto a$. This is just multiplication by $a$, and it is unique as $1$ generates $\mathbb Z/n$. On the other hand, if $\phi: \mathbb Z/n \longrightarrow \mathbb Z/n$ then $\phi$ is totally determined by where $1$ is sent. Indeed, $\phi(m)=m \phi(1)$ so $\phi$ is just the multiplication by $\phi(1)$ map. Thus, we have the following map: $\Phi: End(\mathbb Z/n) \longrightarrow \mathbb Z/n$ via $\phi \mapsto \phi(1)$. If you are unfamiliar with this notation, $End(\mathbb Z/n)$ is the set of endomorphisms of $\mathbb Z/n$, i.e. group homomorphisms $\mathbb Z/n \longrightarrow \mathbb Z/n$.

This map $\Phi$, it turns out, will restrict to the desired map $Aut(\mathbb Z/n)\longrightarrow (\mathbb Z/n)^\times$, and this is a group isomorphism. I'd like to point out that in addition, $\Phi: End(\mathbb Z/n)\longrightarrow \mathbb Z/n$ is actually an isomorphism of rings, where multiplication in $End(\mathbb Z/n)$ is composition. The isomorphism you're looking for arises as the induced map on the unit groups of these rings.

Ring theory aside, here are some ideas to guide you to proving that $\Phi$ is your desired isomorphism.

1. As discussed, if $\Phi(\phi)=a$ then $\phi$ is the multiplication by $a$ map. If $\phi$ is an automorphism, why must $a$ be invertible in $\mathbb Z/n$?

2. You can try to prove that $\Phi$ is an isomorphism by constructing an inverse map. How can you take an element of $\mathbb Z/n$ and get a map $\mathbb Z/n \longrightarrow \mathbb Z/n$?