# Automorphism group of cyclic group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^*$

Prove that the automorphism group $$\operatorname{Aut}(C_n)$$ of a cyclic group of order $$n$$ is isomorphic to $$(\mathbb{Z}/n\mathbb{Z})^*$$

I know that the automorphism group consists of all the homomorphisms from the group to the group. And I know that somehow I need to use the fact that the group is cyclic and has a generator. But I can't quite put the puzzle together.

Let $$\langle g\rangle$$ be the generator, it exists $$m \in \mathbb{N}$$ s.t $$g^m=1$$ so an automorphism must do $$f(1)=1=f(g^m)=f(g)...f(g)=(f(g))^m=1$$ does that mean $$f(g)$$ is also a generator ?

In order to have an isomorphism, I need to consider a homomorphism. Let $$\varphi:\operatorname{Aut}(C_n) \rightarrow \mathbb{Z_n}$$ with $$\varphi(f)=$$ ?

That's how far I go, automorphism are new to me and the whole exercise confuses me a little bit.

• See Theorem 6.5 of Gallian's, "Contemporary Abstract Algebra (Eighth Edition)". Oct 19, 2021 at 20:05
• No, automorphisms must be ISOmorphisms Oct 19, 2021 at 20:06
• @Shaun thanks will do Oct 19, 2021 at 20:07

This is one place where I think it is very helpful to first ponder the problem in slightly more generality. Fix a generator for $$C_n$$ once and for all. I'll call it $$g$$ to meet your notation, so $$C_n = \langle g \rangle$$ throughout.

First, prove that every homomorphism $$f: C_n \to C_n$$ is uniquely determined by the value $$f(g)$$. I think you know how to do this. You just need to prove that given homs $$f,h$$ we have $$f=h$$ if and only if $$f(g)=h(g)$$. This is nothing more than the definition of cyclic group.

Next, prove that every homomorphism $$f: C_n \to C_n$$ can be associated to a unique integer $$m$$ with $$0 \leq m \leq n-1$$. The association is: given a hom $$f$$, we know that $$f(g) \in C_n$$, but $$C_n$$ is generated by $$g$$. Thus $$f(g)$$ is expressible as $$g^m$$ for some unique power $$m$$ with $$0 \leq m \leq n-1$$. Let's call this $$m_f$$ instead of $$m$$ to denote its dependence on $$f$$. In short, $$f(g)=g^{m_f}$$ where $$m_f$$ has been chosen "correctly."

You should be able to check that the association $$f \mapsto m_f$$ defines a perfectly good function from the set of all homs $$f: C_n \to C_n$$ into $$\mathbb{Z}_n$$ (as sets only for the moment).

Here's the first of two crucial moves: prove that given two homs $$C_n \stackrel{f}{\rightarrow}C_n\stackrel{h}{\rightarrow} C_n$$ we have $$m_{h\circ f}=m_h m_f \bmod n$$. That is, our association $$f \mapsto m_f$$ respects composition in a multiplicative way mod $$n$$. This is easier than it looks.

Here's the final bit. Prove that $$f: C_n \to C_n$$ is an automorphism if and only if $$m_f$$ has a multiplicative inverse mod $$n$$. To show one direction, suppose $$f$$ is an automorphism. Then as $$f$$ is surjective there must be an $$a \in C_n$$ with $$f(a)=g$$. This $$a$$, however, is expressible as $$a=g^k$$ for some integer $$k$$, since $$g$$ generates $$C_n$$. Then $$g=f(a)=f(g^k)=f(g)^k=(g^{m_f})^k = g^{m_fk}.$$ Hence $$g^{m_fk}=g^1$$ so you now have that $$m_fk \equiv 1 \bmod n$$, which is what we wanted to show. You can fill in the rest.

Note that the association $$f \mapsto m_f$$ gives your isomorphism $$\operatorname{Aut}(C_n) \to \mathbb{Z}_n^*$$ and it is now clear why the operation must be multiplication mod $$n$$ (because the operation in the automorphism group is composition).

I know that the automorphism group consists of all the homomorphisms from the group to the group.

Notice that automorphisms are not arbitrary homomorphisms, they are isomorphisms (homomorphisms+ bijections).

Let $$$$ be the generator, it exists $$m \in \mathbb{N}$$ s.t $$g^m=1$$ so an automorphism must do $$f(1)=1=f(g^m)=f(g)...f(g)=(f(g))^m=1$$ does that mean $$f(g)$$ is also a generator ?

I want you to first show that an isomorphism preserves the order of an element (infact it preserves every group theoric properties.) Then you can conclude that a generotor must be sent to a generotor.

In the next step, just try to understand when you assign a genorator to another one, actually you assign all elements to one another. Then you can explicitly obtain the rule of your map.

• I will try again tomorrow, it's getting late, and I am tired. I will post here my solution if I figure it out Oct 19, 2021 at 20:28
• If you could verify my answer, I would appreciate it Oct 20, 2021 at 16:20

I think to prove the exercise is enough to show that

1. $$|\text{Aut}(C_n)|=|\mathbb{Z_n}^*|$$

2. $$\text{Aut}(C_n)=$$

For 2. We can check that for a generator $$g$$ of $$C_n$$ we can get every element $$h=g^m ,m\in \mathbb{N}$$ so $$f(h)=f(g^m)=(f(g))^m$$

and for 1. a $$g^i$$ generates $$C_n$$ if and only if $$\operatorname{gcd}(i,m)=1$$.

$$\lvert \operatorname{Aut}(C_n)\rvert=\phi(m)$$ where $$\phi(m)$$ is Euler's function.

• The line $\mathrm{Aut}(C_n)=\langle f(g)\rangle$ makes no sense. Especially since $\mathbb{Z}_n^*$ usually isn't cyclic. Also, $f(g)$ is an elment of $C_n$ by your notation, so how can it generate $\mathrm{Aut}(C_n)$? Oct 20, 2021 at 16:33
• My idea was that this true of every isomorphism $f$ so one must generate the others. Also, it's helpful when you make a comment to explain why this is wrong or give a hint. It's like someone is confused about something and tries to understand it, and you comment "you are wrong" Oct 20, 2021 at 17:01
• I've tried to explain. Since the automorphism group typically isn't cyclic, it would be wrong to write it as a cyclic group generated by some element $f(g)$. Also, if a group $G$ is cyclic it will be generated by one of its own elements, not an element from a foreign group. Oct 20, 2021 at 17:02
• @Randall Your intentions maybe are good, but your comment "sounded" rude to me. (maybe it's my fault because I am confused and can't figure it out. Anyway). I appreciate the effort. Maybe the exercise is too advanced for me since we haven't talk about the group $Z_m^*$ and automorphism yet Oct 20, 2021 at 17:12
• Honestly not trying to be rude. Was trying to ask questions to lead you to the right path. Apologies either way. Oct 20, 2021 at 17:14