Showing that ##Aut ~ Z_p \simeq Z_{p-1}##

[Bashyboy]

Homework Statement

I am trying to show that $Aut~ Z_p$ is isomorphic to $Z_{p-1}$, where $Z_p$ denotes the congruence class of integers $\mod p$ and $p$ a prime.

Homework Equations

The Attempt at a Solution

I have shown $Aut~ Z_p$ consists of $p-1$ elements, using the fact that a homomorphism is uniquely determined by how it maps $[1]_p$, the generator of $Z_p$; the only element it cannot be mapped to is $[0]_p$, otherwise $[1]_p \rightarrow [k]_p \neq [0]_p$ extends to an automorphism.Now I am trying to show that $Aut~ Z_p$ is a cyclic, and show that mapping the generator of $Aut~ Z_p$ to $Z_{p-1}$ defines an isomorphism. However, I am having trouble showing this. I could use some hints.

I just want to add that this is an early exercise in Hungerford, so, for example, the order of an element hasn't even been defined yet.

 

[andrewkirk]

In this exercise is $Z_p$ considered as a group under the operation of addition, or as a field?

 

[andrewkirk]

Having thought about this some more, I concluded that we need to interpret $Z_p$ as a group under addition rather than a field.

With that interpretation, your first sentence in section 3 above is correct.
Given that result, I don't think you need to ask about cyclicality do you (the subject of your next para)? In fact you have already done most of the proof. The remaining part is just to state the isomorphism between $Aut\ Z_p$ and $Z_{p-1}$.

First make a guess as to what map $\theta$ from $Z_{p-1}$ to $Z_p\times Z_p$ (the set of maps from $Z_p$ to itself) might be the required isomorphism, then check that
1. its image is always in $Aut\ Z_p$
2. $\theta$ is a group homomorphism
3. $\theta$ is one-to-one
4. $\theta$ is onto

The hardest part is guessing a form for $\theta$. We need to specify a function $f:Z_{p-1}\times Z_p\to Z_p$ such that
$$\theta([k]_{p-1})([r]_p)=[f(k,r)]_p$$
The properties we require for $f$ are

1. $f(0,r)=r$ ($\theta$ maps $[0]_{p-1}$ to the identity automorphism on $Z_p$)
2. $[f(k,r)]_p\neq [f(k,s)]_p$ if $k\neq 0$, $r\neq s$ and $0\leq r,s<p$ ($\theta(k)$ must be injective)
3. for any $k$, $[f(k,r+s)]_p=[f(k,r)]_p+[f(k,s)]_p$ ($\theta(k)$ must be a homomorphism)

You can expect that $f$ will be fairly simple, although it is not immediately obvious what it should be.

 

[Dick]

Bashyboy has already spelled out what the automorphisms are by considering the possible images of $1$. That's the straightforward part. The hard part is showing the automorphism group is cyclic. This is a standard result, it's showing that there is a primitive root. But I don't know of any proofs that are simple enough that I would assign them as an exercise before I discussed what 'order of an element' means.

 

[aheight]

You may find this Wolfram demonstration interesting although it concerns only the size of automorphism groups.: http://demonstrations.wolfram.com/TheSizeOfAnAutomorphismGroup/

 

[Bashyboy]

Having thought about this some more, I concluded that we need to interpret ZpZpZ_p as a group under addition rather than a field.

Yes, you are right in construing it as a additive group. I am sorry I didn't respond sooner. Also, I forgot to mention that I have already proved that $Z_p$ is multiplicative group whenever $p$ is prime.

First make a guess as to what map θθ\theta from Zp−1Zp−1Z_{p-1} to Zp×ZpZp×ZpZ_p\times Z_p (the set of maps from ZpZpZ_p to itself) might be the required isomorphism, then check that
1. its image is always in Aut ZpAut ZpAut\ Z_p
2. θθ\theta is a group homomorphism
3. θθ\theta is one-to-one
4. θθ\theta is onto

The hardest part is guessing a form for θθ\theta. We need to specify a function f:Zp−1×Zp→Zpf:Zp−1×Zp→Zpf:Z_{p-1}\times Z_p\to Z_p such that

θ([k]p−1)([r]p)=[f(k,r)]pθ([k]p−1)([r]p)=[f(k,r)]p​

\theta([k]_{p-1})([r]_p)=[f(k,r)]_p
The properties we require for fff are

1. f(0,r)=rf(0,r)=rf(0,r)=r (θθ\theta maps [0]p−1[0]p−1[0]_{p-1} to the identity automorphism on ZpZpZ_p)
2. [f(k,r)]p≠[f(k,s)]p[f(k,r)]p≠[f(k,s)]p[f(k,r)]_p\neq [f(k,s)]_p if k≠0k≠0k\neq 0, r≠sr≠sr\neq s and 0≤r,s<p0≤r,s<p0\leq r,sθ(k)θ(k)
   \theta(k) must be injective)

• for any kkk, [f(k,r+s)]p=[f(k,r)]p+[f(k,s)]p[f(k,r+s)]p=[f(k,r)]p+[f(k,s)]p[f(k,r+s)]_p=[f(k,r)]_p+[f(k,s)]_p (θ(k)θ(k)\theta(k) must be a homomorphism)

You can expect that fff will be fairly simple, although it is not immediately obvious what it should be.

So, do you happen to know what $\theta$ and #f## look like? I'm still going to need to think about it.

 

[Dick]

Yes, you are right in construing it as a additive group. I am sorry I didn't respond sooner. Also, I forgot to mention that I have already proved that $Z_p$ is multiplicative group whenever $p$ is prime.
So, do you happen to know what $\theta$ and #f## look like? I'm still going to need to think about it.

You already know what the automorphisms look like. You said it in the first post. If you know where $1$ is mapped you know where everything is mapped. It's a generator.

 

[andrewkirk]

So, do you happen to know what θ\theta and #f## look like? I'm still going to need to think about it.

Whoa, that partial quote facility really doesn't play well with latex does it?

Try $f(k,s)=2^ks$

 

[Dick]

Whoa, that partial quote facility really doesn't play well with latex does it?

Try $f(k,s)=2^ks$

That gives you a homomorphism into $Aut Z_p$. Are you sure it's an isomorphism? Look at $Z_7$.

 

[andrewkirk]

That gives you a homomorphism into $Aut Z_p$. Are you sure it's an isomorphism? Look at $Z_7$.

Great Caesar's Ghost, you're right - it doesn't work!

And I felt so sure of that mapping.

 

[Dick]

Great Caesar's Ghost, you're right - it doesn't work!

And I felt so sure of that mapping.

Right. You can't just use 2 for all values of p. You need a primitive root. There's a number of proofs that one exists. None of them give you a formula for finding one.