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

Question

Okay. This is similar to a question I asked a few days ago. 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. 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.

Note: At this point, I don't know anything about fields (let alone rings), the order of an element, and related concepts.

Edit: I noticed that the question was asked here, but from my cursory reading I didn't see any ideas I could use yet.

The duplicate has also answers without fields. What you should know, in any case, is the order of elements. — Dietrich Burde, Mar 29 '17 at 18:40
@DietrichBurde But the definition of the order of an element comes in the next section. If you open up Hungerford, you will see. — user193319, Mar 29 '17 at 18:41
@MorganRodgers I know what the order of an element is, but Hungerford does not introduce it until the next section. — user193319, Mar 29 '17 at 18:42
Here is the link of your earlier question. Bernard has a hint there as well. — Dietrich Burde, Mar 29 '17 at 18:44
@DietrichBurde I would use that hint, but the fact he cites isn't proven until the next section. — user193319, Mar 29 '17 at 18:47
That's really not the same, unless you know that $\mathbb Z_p^{\times}\cong\mathbb Z_{p-1}$... @DietrichBurde — Thomas Andrews, Mar 29 '17 at 18:48
@DietrichBurde It is obvious that $\mathbb Z_p^\times$ is cyclic? It clearly has $p-1$ elements, but being cyclic is actually one of the world's cool surprises. — Thomas Andrews, Mar 29 '17 at 18:50
@ThomasAndrews Yes. And $Z_p^\times$ hasn't even been introduced yet, so I don't see how this is a duplicate. People I plead: open your copy of Hungerford before your mark a question as a duplicate and you'll see what tools I have available! — user193319, Mar 29 '17 at 18:51
Can I change this somehow? @user193319 I am sorry, the other question is a better duplicate. But you have an answer by Bernard, so everything is fine. — Dietrich Burde, Mar 29 '17 at 18:53
I have successfully changed the duplicate! The answer to that question by Pete L. Clarke is correct. I have never heard of any way of proving the group $Z_p^*$ cyclic that does not involve the result that a polynomial of degree $d$ over a field has at most $d$ roots, but that result is really very elementary - high school level. — Derek Holt, Mar 29 '17 at 20:47

score -3 · Answer 1 · answered Mar 29 '17 at 18:39

-3

Hint:

Let $\varphi$ the automorphism of $\mathbf Z_p$ mapping $[1]$ to $[r]$. What does $\varphi^k$ map $[1]$ to? How can you use Lil' Fermat?

answered Mar 29 '17 at 18:39

Bernard

179,256

I don't see how to finish this line of proof. – Tobias Kildetoft Mar 29 '17 at 18:41
@TobiasKildetoft $\phi(1) = r$, and $\phi^k(1) = r^k$. Specifically, $\phi^{p}(1) = r^p = r$ by Fermat's little theorem/Euler's theorem. So, the two authomorphisms $\phi^1$ and $\phi^p$ are the same. – Mark Schultz-Wu Mar 29 '17 at 18:45
@Mark And how does that answer the question asked by the OP? – Tobias Kildetoft Mar 29 '17 at 18:49
You have the know the multiplicative group $(\mathbf Z/p\mathbf Z)^\times$ is cyclic. – Bernard Mar 29 '17 at 18:52
But that it is cyclic is precisely what the OP is trying to prove. – Tobias Kildetoft Mar 29 '17 at 18:54
The O.P. tries to prove the group $\operatorname{Aut}(\mathbf Z/p\mathbf Z)$ is cyclic. Observing it's isomorphic to the multiplicative group $(\mathbf Z/p\mathbf Z)^\times$ is a way to prove it. – Bernard Mar 29 '17 at 18:58
That just delegates the hard part to another way of considering the group. – Tobias Kildetoft Mar 29 '17 at 18:59
Maybe. The O.P. didn't say what he/she was supposed to know about $\mathbf Z/p\mathbf Z$. – Bernard Mar 29 '17 at 19:31

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

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