# Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$

I'm trying to prove that Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$ (p prime).

I know that Aut $\mathbb Z_p$ has $p-1$ elements because $\mathbb Z_p$ has $p-1$ possiblities of generators, so intuitively I see Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$, but I couldn't prove it formally.

I'm trying to build an isomorphic function, but I don't how to do it.

Am I in the right way? Maybe I'm forgetting some trick or something.

Thanks

• Yes, you are on the right way. You want to show $Aut(Z_p)$ is a cyclic group of p-1 elements. So you have the elements of $Aut(Z_p)$ are f_1,f_2,f_3,..f_{p-1} where f_i maps 1 to i. How do you make this into a group? – Hugo Apr 14 '13 at 20:53
• @LongMai yes I was thinking in this function, but how can I define the functions f_1,...,f_{p-1}? – user42912 Apr 14 '13 at 21:02
• Just in case I wasn't sufficiently clear about this: this is really an important theorem masquerading as an exercise. – Pete L. Clark Apr 14 '13 at 21:04
• @PeteL.Clark strange because it's an item in a question of the first chapter of Hugerford's Book, how can I solve this without knowing the theory of ring and fields? – user42912 Apr 14 '13 at 21:08
• @user42912: existence of primitive roots is usually covered in elementary number theory texts. You could consult one for a proof couched in less algebraic language. (My elementary number theory text does it exactly as I said it in my answer, unfortunately...) – Pete L. Clark Apr 14 '13 at 21:16

What you've said so far is correct, but the fact that the group is cyclic of order $p-1$ is significantly harder. You'll need some real idea here; it's not a matter of just following your nose.

One standard approach is to prove the Cyclicity Criterion: a group $G$ of finite order $n$ is cyclic $\iff$ for each divisor $d$ of $n$ there are at most $d$ elements $x$ with $x^d = 1$ (the identity element). This is generally proved using a counting argument, and it is helpful to know that $\sum_{d \mid n} \varphi(d) = n$, where $\varphi(d) = \# (\mathbb{Z}/d\mathbb{Z})^{\times}$ is the number of integers $1 \leq e \leq d$ with $\gcd(e,d) = 1$.

With that in hand you should observe that $\mathbb{Z}/p\mathbb{Z}$ is a field, and thus the number of roots of the polynomial $x^d - 1 = 0$ in it is at most $d$. This argument ends up showing a little more: any finite subgroup of the multiplicative group of a field is cyclic.

P.S.: If I tell you that the result is often called "existence of primitive roots modulo $p$", that may help you look it up in many standard introductory texts on algebra and/or number theory.

• I noticed my post is essentially a duplicate of yours (I guess I started writing at the same time, but finished later). Is it good form to delete my post? – AlexM Apr 14 '13 at 21:04
• Do you know where I can find this "Cyclicity Criterion"? thank you for your help – user42912 Apr 14 '13 at 21:26
• For instance, see Appendix B of math.uga.edu/~pete/4400FULL.pdf. – Pete L. Clark Apr 14 '13 at 21:39
• thank you again, really useful criterion! – user42912 Apr 15 '13 at 7:12

Take an automoprhism, show it must come from multiplication by nonzero in $\mathbb{Z}_p$ by looking at what it does to 1 and by using that it is a homomorphism. Then show that this group is cyclic (primitive roots and all that jazz). This comes down to existence of primitive roots, which is where the rub is. Then you will have that the automorphism group is cyclic of $p-1$ elements, whence the answer.

Here's an another method. We need to prove that Aut $\mathbb Z_p\simeq \mathbb Z_{p-1}$ (p prime).

We know that Aut $\mathbb Z_p\simeq U(p)$

By Gauss's result, we know that when $p$ is prime, then $U(p^k) \simeq Z_{\phi(p^k)}$

Hence, $U(p) \simeq Z_{p-1}$

Since, isomorphism follows transitivity laws : $=> Aut ~ \mathbb Z_p\simeq \mathbb Z_{p-1}$ when $p$ is prime.