Aut($C_p) = C_{p-1}$ when $p$ is prime, why?
I don't see why this result follows, I'm sure it's obvious though. I appreciate that Aut($C_p)$ will be of order p-1
Asked
Active
Viewed 802 times
1
-
It's easy to show that the order is $p-1$, but it's a little harder to show that is is cyclic. – Derek Holt Feb 14 '15 at 16:33
-
1To show it is cyclic, note that ${\rm Aut}(C_p)$ is isomorphic to the multiplicative group of the finite field ${\mathbb Z}_p$, and multiplicative groups of finite fields are cyclic. – Derek Holt Feb 14 '15 at 16:38
-
possible duplicate of $\operatorname{Aut}(\mathbb Z_n)$ is isomorphic to $U_n$.. See also http://math.stackexchange.com/questions/1070057/automorphisms-in-z-n. – Dietrich Burde Feb 14 '15 at 16:45
-
1@DietrichBurde this is not a duplicat as it does not show the group is cycclic in that case; one might rather use something like http://math.stackexchange.com/questions/807290/proof-of-existence-of-primitive-roots as dupe – quid Feb 14 '15 at 16:49
-
2@DietrichBurde But the earlier post did not ask for a proof that it was cyclic. – Derek Holt Feb 14 '15 at 16:49
-
It is a duplicate, since $U(C_p)=C_p^{\ast}$ is cyclic. – Dietrich Burde Feb 14 '15 at 16:50
-
2@DietrichBurde well yes, but the point of the question is how to prove this. – quid Feb 14 '15 at 16:50
-
Well, OK, I see. Still, I think this question also has appeared at MSE. – Dietrich Burde Feb 14 '15 at 16:51
-
1@DietrichBurde I cannot see any reference to that fact in the earlier post. – Derek Holt Feb 14 '15 at 16:51
-
Here it is: http://math.stackexchange.com/questions/625633/automorphism-group-of-bf-z-p. – Dietrich Burde Feb 14 '15 at 16:52
2 Answers
2
Pick a fixed generating element $e$. Every endomorphism $f$ is uniquely determined by the image of this generating element $f(e)$.
The image of the generating element $f(e)$ must be generating to have an automorphism. There are thus $p-1$ possibilities for $f(e)$; every element but the neutral one.
To show that the group is cyclic it is best to consider $C_p$ as a field. Derek Holt already mentioned a general result that would imply this.
To prove it denote by $t$ the maximal multiplicative order; each non-zero elements multiplicative order is a divisor of it. So each nonzero element is a solution of $X^t=1$. Yet since $X^t-1$ can have at most $t$ roots, we get $t=p-1$, and thus an element of order $p-1$.
quid
- 42,835
-
-
Thank you for pointing this out; I did not read the question sufficiently carefully. – quid Feb 14 '15 at 16:35
1
Write $C_p=\{e,x,x^2,...,x^{p-1}\}$ where $e$ is the identity element. An automorphism is by definition an isomorphism and an endomorphism, which can be thought of as a relableing of $C_p$ such that $f(xy)=f(x)f(y)$ (where $f$ is any given automorphism in question). Let's construct a specific automorphism as follows: Let $f$ be an automorphism. Then by properties of homomorphisms, $f(e)=e$. Now assume $f(x)=x^a$ where $a$ is a primitive root. Then $f(x^n)=f(x)^n=(x^a)^n=x^{an}$. Then by composing $f$ with itself, we get that since $a$ is a primitive root, for all $0<n<p$ there is an $0<m<p-1$, $(f^m)(x)=x^n$, hence this process gives all automorphisms and we can check that all of these compositions of $f$ are automorphisms.
Isaac
- 11
- 1