# Aut($C_p) = C_{p-1}$when $p$ is prime, why?

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

• To 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
• @DietrichBurde this is not a duplicat as it does not show the group is cycclic in that case; one might rather use something like math.stackexchange.com/questions/807290/… as dupe – quid Feb 14 '15 at 16:49
• @DietrichBurde But the earlier post did not ask for a proof that it was cyclic. – Derek Holt Feb 14 '15 at 16:49
• @DietrichBurde well yes, but the point of the question is how to prove this. – quid Feb 14 '15 at 16:50
• @DietrichBurde I cannot see any reference to that fact in the earlier post. – Derek Holt Feb 14 '15 at 16:51

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 $p-1$.