# The number of automorphisms of a finite field

Let $M$ be a finite field and $|M| = p^s$, where $p$ is prime and $s \in \mathbb N$. Prove that the number of different isomorphisms field $M$ to $M$ equal to $s$ and this isomorphisms form a cyclic group under composition of isomorphisms.

We know that the multiplicative group of finite field is cyclic and all cyclic groups the same order are isomorphic. Because of this if I do not mistake the number of different isomorphisms $M^{\times}$ to $M^{\times}$ equal to the number of primitive elements in $M^{\times}$. And it is equal to $\phi(p^s - 1)$ where $\phi(n)$ - Euler function. But we should consider not only $M^{\times}$ but also the addition and I do not know what to do with it and how to connect it with the multiplicative group.

Thanks for the help!

• Hint: As you noted, an automorphism must be in particular an automorphism of the multiplicative group. These have a certain shape, so check which ones respect the addition. – Tobias Kildetoft Dec 24 '14 at 13:29
• The automorphism group is always cyclic, generated by the Frobenius automorphism $x\to x^p$. – Jihad Dec 24 '14 at 13:37
• @TobiasKildetoft Can you explain about what shape do you say? I have no idea. I know that the multiplicative group is cyclic and abelian, no more. – Stanislav Morozov Dec 24 '14 at 15:48
• Have you looked in any books that discuss finite fields or searched on the internet for "automorphism finite field"? – KCd Dec 28 '14 at 17:40