Prove that the group $\operatorname{Aut}(\mathbb{F}_{p^n}^*)$ has order $\varphi(p^n-1)$. We denote $\mathbb{F}_{p^n}^*$ the multiplicative group of the finite field $\mathbb{F}_{p^n}$ and $\varphi$ Euler's arithmotheoretic function.

We know that $\mathbb{F}_{p^n}$ is an extension of the field $\mathbb{F}_p$ with $p^n$ cardinality and the extension is Galois and $\operatorname{Gal}(\mathbb{F}_{p^n}/\mathbb{F}_p)= \langle \sigma \rangle$ where $\sigma$ is the Frobenius automorphism.

Also the multiplicative group of $\mathbb{F}_{p^n}$ is cyclic.

Can someone help me to combine these facts to solve this problem?

Thank you in advance!

  • $\begingroup$ It is not always cyclic, for instance $\operatorname{Aut}(\mathbb{Z}/8\mathbb{Z})$ is elementary abelian of order $4$, as $1^{2} \equiv 3^{2} \equiv 5^{2} \equiv 7^{2} \equiv 1 \pmod{8}$. $\endgroup$ – Andreas Caranti Dec 3 '16 at 23:34
  • $\begingroup$ What can i do for the finite field i mentioned.. $\endgroup$ – Marios Gretsas Dec 4 '16 at 0:33
  • $\begingroup$ The problem is that im not so familiar with group theory! $\endgroup$ – Marios Gretsas Dec 4 '16 at 0:33
  • $\begingroup$ @AndreasCaranti: woof, you're right, thanks. Certainly, it's still of order $\phi(n)$, but I was distracted and thought only of the prime case. $\endgroup$ – Alex Wertheim Dec 4 '16 at 1:16
  • $\begingroup$ You know that $\operatorname{Aut}(\mathbb{F}_{p^n}^*) \cong \mathbb{Z}/(p^n - 1)\mathbb{Z}$. What are the automorphisms of $\mathbb{Z}/m\mathbb{Z}$? Note that a map $\mathbb{Z}/m\mathbb{Z} \to \mathbb{Z}/m\mathbb{Z}$ is determined by the image of $1$. What elements of $\mathbb{Z}/m\mathbb{Z}$ generate the group? $\endgroup$ – Richard D. James Dec 4 '16 at 1:27

$\mathbb{F}_{q}^*$ is a cyclic group of order $q - 1$ where $q = p^n$. Since any two cyclic groups of equal order are isomorphic, we know that $\mathbb{F}_{q}^* \cong \mathbb{Z}/(q - 1)\mathbb{Z}$.

Now it is not too hard to show that $\text{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong U(n)$ where $U(n)$ is the group of units under multiplication mod $n$, which when coupled with the statement above, establishes the claim. In order to understand why $\text{Aut}(\mathbb{Z}/n\mathbb{Z}) \cong U(n)$, think about the image of a generator of $\mathbb{Z}/n \mathbb{Z}$ under an automorphism...


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