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I want to show, that $\mathbb{F}_p^\times \simeq \text{Aut}(\mathbb{F}_p^+)$ holds with $p$ prime.
($\mathbb{F}_p^+$ is the additive group, $\mathbb{F}_p^\times$ multiplicative group)

As hint we got the following homomorphism $$ \Phi :\mathbb{F}_p^+ \rightarrow \text{Aut}(\mathbb{F}_p^+), \ g \mapsto (x \mapsto g+x+(-g)) $$ $\text{kern}(\Phi)=\mathbb{F}_p^+$ because $\mathbb{F}_p^+$ is abelian and $\text{im}(\Phi)=\{\text{id}\}$.

I thought to create another homomorphism $$ \Phi' : \mathbb{F}_p^\times \rightarrow \text{Aut}(\mathbb{F}_p^+), \ g \mapsto (x \mapsto g \cdot x) $$ It is clear that $\text{kern}(\Phi')=\{\overline{1}\}$ and it follows $\mathbb{F}_p^ \times \simeq \text{im}(\Phi') \subseteq \text{Aut}(\mathbb{F}_p^+)$.

How do I know, that there are not more automorphisms?

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    $\begingroup$ You ought to write something like ${\mathbb F}_p^+$ for the additive group. By default, ${\rm Aut}({\mathbb F}_p)$ denotes the automorphism group of the field of order $p$. If $p$ is prime, then this group is trivial. $\endgroup$
    – Derek Holt
    Aug 15, 2014 at 16:19
  • $\begingroup$ @DerekHolt : thanks for your comment. I also forgot to mention, that $p$ have to be prime. I also fixed the notation. $\endgroup$
    – DerJFK
    Aug 15, 2014 at 18:24

2 Answers 2

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HINT: The additive group $\Bbb F_p$ is cyclic, generated by any non-zero class. Now, an automorphism of $\Bbb F_p$ is a homomorphism $$ f:\Bbb F_p\longrightarrow\Bbb F_p $$ and a homomorphism between cyclic groups is completely determined by the image of a single chosen generator.

This should convince you that the map $\Phi^\prime$ you defined is surjective.

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$\mathbf{F}_p$ is a cyclic group. You know a lot about group homomorphisms whose domain is a cyclic group.

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