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Let $\sigma : \mathbb{F}_{243} \rightarrow \mathbb{F}_{243}$ be define by $$\sigma(x)=x^{80}$$

I have to check whether $\sigma$ is an automorphism of $\mathbb{F}_{243}$.

Since $243=3^{5}$, I think that it is not, as $80$ cannot be expressed as a power of 3. What would be a legit proof about this?

Thanks for any help.

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It is enough to show that $2^{80} = \sigma(1+1) \neq \sigma(1) + \sigma(1) = 2$ in $\mathbb{F}_{243}$. If it is, we have $2^{79} = 1$ mod 243. However, we also have $2^{242} = 1$, and since $79$ and $242$ are coprime, we may get $2 = 1$. This is not true!

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  • $\begingroup$ Thank you for the answer. Forgive if these are trivial questions, but why does $2^{242}=1$ and how do we get $2=1$ from $79$ and $242$ being coprime? $\endgroup$ – gauss134 Jul 20 at 9:37
  • $\begingroup$ Never mind, I figured it out. Thanks again. $\endgroup$ – gauss134 Jul 20 at 10:25

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