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I'm trying to find the order and describe a generator of the group $$\mathrm{Aut}_{\mathrm{GF}(2^3)}(\mathrm{GF}(2^{12}))$$

It's clear that the order is 4, but how would you describe the generator? Thanks!

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The generator is the Frobenius automorphism $x\mapsto x^{2^3}$.

In general, the generator of the automorphism group $\mathrm{Aut}_{\mathrm{GF(p)}}(\mathrm{GF}(p^n))$ is the Frobenius map $\mathrm{Frob}$ that maps $x$ to $x^p$. By the Galois correspondence, if you are looking at the automorphism group over $\mathrm{GF}(p^k)$, $k\leq n$, then the generator maps $x$ to $x^{p^k}$.

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Restated and slightly generalized, for any algebraic extension of any $GF(q)$, the generator of the Galois group is $x \mapsto x^q$ -- even for infinite extensions! (but there we have to think in terms of "profinite groups" instead of merely groups) – Hurkyl May 7 '12 at 1:13
For proofs, see Section 3 of this this handout of KCd's. [The proofs there are for a base field of prime order, but the case of order $p^n$ is no harder.) – Dylan Moreland May 7 '12 at 17:26
@Dylan: Or you get them from the Galois correspondence out of the case where the base field is of prime order, like I mention in the second paragraph. – Arturo Magidin May 7 '12 at 17:49
@Arturo Quite right! – Dylan Moreland May 7 '12 at 17:51

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