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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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    @Arturo Quite right!2012-05-07