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This question already has an answer here:

I am looking for an explicit description of the additive, multiplicative group structure and automorphism group of the field with 729 elements.

Sorry for not being clear. I have no clue what does "explicit description" mean, but I am guessing that the underlying group might be cyclic, but I do not know how to give a proof.

For its automorphism group, What I have tried: 1). Comparing this with the Galois group of this field over $F_3$ which is cyclic. 2). I also know that Aut(Z_3) = Z_2, which is, in general, true for or prime fields. That is why I am guessing that the underlying additive group should be cyclic, then by the same method, I can show that the automorphism group is a cyclic group of order 728 probably.

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marked as duplicate by Somos, Travis, Eevee Trainer, Lord Shark the Unknown abstract-algebra Mar 21 at 6:16

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    $\begingroup$ Please be explicit by what you mean by 'explicit' ...Namely, not the basic structures of the groups, but generators? $\endgroup$ – peter a g Mar 21 at 0:58
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    $\begingroup$ Also what have you tried, and what is the context for this question? $\endgroup$ – jgon Mar 21 at 1:07
  • $\begingroup$ The additive group will not be cyclic; the field will have characteristic $3$, i.e. $x + x + x = 0$ for any $x$ in the field. The multiplicative group will be cyclic, as is always the case for finite fields. $\endgroup$ – Theo Bendit Mar 21 at 1:45
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A field $F$ with $729$ elements is a vector space of dimension $6$ over $\mathbb F_3$ because $729=3^6$. Therefore, the additive group of $F$ is isomorphic to $\mathbb F_3^6$, the cartesian product of six copies of $\mathbb F_3$.

The multiplicative group of every finite field is cyclic. Therefore, the multiplicative group of $F$ is cyclic of order $728$.

$F$ is a simple extension of $\mathbb F_3$ generated by any root $\theta$ of any polynomial of degree $6$ that is irreducible mod $3$. The simplest one if $x^6+x+2$. An automorphism of $F$ must send $\theta$ to another root of the same polynomial. Therefore, the automorphism group of $F$ has order $6$. You just have to decide whether it is $C_6$ or $S_3$.

Alternatively, $F$ is a splitting filed (of $x^{729}-x$ for instance) and so is a Galois extension of $\mathbb F_3$. Since its only proper subfields are $\mathbb F_{3^2}$ and $\mathbb F_{3^3}$, the Galois group must be $C_6$.

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  • $\begingroup$ Why must the automorphism group be of order $6$? $\endgroup$ – zach Mar 21 at 2:31
  • $\begingroup$ @zach, because there are $6$ roots of $x^6+x+2$ in $F$ to choose. $\endgroup$ – lhf Mar 21 at 10:44

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