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Considering the fields that are described by: $$\mathbb{F}_i = \mathbb{Z}_2[x] /\langle\mkern 1.5mu p_i(x)\mkern1.5mu\rangle, \enspace i=1,2 $$ where $p_1(x)=x^3 + x + 1$ and $p_2(x)=x^3+x^2+1$

I have to prove that the multiplicative groups $\mathbb{F_i^*}=\mathbb{F_i} \backslash \{0\}$ are isomorphic.

So by substituting 0 and 1 into the two polynomials, we see that they are both irreducible and are the maximal ideal in $\mathbb{Z_2}$. So $\mathbb{F_i}$ is a field.

Elements of $\mathbb{F_i}$ have the form of: $a_0+a_1 x + a_2x^2 + \langle\mkern 1.5mu p_i(x)\mkern 1.5mu\rangle$. So $|\mathbb{F_i}|=2^3=8$ and so is $|\mathbb{F_i^*}|=8-1=7$.

Now the actual question. Can I just say: since both fields have the same number of elements that they are isomorphic? Or do I need to consider something else too?

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    $\begingroup$ Yes, finite fields of the same order are isomorphic. But since you are interested in the multiplicative groups, it might be easier to note that there is only one group of order 7, up to isomorphism. $\endgroup$ – carmichael561 Jul 15 at 14:35
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    $\begingroup$ @carmichael561 Thanks! But now I'm wondering, why would there be only one group of order 7, up to isomorphism? $\endgroup$ – user688708 Jul 15 at 14:47
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    $\begingroup$ For any prime $p$ there is only one group of order $p$, namely the cyclic group. $\endgroup$ – carmichael561 Jul 15 at 15:08
  • $\begingroup$ This thread descsribes an explicit isomorphism between these two fields. Of course, as Carmichael561 said, your question needs less, and technology from the first course on groups suffices. $\endgroup$ – Jyrki Lahtonen Jul 16 at 6:25
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There is only one group of order 7 bcs every element of group has rank which divides order of group when group has finite order. So every element must be order 1 or 7. But only 1 element can be order 1, so all other elements must be order 7, so there is only one group of order 7.

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