TITLE: Proving that multiplicative groups of two fields F* are isomorphic
QUESTION [3 upvotes]: 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?

REPLY [1 votes]: 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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