Setting up a Galios Group with the roots of $x^7 - 1$ in $\mathbb{Q}[x]$. From Pinter's A Book of Abstract Algebra, Chapter 32.
Let: $p(x)=x^7 - 1$
$x^7 - 1 = (x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.
Let: $q(x) = x^6+x^5+x^4+x^3+x^2+x+1$.
Let: $\omega$ be any root of $q(x)$.
The splitting field ("root field" in this book) of $q(x)$ is $\mathbb{Q}$($\omega$).
FIND the members of $\;Gal(\Bbb Q(\omega)/\Bbb Q)$ (or "$Gal(\Bbb Q(\omega):\Bbb Q)$" in the book I'm using).
Then FIND the subgroups of $\;Gal(\Bbb Q(\omega)/\Bbb Q)$ and FIND the associated fixfields.
$q(x)$ is the minimal polynomial of $\Bbb Q(\omega)$ over $\Bbb Q$. $p(x)$ is seventh order and is prime, so the degree of $\Bbb Q(\omega)$ over $\Bbb Q$ is $7-1=6$. Therefore, there exist six automorphisms h:$\Bbb Q(\omega)$ to $\Bbb Q(\omega)$ which fix $\Bbb Q$. Therefore $\Bbb Gal(\Bbb Q(\omega)/\Bbb Q)$ has six elements.
All members of $\;Gal(\Bbb Q(\omega)/\Bbb Q)$ are automorphisms which permute $\;\omega$, $\;\omega^2, \omega^3, \omega^4, \omega^5, \omega^6$ and fix $\Bbb Q$. Let h be any member of $\;Gal(\Bbb Q(\omega)/\Bbb Q)$. Each automorphism is fully determined by the choice of h($\omega$) (this is true only because $x^7-1$ has a prime order). The members of $\;Gal(\Bbb Q(\omega)/\Bbb Q)$:
\begin{matrix} \epsilon\\ \end{matrix} \begin{bmatrix} \omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6\\ \omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6\\ \end{bmatrix}
\begin{matrix} \alpha\\ \end{matrix} \begin{bmatrix} \omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6\\ \omega^2&\omega^4&\omega^6&\omega&\omega^3&\omega^5\\ \end{bmatrix}
\begin{matrix} \beta\\ \end{matrix} \begin{bmatrix} \omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6\\ \omega^3&\omega^6&\omega^2&\omega^5&\omega&\omega^4\\ \end{bmatrix}
\begin{matrix} \gamma\\ \end{matrix} \begin{bmatrix} \omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6\\ \omega^4&\omega&\omega^5&\omega^2&\omega^6&\omega^3\\ \end{bmatrix}
\begin{matrix} \delta\\ \end{matrix} \begin{bmatrix} \omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6\\ \omega^5&\omega^3&\omega&\omega^6&\omega^4&\omega^2\\ \end{bmatrix}
\begin{matrix} \zeta\\ \end{matrix} \begin{bmatrix} \omega&\omega^2&\omega^3&\omega^4&\omega^5&\omega^6\\ \omega^6&\omega^5&\omega^4&\omega^3&\omega^2&\omega\\ \end{bmatrix}
$\;Gal(\Bbb Q(\omega)/\Bbb Q)$ is cyclic, with $\beta$ as a generator (maybe some others can act as generator as well, I didn't check them all).
The proper subgroups of $\;Gal(\Bbb Q(\omega)/\Bbb Q)$ consist of one isomorphic to $\Bbb Z_2$ and one isomorphic to $\Bbb Z_3$. The degree of $\Bbb Q(\omega)$ over $\Bbb Q$ is six. The number of elements in any subgroup $H$ is equal to the degree of $\Bbb Q(\omega)$ over the fixfield of $H$.
And vice-versa, so since $\Bbb Z_2$ has two elements, the degree of $\Bbb Q(\omega)$ over the fixfield associated with the subgroup isomorphic to $\Bbb Z_2$ is two. Similarly for the other fixfield, with a degree of three.
What I'm not understanding is via the definion of a fixfield:
Letting $H$ be a subgroup of $\;Gal(\Bbb Q(\omega)/\Bbb Q)$, and letting h be any member of $H$, the fixfield associated with $H$ is defined as:
{$x\in\Bbb Q(\omega) : h(x)=x\;\forall\;h\in\;H$}
$\alpha(x)$≠$x$ for any $x=\omega^n$ (n=1,2,3,4,5,6)
$\beta(x)$≠$x$ for any $x=\omega^n$ (n=1,2,3,4,5,6)
$\gamma(x)$≠$x$ for any $x=\omega^n$ (n=1,2,3,4,5,6)
$\delta(x)$≠$x$ for any $x=\omega^n$ (n=1,2,3,4,5,6)
$\zeta(x)$≠$x$ for any $x=\omega^n$ (n=1,2,3,4,5,6)
Only the identity element fixes $\;\omega$, $\;\omega^2, \omega^3, \omega^4, \omega^5, \omega^6$.
That leads to my question.
Since $h(x)$≠$x\;\forall\;h\in\;H$ except $h=\epsilon$, then as long as any subgroup contains any h≠$\epsilon$ then the fixfield would be $\;\Bbb Q$. That doesn't seem right though, and it contradicts the above, as [$\Bbb Q(\omega)$ :$\Bbb Q$)] is six.
I've done a similar problem with $Gal(\mathbb{Q}(\sqrt2, \sqrt3, \sqrt5)/\mathbb{Q})$ but in that case, I was able to multiply the extension elements to obtain elements such as $\sqrt15$ which are members of $\mathbb{Q}(\sqrt2, \sqrt3, \sqrt5)$ but not members of $\Bbb Q$. In the current problem, $\omega^n\omega^m$ = $\omega^k$ (k=0,1,2,3,4,5 or 6) for any integers n and m.
I tried setting up intermediate subfields such as $Q(i)$ or $Q(cos(2\pi/7))$ but those don't even seem to be actual subfields. I can't isolate $i$ from $Q(cos(2\pi/7)-sin(2\pi/7))$ without assuming square roots exist for elements.