Finding the fixed subfield (Galois theory) Let's say we are working with the field extension $\mathbb{Q}(\gamma)$, where $\gamma$ is the seventh root of unity.  I know my basis for this extension will thus be: 
$\{1, \gamma, \gamma^2, \gamma^3, \gamma^4, \gamma^5, \gamma^6 \}$ 
And the Galois group will consist of automorphisms that take the generator $\gamma$ and map it to it's different powers.  Since this group has order $6$,  I know I will have a subgroup of order $2$ and a subgroup of order$3$.  The subgroup of order $2$ will be: 
$<\sigma_2>$ where $\sigma_2$ sends $\gamma$ to $\gamma^2$
And the subgroup of order 3 will be: 
$<\sigma_6>$ where $\sigma_6$ sends $\gamma$ to $\gamma^6$
My goal is to find the subfield that is fixed by, say, the subgroup $<\sigma_2>$
What I did was take an arbitrary element of $\mathbb{Q}(\gamma)$ and acted on it with $\sigma_2$.  The arbitrary element $x$ will just be a linear combination of the basis elements of $\mathbb{Q}(\gamma)$, thus: 
$x = a + b\gamma + c\gamma^2 + d\gamma^3 + e\gamma^4 + d\gamma^5 + e\gamma^6$ 
Then: 
$\sigma_2(x)$ = $\sigma_2(a) + \sigma_2(b\gamma) + \sigma_2(c\gamma^2) + \sigma_2(d\gamma^3) + \sigma_2(e\gamma^4) + \sigma_2(d\gamma^5) + \sigma_2(e\gamma^6) = a + b\gamma^2 + c\gamma^4 + d\gamma^6 + e\gamma + f\gamma^3 + g\gamma^5$ 
Comparing coefficients, we see that $a = a$ and $b = e, c = b, d = f, e = c, d = g, e = d$
When we did a similar exercise in class, I remember things simplified much more nicely.  From the looks of it, my calculation shows that things get shuffled around, so the only things fixed are the coefficients from $\mathbb{Q}$, but I know that's not right.  What am I doing wrong?  

Your above writings imply that $\mathbb{Q}(\zeta)$ has degree $7$ over $\mathbb{Q}$. What is wrong?
Try writing $\zeta^{6}$ in terms of $1,\dots,\zeta^{5}$. Remember the minimal polynomial for $\zeta$.
