Find generators of a subgroups of a Galois group of a cyclotomic field over rational numbers My homework asks me to match up the subgroups and the subextensions of the field extension $\mathbb{Q}(e^\frac{2\pi i}{72})/\mathbb{Q}$. 
It's easy to find out all the subgroups of $$\mathrm{Gal}(\mathbb{Q}(e^\frac{2\pi i}{72})/\mathbb{Q})\cong(\mathbb{Z}/72\mathbb{Z})^\times\cong (\mathbb{Z}/2\mathbb{Z})^3\times \mathbb{Z}/3\mathbb{Z}$$.
But I am stuck when considering the generators of the corresponding subfields.
Take the subgroup $\{1,\phi_{17}\}$ for example, in which $\phi_{17}(e^\frac{2\pi i}{72})=e^{17\frac{2\pi i}{72}}$.
Denote $\zeta=e^\frac{2\pi i}{72}$; we have $\phi_{17}(\zeta)=\zeta^{17}$ and 
$\phi_{17}^2(\zeta)=\phi_{17}(\zeta^{17})=(\zeta^{17})^{17}=\zeta$.
Here are some fixed points, like $\zeta+\zeta^{17},\zeta^9$. But I have no idea whether it can generate the subextension just with $\zeta+\zeta^{17}$. I know that the 72-nd cyclotomic polynomial is of degree $\phi(72)=\phi(8)\phi(9)=24$ and $1,\zeta,\dots,\zeta^{23}$ can serve as a basis for $\mathbb{Q}(\zeta)$ while the set $\{\zeta^k:\gcd(k,72)=1\}$ can not since $72$ is not square-free.
Here the task is 

Please find the generators of the field $\mathbb{Q}(\zeta)^{\{1,\phi_{17}\}}$ and 
  $\mathbb{Q}(\zeta)^{\{1,\phi_{25},\phi_{49}\}}$.

Thanks in advance.
 A: Notice that $\zeta+\zeta^{17}$ both $\zeta\cdot\zeta^{17}=i$ are fixed by $\phi_{17}$. Since $\mathbb{Q}(\zeta)\supseteq\mathbb{Q}^{\{1,\phi_{17}\}}\supseteq\mathbb{Q}(\zeta+\zeta^{17},i)$ is an extension of degree 2 (characteristic polynomial of $\zeta$ is $x^2-(\zeta+\zeta^{17})x+i$). Since the first two fields can't be equal, we have that $\mathbb{Q}^{\{1,\phi_{17}\}}=\mathbb{Q}(\zeta+\zeta^{17},i)$, because 2 is prime and no such intermediate extension could occour.
Now, for a candidate of generator of $\mathbb{Q}^{\{1, \phi_{25},\phi_{49}\}}$, consider $\zeta\cdot\zeta^{25}\cdot\zeta^{49}=\zeta^3$, which is fixed by $\phi_{25}$, the generator of the subgroup $\{1, \phi_{25},\phi_{49}\}$ (since $\phi_{25}$ just switch things around). So $\mathbb{Q}(\zeta)\supseteq\mathbb{Q}^{\{1,\phi_{25},\phi_{49}\}}\supseteq\mathbb{Q}(\zeta^3)$. Similarly, this extension is of degree 3 and $\mathbb{Q}^{\{1,\phi_{25},\phi_{49}\}}=\mathbb{Q}(\zeta^3)$.
A: Do you really care of primitive elements ?
Given a Galois extension $K/\Bbb{Q}$ and a $\Bbb{Q}$-basis $$K= \sum_{n=1}^N b_n \Bbb{Q}$$ then the subfield fixed by $H$ is $$K^H =  \sum_{n=1}^N Tr_{K/K^H}(b_n) \Bbb{Q}=\Bbb{Q}(Tr_{K/K^H}(b_1),\ldots,Tr_{K/K^H}(b_N))$$

Proof : $Tr_{K/K^H}(x)=\sum_{h\in H} h(x)$ is surjective $K\to K^H$.

Here $b_n = \zeta^{n-1}, N=  24$.
Then you can use the primitive element theorem $$K^H=\Bbb{Q}(\sum_{n=1}^N c_n Tr_{K/K^H}(b_n))$$ for some $c_n\in \Bbb{Q}$ that you can find by checking that no element $\not \in H$ fixes $\sum_{n=1}^N c_n Tr_{K/K^H}(b_n)$.
