I am trying to understand the applications of the Galois correspondence and am faced with the following problem:
let $F=\mathbb{Q}(\zeta)$ where $\zeta = \exp(2\pi i/24)$, be a cyclotomic field. First we determine the Galois group of $F$ which I calculated to be $\mathbb{Z}_{24}^\times \cong C_2\times C_2\times C_2$. Then we are asked to find all quadratic and quartic extensions of $\mathbb{Q}$ contained in $F$. By the Galois correspondence this is equivalent to finding subgroups of index $2,4$ and then looking at their fixed fields.
Take for instance the index $4$ subgroup $C_2\times 1\times 1$. I am very confused how to in practice find the fixed field of this subgroup. My idea was to write $$\mathbb{Z}_{24}^\times=(1,5,7,11,13,17,19,23)$$ and then $C_2\times 1\times 1$is generated by $5$, say. So we look for all elements $\zeta^a $ such that $$\zeta^{5a}= \zeta^a$$ I.e. $4a = 0$ mod$(24)$. I.e. $a=6$. But I am unsure if this is even right, or how to apply this in the $C_2\times C_2\times 1$ case.
Your way of reasoning seems not to work correctly. Applying your method to a group $\{1,\tau_2\}$ would yield $a=12$, but $\mathbb{Q}[\zeta^{12}]=\mathbb{Q}$.
However there is another way to find an intermediate field. The idea to find the fixed field of a given subgroup $H$ is interestingly easy: consider $\alpha_H=\sum_{\sigma\in S} \sigma(\zeta)$. Then you can check that $\alpha_H$ is fixed by elements of $H$. Moreover, we can show that $\operatorname{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q}(\alpha_H))=H$ by checking automorphisms that are not of $H$ does not fix $\alpha_H$.
For example, you can see that the fixed field of a group generated by $\zeta\mapsto \zeta^5$ is $\mathbb{Q}(\zeta+\zeta^5)$.