For which $p$ does $\mathbb{Q}(\zeta)$ have a unique maximal proper subfield 
Let $p$ be a prime number, $\zeta\in\mathbb{C}$ a primitive $p^\text{th}$ root of unity and $K = \mathbb{Q}(\zeta).$ Determine those $p$ for which $K$ has a unique maximal proper subfield $k\subsetneq K.$

I'm not confident on this, but here are my thoughts.
We know that $K$ is the splitting field for $x^p-1$. Let $G = \mathrm{Gal}(K/\mathbb{Q}).$ Then $$|G| = [K:\mathbb{Q}] = \varphi(p) = p-1.$$ From the Galois correspondence, I believe we are looking for groups $G$ such that $G$ has a minimal nontrivial subgroup. For $p = 2,$ $G = \{e\}$, and for $p=3$, $G\cong\mathbb{Z}_2$ whose only proper subgroup is $\{e\}$. Otherwise, for $p\geq 5,$ $2\mid|G|,$ so $G$ has a subgroup of order $2$ by Cauchy. And since all groups of order $2$ are isomorphic, they correspond to subfields of $K$ which are isomorphic. Hence there is a unique maximal subfield of $K$ (up to isomorphism).
 A: Your analysis is good up until the sentence "since all groups of order $2$ are isomorphic...". The question asks about unique maximal subfields, not unique maximal subfields up to isomorphism, so though you're right that the corresponding field extensions will be isomorphic, $G$ may yet contain distinct subgroups of order $2$, which will correspond to distinct subfields of index $2$. Consider, for example, the Galois extension $K := \mathbb{Q}(\xi, \sqrt[3]{2})$ over $\mathbb{Q}$, where $\xi$ is a primitive third root of unity. 
This extension has Galois group $S_{3}$, and $S_{3}$ has $3$ elements (and hence distinct subgroups) of order $2$. This corresponds to the isomorphic (but distinct!) maximal subfields $\mathbb{Q}(\sqrt[3]{2}), \mathbb{Q}(\xi\sqrt[3]{2}), \mathbb{Q}(\xi^{2}\sqrt[3]{2}) \subset K$. 
More data is needed, and it comes from the fact that the $G := \mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ is cyclic of order $p-1$, so the subgroup of order $2$ is in fact unique. The corresponding subfield is the so-called maximal real subfield, $\mathbb{Q}(\zeta + \zeta^{-1}) \subset \mathbb{Q}(\zeta)$. It is a nice, short exercise to prove that $[\mathbb{Q}(\zeta):\mathbb{Q}(\zeta + \zeta^{-1})] = 2$ when $p \geqslant 5$, as you describe.
Edit: Prof Lubin points out in the comments that this answer assumes the interpretation that maximal = ''of maximal degree'' and not ''maximal with respect to inclusion''. The latter is likely the intended interpretation of the question, so I will summarize the comments below which give a more complete answer to the question. 
As noted above $G := \mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ is cyclic of order $p-1$, so there is a unique (cyclic) subgroup of order $d$ for each divisor $d$ of $p-1$. This subgroup of order $d$ corresponds with a subfield $k$ of $K$ of index $[K:k] = d$. For two subgroups $H_{1}, H_{2}$ of $G$ of order $d_{1}, d_{2}$ respectively with $d_{1} \mid d_{2}$, we have an inclusion of subgroups $H_{1} \subset H_{2}$, we have a (reverse) proper inclusion $k_{1} \supset k_{2}$ of the corresponding subfields. It is clear, then, that there is a maximal subfield of $K$ of index $q$ for each distinct prime $q$ dividing $p-1$, and these are all such maximal subfields. Since $p-1$ is always divisible by $2$, the only time $K$ has a unique maximal subfield is when $p-1$ is a power of $2$, i.e. $p = 2^{n}+1$ for some $n \in \mathbb{N}$.  
