= Solution
The <eleventh cyclotomic field> $L=\mathbb Q(\xi_{11})$ has
$$
\operatorname{Gal}(L/\mathbb Q)
\cong(\mathbb Z/11\mathbb Z)^\times\cong C_{10}.
$$
The cyclic group $C_{10}$ has exactly one subgroup of each order $1,2,5,10$. The <Galois correspondence> therefore gives exactly four subfields:
$$
\begin{array}{c|c|c}
M&[M:\mathbb Q]&\text{an explicit generator}\\ \hline
\mathbb Q&1&-\\
\mathbb Q(\sqrt{-11})&2&
\eta=\xi_{11}+\xi_{11}^3+\xi_{11}^4+\xi_{11}^5+\xi_{11}^9\\
\mathbb Q(\xi_{11}+\xi_{11}^{-1})&5&
\theta=\xi_{11}+\xi_{11}^{-1}\\
L&10&\xi_{11}.
\end{array}
$$
The <Quadratic Gaussian period in the eleventh cyclotomic field> satisfies $\eta=(-1+\sqrt{-11})/2$, while $\theta$ generates the <maximal real subfield of the eleventh cyclotomic field>. Every subgroup of the abelian Galois group is normal, so all four subfields are Galois over $\mathbb Q$.
Each nontrivial proper subfield has prime degree, two or five. If $a\in M\setminus\mathbb Q$, then $\mathbb Q\subsetneq\mathbb Q(a)\subseteq M$, and the tower law forces $\mathbb Q(a)=M$. Thus every such element is a <primitive element of a field extension>[primitive element]. The displayed $\eta$ and $\theta$ are the requested examples. Finally, $\xi_{11}$ itself is a primitive element for $L/\mathbb Q$.
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