 Solution

The eleventh cyclotomic field $L=\mathbb{Q}({\xi }_{11})$ hasThe cyclic group $C_{10}$ has exactly one subgroup of each order $1,2,5,10$. The Galois correspondence therefore gives exactly four subfields: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 $\mathbb{Q}⊊\mathbb{Q}(a)\subseteq M$, and the tower law forces $\mathbb{Q}(a)=M$. Thus every such element is a primitive element. The displayed $\eta$ and $\theta$ are the requested examples. Finally, ${\xi }_{11}$ itself is a primitive element for $L/\mathbb{Q}$.