Find a normal extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}_5$ Q) Find a normal extension $E$ of $\mathbb{Q}$ with $\operatorname{Gal}(E/\mathbb{Q}) = \mathbb{Z}_5$ and a primitive element $a\in E$ s.t. $E = \mathbb{Q}(a)$. Hint: Realize $E$ as a subextension of some cyclotomic extension. 
I'm not sure how to start constructing subextension of a cyclotomic extension with the constraint that $\operatorname{Gal}(E/\mathbb{Q}) = \mathbb{Z}_5$. A hint is appreciated.
 A: Recall that, for any $n > 2$, the cyclotomic extension $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ has degree $\varphi(n)$ where $\varphi$ is Euler's totient function, and its Galois group is isomorphic to $(\mathbb{Z}/n\mathbb{Z})^\times$ (by sending an automorphism $g \in Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ to $m$ if $g(\zeta_n) = \zeta_n^m$).
In particular, the Galois group is cyclic (for example) whenever $n$ is prime. Your problem reduces to finding a number $n$ such that $(\mathbb{Z}/n\mathbb{Z})^\times$ is cyclic and with  $5 \vert \varphi(n)$. 
(Do you see how you can find your normal extension from there, and why it must be normal? Use the Fundamental Theorem of Galois Theory).
As for finding the primitive element -- invoke the Fundamental Theorem once again: the element must be invariant under any $\mathbb{Q}$-automorphism of $\mathbb{Q}(\zeta_n)$ of order $\varphi(n)/5$. Choose your $n$ wisely (prime is always nice) and the calculations will be straightforward.
Edit: Elaborating with regard to your questions below. I'll use $F := \mathbb{Q}(\zeta_{11})$.
Now, $F$ is Galois over $\mathbb{Q}$ because it is the splitting field of the minimal polynomial of $\zeta_{11}$. Its Galois group (I'll call it $G$) is isomorphic to $C_{10}$. By the fundamental theorem, there exists an inclusion-reversing bijection between the subfields of $F$ over $\mathbb{Q}$ and the subgroups of $G$.
A subfield of $F$ is normal if and only if the corresponding subgroup $U \le G$ is normal in $G$. (I'm sure you can look up the proof in your notes or textbook, if needed). In any case, that proves normality of all subfields of $F$, since $G$ is abelian and thus every subgroup is normal.
Now for the primitive element: in accordance with what I wrote above, we're not looking for an element of multiplicative order $2$, (that would just be $-1$), we are looking for the fixed field $E$ of the unique subgroup $H \le G$ with $H \cong C_2$, because the fundamental theorem of Galois theory tells us that this subfield will satisfy $[F : E] = |H| = 2$, and then $[E:\mathbb{Q}] = 5$ follows using the tower rule. Furthermore, $Gal(E/\mathbb{Q}) \cong G/H \cong C_{10}/C_2 = C_5$.
We can see that $\zeta_{11} + \zeta_{11}^{-1}$ works, for example by showing that the minimal polynomial of $\zeta_{11}$ over $\mathbb{Q}(\zeta_{11} + \zeta_{11}^{-1})$ has degree $2$, but I'll show you a more structural approach: remember we are searching the fixed field of $H \le G$, $H\cong C_2$. What is the automorphism of order $2$ in $G$? Well, all automorphisms of $F$ are given by $\zeta_{11} \mapsto \zeta_{11}^m$ for some $m \in \mathbb{Z}$. One can thus easily see that 
$$
\sigma: F \to F, \zeta_{11} \mapsto \zeta_{11}^{-1}
$$
is the automorphism we're looking for. To get an invariant element we can take any element in $F$ and take the sum of all its images under automorphisms in $\langle \sigma \rangle$; this is the heuristic for finding the element $\xi := \zeta_{11} + \zeta_{11}^{-1}$. To show that $\xi$ is not contained in any subfield of $E$ it will suffice to show that $\tau(\xi) \ne \xi$ for any automorphism $\tau$ of order $5$. Since $\sigma$ is a non-trivial automorphism which does fix $\xi$, we cannot have $\mathbb{Q}(\xi) = F$ and the only remaining possibility is $\mathbb{Q}(\xi) = E$.
A: The Galois group of  $x^5+x^4-4x^3-3x^2+3x+1$ over $\Bbb Q$ is isomorphic to $C_5$, see here:
Prove that the Galois group of $x^5+x^4-4x^3-3x^2+3x+1$ is cyclic of order $5$
Actually, we have $[\mathbb Q(\zeta_{11}+\zeta_{11}^{-1}) : \mathbb Q] = 5$, and the Galois group is $C_5$.
The Galois group of $x^5-3$ over $\Bbb F_{11}$ is isomorphic to $C_5$, see here:
How to find irreducible polynomials over $\mathbb{Q}(i)$ with prescribed Galois group?
A: Let $\zeta=\zeta_{25}$ be a primitive $25$-th root of unity. The cyclotomic field $K=\mathbf Q(\zeta)$ is a cyclic extension of $\mathbf Q$ of degree $\phi(25)=20 = 5.4$. A generator of $G=Gal(K/\mathbf Q)$ is any $\mathbf Q$-automorphism $s_m$ which sends $\zeta$ to $\zeta^m$ with $m$ coprime to $20$, e.g. $s_3$. Then ${s_3}^5=s_{3^5}$ has order $4$ and fixes the unique (cyclic) subextension $k/\mathbf Q$ of degree $5$.
To get hold of a primitive element $\alpha$ of $k$ (i.e. $k=\mathbf Q(\alpha)$), introduce the subfield $K^+=\mathbf Q(\zeta + {\zeta}^{-1})$ of $K$ fixed by the complex conjugation, i.e. by ${s_3}^{10}=s_{3^{10}}$. Introduce also the subfield ${\mathbf Q(\zeta_5)}^+=\mathbf Q(\zeta_5 + {\zeta_5}^{-1})$ of $\mathbf Q(\zeta_5)$, which is classically known to coincide with $\mathbf Q(\sqrt 5)$ because $5 \equiv 1 $ mod $4$. Composing with $k$ and writing $\theta =\zeta + {\zeta}^{-1}$ for simplicity, we get $\mathbf Q(\theta)=\mathbf Q(\alpha, \sqrt 5)$, so that $\alpha=x\theta + y\sqrt 5$, with $x,y\in \mathbf Q$. Reciprocally, any such $\alpha$ with non null $x,y$ will be a primitive element because the degree $5$ is prime.
NB. (1) The approach works for any prime $p$>3 in place of $5$ ; (2) Using a bit of Kummer theory, one can characterize all the cyclic extensions of $\mathbf Q$ which are cyclic of degree $p$.
