$K$ be a finite extension of $\mathbb Q$ then there exists a finite extension $E$ over $K$ such that $E/K$ is not normal. 
Let $K$ be a finite extension of $\mathbb Q$. Then I have to show that there exists a finite extension $E$ over $K$ such that $E/K$ is not normal.

It is clear that $E/\mathbb Q$ is not normal as well.
I tried taking $K=\mathbb Q(\alpha)$, $|K:\mathbb Q|=n$ and consider $L=\mathbb Q(\beta)$,  where $\beta$ is root of the irreducible polynomial $X^m-2$. $m$ is chosen to be relatively prime with $n$, so that the compositum has degree $mn$, i.e., $|KL:\mathbb Q|=mn$, where $KL=\mathbb Q(\alpha,\beta)$. Then by natural irrationality $KL/L$ is Galois of degree $n$. I tried to show $KL$ play the role for $E$, but I couldn't pull it back to the other chain.
Any help will be highly appreciated.Thank You.
 A: Let $n := [K : \Bbb Q]$. Let $p$ be a prime with $p - 1 > n$.
Let $\alpha = 2^{1/p} \in \Bbb R$. Note that $[\Bbb Q(\alpha) : \Bbb Q] = p > n$ and thus, $\alpha \notin K$. Let $E := K(\alpha)$. Note that in particular, we have
$$p \mid [E : \Bbb Q]. \tag{1}$$
Claim. $E/K$ is not normal.
Proof. For the sake of contradiction, assume that $E/K$ is normal.
Consider the polynomial $f = X^p - 2 \in K[X]$. Note that $\alpha$ satisfies $f$.
Let $g \in K$ be the minimal (irreducible) polynomial satisfied by $\alpha$ over $K$. Then, we can write
$$f = g h$$
for some $h \in K[X]$.
Let $d := \deg(g) \le p$. Then, we have $[E : K] = d$.
Since $\alpha \notin K$, it follows that $g$ is not linear. By assumption of normality, we see that $g$ has another root $\beta \in E$.
Define $\gamma := \beta/\alpha \in E$. Then, we have $\gamma \neq 1$ and $\gamma^p = 1$. Thus, $\gamma$ is a primitive $p$-th root of $1$ and hence, $$(p - 1) \mid [E : \Bbb Q]. \tag{2}$$
(This is because $[\Bbb Q(\gamma) : \Bbb Q] = p - 1$.)
From $(1)$ and $(2)$, we conclude that $$p(p - 1) \mid [E : \Bbb Q]. \tag{3}$$
On the other hand, we have
$$[E : \Bbb Q] = [E : K][K : \Bbb Q] = dn < p(p - 1),$$
contradicting $(3)$. $\Box$
