infinitely many Abelian extensions of $K$ whose Galois group is $G$. Let $K$ be a finite extension of $\mathbb{Q}$ and $G$ be a finite Abelian group. Show that there exist infinitely many Abelian extensions of $K$ whose Galois group is $G$.
This a exercise from Lang and I'm stuck on it without any idea, can one please help ? 
 A: We can write $G= \prod_{i=1}^{k} Z_{n_i}$. Now using the fact that there are infinitely many primes $p$ of form $(n_i-1)|p$ for all $i$, we get infinitely many tuples of primes $S_j=(p_{j,1},p_{j,2},..,p_{j,k})$ such that $n_i-1|(p_{j,i}-1)$ for all $1\leq i\leq k$. Now we can make a infinite set $I$ such that for every distinct pair $i,j\in I$ we've $S_i\cap S_j=\mathbb{Q}$. Now, there is $E_j \subset \mathbb{Q}(\omega_{\prod p_{j,i}})$ such that $Gal(E_{j}/\mathbb{Q})=G$. 
To do this you need to use the Galois correspondence and the fact that $Gal(\mathbb{Q}(w_n)/Q)$ is Abelian for all $n$ so, any intermediate extension is normal and hence Galois over $Q$. Now note that $E_i \cap E_j=\mathbb{Q}$ for any distinct $i,j \in I$. 
Now $Gal(KE_i/K)=Gal(E_i/K\cap E_i)$ for all $i\in I$. Now since $Q \subset E_i \cap K \subset K$ and $[K:Q] <\infty$ so exist a infinite subset $I'$ of $I$ such that $E_i\cap K=E_j\cap K$ for any $i,j \in I'$. Bug then $E_i\cap K= (E_i\cap K)\cap (E_j\cap K)=K\cap (E_i\cap E_j)=K \cap \mathbb{Q}=\mathbb{Q}$ for any $i\neq j \in I'$. 
So, then for any $i\in I'$ we've $Gal(KE_i/K)=Gal(E_i/\mathbb{Q})=G$. Now as $KE_{i}\cap KE_j=K$ and $I'$ is infinite so we're done.
