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Let $E$ be a finite Galois extension of $\mathbb{Q}$ of degree $pq^m$, where p and q are prime such that $p<q$. I need to prove that every irreducible polynomial over $\mathbb{Q}$ that splits in $E$ is solvable by radicals.

For a given $f$ that's irreducible and splits in $E$, we know that $f(x)=0$ is solvable by radicals if and only if its Galois group is solvable. Since $E/\mathbb{Q}$ is of degree $pq^m$ and Galois, we know $|Gal(E/\mathbb{Q})|=pq^m$, and its easily shown that any group of such order is solvable. Is this problem really that simple or am I forgetting something?

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Ah, I believe I was one step in logic short. To show that $f$ is solvable by radicals, I need the splitting field of $f$ over $\mathbb{Q}$, let's call it $K$, to be solvable by radicals. So we have the tower of power $\mathbb{Q}\subset K\subset E$. Now we know that $E/\mathbb{Q}$ is Galois, and the Galois group is solvable. Furthermore, since our prime field is char $0$, and $K$ is the splitting field over $\mathbb{Q}$, then $K/\mathbb{Q}$ is Galois. Since $[E:\mathbb{Q}]=[K:\mathbb{Q}][E:K]$, one can see that regardless of the degree $[E:K]$, the resulting group $Gal(K/\mathbb{Q})$ has order $p^iq^n$, where $i\in \{0,1\}$ and $n\leq m$. This will still be solvable however, so we're done.

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