I am looking for an English reference or a proof of the following result due to Holder:
"Let $ f(x) $ be an irreducible polynomial over a real field $ K $. If all the roots of $ f(x) $ are real and expressible by real radicals (in the sense that they lie in some real radical extension of $ K $), then the Galois group of $ f $ over $ K $ is a $ 2 $-group." (quoted verbatim from this article.)
In the linked article, there is a reference, but it is German. I have found a page on this result on planetmath, but their argument depends on this proof, where I do not see how $ L' = F'(\sqrt[n]{\beta}) $ implies that $ L = F(\sqrt[n]{\beta}) $, or even that $ \beta \in F $ in the first place.
I would appreciate it if someone provided an English reference for this result, or even better, posted a proof sketch as an answer.
Note: The planetmath references have made a previous appearance on the site in the comments section of this question, but Gerry Myerson's proof sketch is extremely lacking. (I do not see how to make his argument work, concretely.)
I found a proof of this statement, so I am posting it as an answer for the sake of completion.
The key idea of the proof is to reduce to the case where the splitting field of $ f $ has prime degree, then use the following lemma:
Lemma. Let $ L/K $ be a Galois extension of real fields with odd prime degree $ p $. Then, $ L $ is contained in no real radical extension of $ K $.
Proof. Let $ \alpha \in \mathbb R $ such that $ \alpha^m \in K $ and $ \alpha \notin K $. Since any real radical extension is obtained by successively adjoining prime radicals, we may assume that $ m $ is prime for our argument. If $ \alpha \in L $, then all roots of $ X^m - \alpha^m $ are in $ L $ by normality, thus so are all of the $ m $th roots of unity. Since $ L $ is real, this implies $ m = 2 $, a contradiction since $ 2 $ does not divide the odd prime $ p = [L : K] $.
Thus, $ \alpha \notin L $. It follows that no root of $ X^m - \alpha^m $ can be contained in $ L $ by normality, and thus the polynomial is irreducible in $ L[X] $. Hence $ [L(\alpha) : L] = [K(\alpha) : K] = m $, and by the tower law it follows that $ [L:K] = [L(\alpha) : K(\alpha)] = p $. Thus, for any (finite) real radical extension $ M/K $, we have $ [LM : KM] = p \neq 1 $, which implies $ L \not\subset KM $. QED.
Now, let $ L/K $ be the splitting field of $ f \in K[X] $, with degree not a power of $ 2 $. The degree is then divisible by an odd prime $ p $, and $ G = \textrm{Gal}(L/K) $ contains an element $ \sigma $ of order $ p $. We may assume that $ \sigma(\alpha) \neq \alpha $ for any particular $ f(\alpha) = 0 $ by conjugating by an element of $ G $ if necessary. ($ f $ is irreducible, thus $ G $ acts transitively on its roots.) Looking at the fixed field $ M $ of $ \langle \sigma \rangle $, we find that $ L/M $ is a Galois extension of real fields with odd prime degree $ p $, and we have $ L = M(\alpha) $. By the lemma, $ L $ is contained in no real radical extension of $ M $, thus neither is $ \alpha $. Since $ K \subset M $, this is also the case for real radical extensions of $ K $, and the result follows.