Galois Group of the Hilbert Class Field Let $K/\mathbb{Q}$ be a number field with Galois group G and let $L/K$ be the Hilbert class field of $K.$ It is easy to show that $L$ is Galois over $\mathbb{Q}$ and I am interested in knowing this Galois group. 
We know that the class group of $K$ is isomorphic to the Galois group $L/K$ so I would assume that $\operatorname{Gal}(L/\mathbb{Q})\cong cl K \rtimes G.$ However this (somewhat unrelated) MO post (esp. Kevin Buzzards comment on Felipe Voloch's answer) seems to indicate otherwise. Can somebody shine light on the situation?
Thanks.
 A: C. S. Herz (Construction of Class Fields,  Seminar on complex multiplication (Chowla et al. eds.) (1957), Lecture Notes Math. 21, Springer Verlag) has claimed  that the answer to your question is positive. The first counterexample was provided by Wyman (Hilbert class fields and group extensions,  Scripta math. 29 (1973), 141-149). Simpler proofs and more examples were given later by


*

*R. Gold, Hilbert class fields and split extensions, 
Ill. J. Math. 21 (1977), 66-69

*G. Cornell, M. I. Rosen, A note on the splitting of the Hilbert class field, 
J. Number Theory 28 (1988), 152-158

*R. Bond, On the splitting of the Hilbert class field, 
J. Number Theory 42 (1992), 349-360

A: $\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$In general, it is not true that $Gal(L/\QQ) \cong Gal(K/\QQ) \ltimes \mathrm{Cl}(K)$. 
What is true
Suppose that we have a short exact sequence of groups 
$$0 \to A \to H \to G \to 0$$
where $A$ is abelian. Since $A$ is normal, the group $H$ acts on $A$ by conjugation; since $A$ is abelian, this action factors through $H/A \cong G$. So, in any such setting, we get an action of $G$ on $A$. 
In your notation, Galois theory gives us 
$$0 \to \mathrm{Cl}(K) \cong Gal(L/K) \to Gal(L/\QQ) \to Gal(K/\QQ) \to 0.$$
It turns out that this action of $Gal(K/\QQ)$ is the same as the natural action of $Gal(K/\QQ)$ on the class group. I don't know a reference for this, but it's not hard to see that $\sigma \cdot \mathrm{Frob}_{\mathfrak{p}} \cdot \sigma^{-1} = \mathrm{Frob}_{\sigma \mathfrak{p}}$ for an unramified prime $\mathfrak{p}$ of $K$ and an element $\sigma \in G$. The result follows from this plus an extremely weak version of Cebataorv density (just enough to know that the values of Frobenius generate the Galois group).
However, this does not mean that the short exact sequence is semidirect and, as we will see below, it need not be.
A strategy for finding a counterexample
Suppose that we could construct a Galois extension $F/\QQ$, with Galois group $H$, and an abelian subgroup $A$ of $H$ such that
(1) The sequence $0 \to A \to H \to H/A \to 0$ did not split and
(2) For every prime $\mathfrak{p}$ of $F$ (including the infinite primes), the inertia group $I_{\mathfrak{p}}$ is disjoint from $A$.
I claim that taking $K$ to be the fixed field of $A$ will give a counterexample. 
The extension $F/K$ is abelian (since $A$ is abelian) and unramified (by the condition on inertia groups). So we have $F \subseteq L$ and we have a commuting diagram
$$\begin{matrix}
0 &\to& \mathrm{Cl}(K) &\longrightarrow& Gal(L/\QQ) &\longrightarrow& Gal(K/\QQ) &\to & 0 \\
& & \downarrow & & \downarrow & & \| & & \\
0 &\to& A &\longrightarrow& Gal(F/\QQ) &\longrightarrow& Gal(K/\QQ) &\to & 0 \\
\end{matrix}$$
Suppose for the sake of contradiction that we had a map $Gal(K/\QQ) \to Gal(L/\QQ)$ splitting the top sequence. Then the composite map $Gal(K/\QQ) \to Gal(L/\QQ) \to Gal(F/\QQ)$ would split the bottom sequence, contradicting (1). 
This makes it clear why counterexamples are hard to find. Since $\QQ$ has no unramified extensions, one of the $I_\mathfrak{p}$ must be nontrivial. So, just on the level of group theory, we need to find a non-semidirect short exact sequence $0 \to A \to H \to G \to 0$, with $A$ abelian and a nontrivial subgroup $I \subset H$ so that $I \cap A = \{ e \}$. If $I \to G$ is surjective, this implies that the sequence is semidirect after all. Just on the group theory level, we see that there are no examples with $G$ a cyclic group of prime order. 
A counterexample
Let $\zeta$ be a primitive $85$-th root of unity. Recall that $$Gal(\QQ(\zeta)/\QQ) \cong (\ZZ/85)^{\ast} \cong (\ZZ/5)^{\ast} \times (\ZZ/17)^{\ast} \cong (\ZZ/4) \times (\ZZ/16).$$ 
We will always write elements of this group as ordered pairs in $(\ZZ/4) \times (\ZZ/16)$. 
Let $F$ be the fixed field of $\{ 0 \} \times (4 \ZZ/16)$. So $Gal(F/\QQ) \cong (\ZZ/4) \times (\ZZ/4)$. The only ramified primes are $5$, $17$ and $\infty$ with inertia groups $\ZZ/4 \times \{0 \}$, $\{ 0 \} \times \ZZ/4$ and $(2,0)$ respectively. Let $A$ be the order $2$ subgroup generated by $(2,2)$. We see that $A \cap I_{\mathfrak{p}}$ is trivial for every $\mathfrak{p}$. Finally, we have $A \cong \ZZ/2$, $H \cong (\ZZ/4) \times (\ZZ/4)$ and $H/A \cong (\ZZ/4) \times (\ZZ/2)$, so the sequence is not semidirect.
I have not actually computed $\mathrm{Cl}(K)$ for this example. 
One can build similar counterexamples whenever $H \cong (\ZZ/q^2)^2$ for some prime $q$, being a little careful about the prime at $\infty$ if $q=2$.
Smaller counterexample 
I messed around a bit more, and concluded that there is nothing stopping me from taking $H$ to be the dihedral group of order $8$ and $A$ to be its (two-element) center. For example, let $F$ be the splitting field of $\QQ(\sqrt{1+8 \sqrt{-3}})$ (so $K = \QQ(\sqrt{-3}, \sqrt{193})$). If I haven't made any dumb errors, the only ramified primes are $3$, $193$ and $\infty$, and the inertia groups are all noncentral two-element subgroups. Even if I got this particular example wrong, I'm pretty sure there is no general obstacle to making an example like this.
A: Although by itself this doesn't directly answer the question, it might be worth mentioning that the extension
$$1 \to Cl_K \to Gal(L/\mathbb Q) \to Gal(K/\mathbb Q) \to 1$$
is classified by an element in
$$H^2(Gal(K/\mathbb Q), Cl_K),$$ and this element is precisely the image
under the natural surjection $K^{\times}\setminus \mathbb A^{\times}_K \to Cl_K$ of the fundamental class (as constructed in global class field theory) in
$$H^2(Gal(K/\mathbb Q), K^{\times} \setminus \mathbb A^{\times}_K).$$
(Here $K^{\times} \setminus \mathbb A^{\times}_K$ is the idele class group of $K$.)
Thus the possibility of the first-mentioned extension being non-split is related to the fact that the fundamental class is non-trivial.

In principle you can use the above to compute examples.  If $K/Q$ has degree $n$, then class field theory tells us that $H^2(Gal(K/Q), K^{\times}\setminus \mathbb A_K^{\times})$ is cyclic of order $n$, generated by the fundamental class.  Now in an explicit example, we could compute this $H^2$ directly (this is a finite computation, since the Galois group is finite), and then taking a generator, consider whether or not it becomes trivial after we map the coefficients down to $Cl_K$.  
This is easiest when $K$ is cyclic, since then $H^2 = \widehat{H}^0$, and so the map on $H^2$ can be rewritten as $$\mathbb Q^{\times}\setminus \mathbb A_Q^{\times}/N_{K/\mathbb Q}\mathbb A_K^{\times} \to
Cl_K^{Gal(K/\mathbb Q)}/N_{K/\mathbb Q} Cl_K.$$
In fact, since this map factors through the class group of $\mathbb Q$, which is trivial, we see that it is necessarily trivial, and so the original extension is split when $K/\mathbb Q$ is cyclic.  (In the case of prime degree cyclic extensions, this was observed in David Speyer's answer; there is probably a simple argument using inertia groups in this more general case as well.)
I didn't try to compute a non-trivial example this way, though ... .
