Put $s=\sqrt{-6}$, so $\mathcal O_K=\mathbb Z[s]$ and $\mathcal O_K^\times=\{\pm1\}$. The primes $2$ and $3$ are ramified:
$$
(2)=\mathfrak p_2^2,
\qquad
(3)=\mathfrak p_3^2.
$$
Thus the finite modulus in the question is
$$
\mathfrak m=\mathfrak p_2^2\mathfrak p_3^2=(6).
$$

The <ray class exact sequence>, with no real component in the modulus, gives
$$
\mathcal O_K^\times\longrightarrow
(\mathcal O_K/(6))^\times\longrightarrow
\operatorname{Cl}_{\mathfrak m}(K)\longrightarrow
\operatorname{Cl}(K)\longrightarrow1.
$$
The <Chinese remainder theorem for unit groups> and the ramification relations yield
$$
\begin{aligned}
(\mathcal O_K/(6))^\times
&\simeq(\mathcal O_K/(2))^\times\times(\mathcal O_K/(3))^\times\\
&\simeq\bigl(\mathbb F_2[\epsilon]/(\epsilon^2)\bigr)^\times
\times\bigl(\mathbb F_3[\epsilon]/(\epsilon^2)\bigr)^\times\\
&\simeq C_2\times C_6.
\end{aligned}
$$
The image of $-1$ kills the $C_2$ coming from $\mathbb F_3^\times$. Consequently the kernel of the map from the ray class group to the ordinary class group is
$$
(\mathcal O_K/(6))^\times/\{\pm1\}\simeq C_2\times C_3\simeq C_6.
$$
Since the given <class number> is two, $|\operatorname{Cl}_{\mathfrak m}(K)|=12$.

It remains to distinguish $C_{12}$ from $C_6\times C_2$. Let
$$
\mathfrak q=(5,s-2).
$$
The prime $5$ splits in $K$, and $\mathfrak q$ represents the nontrivial ordinary ideal class because no element of $\mathbb Z[s]$ has norm $5$. Direct multiplication, or comparison of norms and valuations at the two primes over $5$, gives
$$
\mathfrak q^2=(1+2s).
$$
Modulo $2$, the element $1+2s$ is $1$. Modulo $3$, its class $1+2\epsilon$ is nontrivial and has order three because $\epsilon^2=0$. Hence the ray class of $\mathfrak q$ has order six: its square is a nontrivial element of order three in the congruence kernel. The remaining order-two factor of that kernel, coming from $(\mathcal O_K/(2))^\times$, is independent of $\langle[\mathfrak q]\rangle$. Therefore
$$
\boxed{\operatorname{Cl}_{\mathfrak m}(\mathbb Q(\sqrt{-6}))\simeq C_6\times C_2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 123 / 4 / ii / Solution

For a finite modulus $\mathfrak m$, there is an exact sequence
$$
\mathcal O_K^\times\longrightarrow
(\mathcal O_K/\mathfrak m)^\times\longrightarrow
\operatorname{Cl}_{\mathfrak m}(K)\longrightarrow
\operatorname{Cl}(K)\longrightarrow1.
$$
With real places in the modulus, the residue-unit term and the unit map also record the required signs.


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 Incoming links: Ray class exact sequence