 Solution

Put $w=\sqrt{-14}$. The ring of integers is $\mathbb{Z}[w]$, with discriminant $-56$. The Minkowski bound for ideal classes for an imaginary quadratic field says that every ideal class has an integral representative of norm at most $(2/\pi )\sqrt{56}<5$. Therefore the ideal class group is generated by primes above two and three: norm-four representatives contribute only products of primes above two.

The prime two ramifies: ${\mathfrak{p}}_2=(2,w)$ has norm two and ${\mathfrak{p}}_2^2=(2)$. It is not principal, since an element of norm two would require $a^2+14b^2=2$, which has no integral solution. Thus $[{\mathfrak{p}}_2]$ has order two. Since $X^2+14=(X-1)(X+1)$ modulo three, three splits into ${\mathfrak{p}}_3=(3,w-1)$ and its conjugate, whose classes are inverses.

The element $2+w$ has norm $18$. It belongs to ${\mathfrak{p}}_3$ but not its conjugate, and its norm has exactly one factor two. Unique prime ideal factorization therefore givesHence $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]\ne 1$. Since all generating classes are powers of $[{\mathfrak{p}}_3]$, $\boxed{\mathrm{Cl}(\mathbb{Q}(\sqrt{-14}))\cong \mathbb{Z}/4\mathbb{Z}}$.