= Solution
Modulo five, $X^2+26=(X-2)(X+2)$. Choose $\mathfrak p_5=(5,2+\alpha)$; then $(5)=\mathfrak p_5\overline{\mathfrak p}_5$, and $N(2+\alpha)=4+26=30$. The element $2+\alpha$ belongs to $\mathfrak p_2$, $\mathfrak p_3$ and $\mathfrak p_5$. Their product has norm $2\cdot3\cdot5=30$, so
$$
(2+\alpha)=\mathfrak p_2\mathfrak p_3\mathfrak p_5,
\qquad[\mathfrak p_5]=([\mathfrak p_2][\mathfrak p_3])^{-1}.
$$
The imaginary-quadratic <Minkowski bound for ideal classes> is $(2/\pi)\sqrt{104}<7$. Every <ideal class> has an integral representative of norm at most six, whose prime ideal factors lie over two, three or five. Thus these prime ideals generate the <ideal class group>. Conjugate classes are inverses, and the displayed relation eliminates the class over five. The remaining generators have orders two and three, so their product has order six and generates the whole group:
$$
\boxed{\operatorname{Cl}(\mathbb Q(\sqrt{-26}))\cong C_6.}
$$
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