I'm practicing finding class groups, in this case for $K = \mathbb{Q}(\sqrt{-69})$, and found the class number $h_K$ to be 16, with $C_K \cong C_2 \times C_2 \times C_4$ (cyclic product) whereas https://www.lmfdb.org/NumberField/2.0.276.1 says it's 8, with one less $C_2$ factor. I'm wondering if I could get help seeing where I made a mistake. (Also if anyone has tips to speed up/skip steps I'd be more than happy to hear.)
We note $-69 \equiv 3 \mod 4$, hence $\mathcal O_K = \mathbb Z[\sqrt{-69}]$, and we'll write $m(x) = x^2 + 69$.
I think the mistake is probably in one of the paragraphs checking relations between generating elements, but I'll include details before then just in case. So as usual for approaching these problems, we find Minkowski's bound, noting 1 pair of complex embeddings: $$ c_K = \dfrac{4}{\pi} \cdot \dfrac{2!}{2^2} \cdot 2\sqrt{69} \approx 10.58 < 11 $$ so classes of primes lying over $(2), (3), (5), (7)$ generate $C_K$, and we can apply Dedekind's as $[O_k : Z[\sqrt{-69}]] = 1$. Finding $m(x) \mod p$ for corresponding primes we get $$ \begin{align} (2) = P_2^2 & \ \mbox{ where } & P_2 := (2, 1+\sqrt{-69}) \\ (3) = P_3^2 & \ \mbox{ where } & P_3 := (3,\sqrt{-69}) \\ (5) = P_5P'_5 & \ \mbox{ where } & P_5 := (5, 1+\sqrt{-69}) \\ && P'_5 := (5, 1-\sqrt{-69}) \\ (7) = P_7P'_7 & \ \mbox{ where } & P_7 := (7, 1+\sqrt{-69}) \\ && P'_7 := (5, 1-\sqrt{-69}) \\ \end{align} $$
Now for a general element of $\mathcal O_K$, ($a,b \in \mathbb Z$) we have $ N(a+b\sqrt{-69}) = a^2 + 69b^2$ which immediately tells us that each $P_p$ with ideal norm $p \in \{2,3,5,7\}$ is not principal, otherwise $b=0$ and $a^2 = p$, hence the resp. classes of $P_2$ and $P_3$ have order 2. Equally, $[P'_p] = [P_p]^{-1}$ for $p \in \{5,7\}$.
Also, $N(1+\sqrt{-69}) = 70$ and $1+\sqrt{-69} \in P_2,P_5,P_7$, which are all primes (hence coprime) so $(1+\sqrt{-69}) \subseteq P_2P_5P_7$ and noting $N\left(P_2P_5P_7\right) = 70$ we have that $(1+\sqrt{-69}) = P_2P_5P_7$ hence $[P_7] = [P_2]^{-1}[P_5]^{-1}$, so $C_K$ is generated by classes of $P_2$, $P_3$, and $P_5$.
We find the order of $[P_5]$ (which I think is correctly 4, since we need some element of order 4): write $\alpha = a+b\sqrt{-69}$ an arbitrary element.
If $P_5^2 = (\alpha)$, then $N(\alpha)$ = 25 so $b=0$, $a = \pm 5$, so $P_5^2 = (5) = P^5P'_5$, so $P_5 = P'_5$ contradicting that these are different (Dedekind's). If $P_5^3 = (\alpha)$ then $N(\alpha) = 125 = 56+ 69 \not = a^2+69b^2$, contradiction. Now note $N(2+3\sqrt{-69}) = 4+9*69 = 625$, so by unique factorisation of ideals $$ (2+3 \sqrt{-69}) \in \{P_5^4, P_5'^3 P_5, P_5^2 P_5'^2, P_5 P_5'^3, P_5'^4 \} $$ and that $(2+3 \sqrt{-69}) \not = (25) = P_5^2P_5'^2$, so taking classes $$ [\mathcal O_K] \in \{[P_5]^4, [P_5]^2, [P_5]^{-2},[P_5]^{-4}\} $$ and since we showed $[P_5]$ doesn't have order 2, it must have order 4.
Checking for relations
Now comes the part where I probably made a mistake, but don't know where. $P_2P_3$ has norm 6, and $6 \not = a^2+69b^2$, so is not principal. Hence $$ H := \{[\mathcal O_K], [P_2], [P_3], [P_2][P_3]\} $$ is a subgroup of $C_K$ of 4 distinct elements (since if $[P_2] = [P_3]$ then $[P_3] = [P_2]^{-1}$ hence $[P_2][P_3] = [\mathcal O_K]$, and if $[P_2][P_3] = [\mathcal O_K]$ then $P_2P_3$ is principal).
Similarly, $$ \begin{align} P_2P_5 & \mbox { has norm 10} \\ P_3P_5 & \mbox { has norm 15} \\ P_2P_3P_5 & \mbox { has norm 30} \\ P_2P_5^2 & \mbox { has norm 50} \\ P_3P_5^2 & \mbox { has norm 75} \\ \end{align} $$ And none of these norms can be of the form $a^2 + 69b^2$ so none of these are principal, so $[P_5]^2 \not \in H$, so as before we must have a larger subgroup containing $[P_5]^2$ then a larger one containing $[P_5]$ yielding $$ \begin{align} C_K = \langle [P_2] \rangle \times \langle [P3] \rangle \times \langle [P_5] \rangle \cong C_2 \times C_2 \times C_4 \end{align} $$