Show that $\operatorname{Cl}(\mathbb{Z}[\sqrt{-5}])$ is of order 2 using a specific lemma. Show that $\operatorname{Cl}(\mathbb{Z}[\sqrt{-5}])$ is of order 2 using a specific lemma.
$\textbf{Lemma:}$ For a ring of integers $O_k$, $\exists$ positive integer $M$ only depending on $K$ with the following property. Given $\alpha, \beta \in O_k, \beta \neq 0$, there is an integer $t$, $1 \leq t \leq M$ and element $\omega \in O_k$ such that $|N(t\alpha-\omega\beta)| < |N(\beta)|$.
I want to show that using this lemma (proof of this lemma is not needed), that the ideal class group $\operatorname{Cl}(\mathbb{Z}[\sqrt{-5}])$ is of order 2. 
This is similar to the question asked here however the answer did not incorporate the lemma. I would have commented for a clarification but the OP already commented and received no response. Any help is much appreciated.
 A: In GerryMyerson's answer it is shown that for $I$ an ideal there is $\alpha \in I$ such that $N(\alpha) \le 6 N(I)$, thus $(\alpha) = IJ$ for some ideal of norm $N(J) \le 6$.
$J$ is inverse to $I$ in the class group.
Whence it suffices to enumerate the (prime) ideals of norm $\le 6$ to find the (generators of the) class group.
Norm $2$ : $(2,1 + \sqrt{-5})$, norm $3$ : $(3,1+\sqrt{-5}),(3,1-\sqrt{-5})$, norm $5$ : $(\sqrt{-5})$.
And since $(2,1 + \sqrt{-5})^2 = (2)$ and $N(1+\sqrt{-5}) =6$ the class group is $(1),(2,1+\sqrt{-5})$.

If $I$ is an ideal and $t = \lfloor N(I)\rfloor^{1/2}$ then the $a_1+a_2\sqrt{-5}, a_i \in [0,t]$ are $(t+1)^2$ distinct elements so two of them must be equal modulo $I$ ie. $\alpha = b_1+b_2 \sqrt{-5} \in I$ with $|b_1| \le t, |b_2| \le t$ so that $N(\alpha) \le t^2+5t^2 \le 6 N(I)$.
The same argument works in any number field with $t = \lfloor N(I)\rfloor^{1/n},n = [K:\Bbb{Q}]$ and looking at the $\sum_{j=1}^n a_i \gamma_j, a_i \in [0,t]$ such that the $\gamma_j\in O_K$ are $\Bbb{Z}$-linearly independent,  obtaining that $I \ni \alpha = \sum_{j=1}^n b_i \gamma_j,|b_i| \le t$ so $N(\alpha) \le M_K N(I)$, where a $M_K$ can be found from $N_{K/\Bbb{Q}}(\sum_{j=1}^n b_j \gamma_j) = \prod_{i=1}^n (\sum_{j=1}^n b_j \sigma_i(\gamma_j)) $ with $\sigma_i$ the embeddings $K \to \Bbb{C}$ and $|N(\alpha)| = |N_{K/\Bbb{Q}}(\sum_{j=1}^n b_j \gamma_j)| \le (t n\sup_{i,j}|\sigma_i(\gamma_j)|)^n\le M_K N(I), M_K = n^n \sup_{i,j}|\sigma_i(\gamma_j)|^n$

