Is SL_n of an order in a number ring finite-index in SL_n of the number ring?

Question

Let $\mathcal{O}$ be the ring of integers in an algebraic number field and let $R \subset \mathcal{O}$ be an order. For instance, we might have $\mathcal{O} = \{\text{$x+i y$ $|$ $x,y \in \mathbb{Z}$}\}$ and $R = \{\text{$x+i y \in \mathcal{O}$ $|$ $y$ even}\}$. Question : Is $\text{SL}_n(R)$ a finite-index subgroup in $\text{SL}_n(\mathcal{O})$? I might be missing something obvious here, but I'm having trouble proving this (maybe because it isn't true?).

Answer 1

Yes. If $f = (\mathcal O : R)$, then $f\mathcal O \subset R$ and consequently $\text{SL}_n(R)$ contains $\ker (\text{SL}_n(\mathcal O) \to \text{SL}_n(\mathcal O/f\mathcal O))$, which has finite index because $\mathcal O/f\mathcal O$ is finite.