Let $D$ be a central division algebra of finite degree over $\mathbb{Q}$. Let $\mathcal{O}$ be a maximal order of $D$. I am interested what happens to the order if we intersect it with a subalgebra. More precisely, say $\mathbb{Q} \subset A \subset D$ is a $\mathbb{Q}$-subalgebra. Is $\mathcal{O} \cap A$ a maximal order in $A$?
First of all, I believe $\mathcal{O} \cap A$ is an order of $A$ since it is still a ring containing $1$, it generates $A$ over $\mathbb{Q}$ ($\mathbb{Q}(\mathcal{O} \cap A)=\mathbb{Q} \mathcal{O} \cap \mathbb{Q} A = D \cap A = A$), and it is still a finitely generated $\mathbb{Z}$-module as a submodule of $\mathcal{O}$. But I am not sure if maximality is preserved.
For instance, if $\mathbb{Q} \subset A \subset D$ is a tower of number fields with rings of integers $\mathcal{O}_A$ and $\mathcal{O}_D$, then we would certainly have $\mathcal{O}_D \cap A = \mathcal{O}_A$. This would be based on the description of the maximal order as the ring of all integral elements. Yet noncommutative algebras might not have such unique maximal orders.
If $\mathcal{O} \cap A$ in the original setting is not maximal, then what could we say about its index inside a maximal order?
Maximality is certainly not preserved in general. Take $D$ to be the Hamilton quaternions and $\mathcal O$ the quaternionic integers. Any element $d \in D$ whose square is integer but is itself not a quaternionic integer, generates a subfield whose maximal order contains $d$. Because $(i+j)^2 = -2$, we can take $d$ to be an appropriate conjugate of $i+j$: $$\begin{align*} d &= (i+2j) (i+j) \frac1{i+2j} \\ &= \frac1{5}(i+2j)(i+j)(-i-2j) \\ &= \ldots\end{align*}$$ should not lie in $\mathbb Z[i, j]$. Or a simpler construction: any non-integral rational solution $(a, b, c)$ to $a^2+b^2+c^2 = 2$ gives an element $z = ai+bj+ck$ (conjugate to $i+j$) with $z^2=-2$. So for example $(\frac43, \frac13, \frac13)$.
In the degree 4 (quaternion algebra) case, given an order $\mathcal O \subset D$, a quadratic subfield $F$ is called optimally embedded if its maximal order $\mathcal O_F$ lies in $\mathcal O$. There are only finitely many $\mathcal O^\times$-conjugacy classes of optimally embedded fields of given discriminant. So "most" subalgebras $A$ will not intersect $\mathcal O$ in a maximal order.
A good reference is John Voight's book (online) on quaternion algebras: https://math.dartmouth.edu/~jvoight/quat-book.pdf