Let $M_n(\mathbb{H})$ be the set of all $n \times n$ matrices with entries in the quaternions $\mathbb{H}$.
For $A=(a_{ij} ) $ let $ A^*=(a^*_{ij} ) $ be the matrix with $a^*_{ij}=\bar{a}_{ij} $, where $ \bar{x} $ denotes the quaternion conjugate of $x \in \mathbb{H} $. Let:
$$ \operatorname{Sp}(n)=\{A \in M_n(\mathbb{H})\mid AA^*=I=A^*A\} $$
Show that $ \operatorname{Sp}(n) $ is a compact topological group
A compact topological space if one such that any open cover $\bigcup_\alpha \cup_\alpha$ of $X$ has a finite subcover $\bigcup_{i=1, \ldots, n} U_{\alpha_i}$ where finitely many indices $\alpha_i$ are from the same index set $A$
Also the Heine Borel criterion states that a space is compact if it is closed and bounded.
I also note that there is no determinant function $M_n(\mathbb{H}) \rightarrow \mathbb{H}$, so how else can we show this