When considering the set of matrices:
$$Sp(n) = \{ S \in \text{GL} (2n, \mathbb{R}) \hspace{2mm} \text{s.t.} \hspace{2mm} S^T \Omega S= \Omega\} \tag{1}$$
where
$$\tag{2} \Omega = \begin{pmatrix} 0 & \mathbb{I}_{n} \\ - \mathbb{I}_{n} & 0\end{pmatrix}$$
with $\mathbb{I}_{n}$ being the $n \times n$ identity matrix and $0$ denotes the zero matrix of appropriate dimension.
Show that $Sp(n)$ forms a group under matrix multiplication (assuming associativity).
When trying to prove closure I took two matrices into account, labeled $S_1$ and $S_2$.
$$S_1 \in GL(2n, \mathbb{R}) \hspace{5mm} ; \hspace{5mm} S_2 \in GL(2n, \mathbb{R})$$
and therefore $S_1 \cdot S_2 \in GL(2n, \mathbb{R})$
But then I know I must show that these matrices respect the equation above, $S^T \Omega S= \Omega$. How do I do this?
I thought that I needed a representation (general) of an $S$ matrix that had a $det(S) \neq 0 $ but how do I make a general matrix (using letter and not numbers) that clearly shows having a $det \neq 0 $?
Must I also show that $\Omega$ belongs to $GL(2n, \mathbb{R}$? Or is that assumed?
It seems your question is: show that $Sp(n)$ forms a group under matrix multiplication.
However, when you want to prove the "closure" you do not have to take $S_1$ and $S_2$ in $GL(2n)$, but in $Sp(n)$.
How to do it: take $S_1 \in Sp(n)$ and $S_2 \in Sp(n)$. Is it true that $S_1 S_2 \in Sp(n)$? Let's try:
$$ (S_1 S_2)^T \Omega (S_1 S_2) = S_2^T S_1^T \Omega S_1 S_2 = S_2^T (S_1^T \Omega S_1) S_2 = S_2^T \Omega S_2 = \Omega $$
This proves that $S_1 S_2 \in Sp(n)$.