I'm trying to prove that the symplectic group is, indeed, a group. It was easy to show that the operation is closed and the neutral element is inside the group itself, but I'm stuck trying to show that the inverse belongs to $\operatorname{Sp} (n)$.
The group is defined as follows:
$$\text{Sp}(n,\mathbf{R})=\{M\in M_{n\times n}(\mathbf{R}) : M^T\Omega M=\Omega\},$$
where $\Omega$ is defined as an $n \times n$ matrix with $I_{\frac{n}{2}}$ and $-I_{\frac{n}{2}}$ in the non-diagonal inputs, and 0 elsewhere.
The inverse of any element in the group should be
$M^{-1} = \Omega^T M^T J$.
But when computing
$$(M^{-1})^T \Omega M^{-1} = \Omega$$
it follows that
$$(M^{-1})^T \Omega M^{-1} = \Omega^T M J M^T J,$$
and I don't know how to continue. Any hint?
As pointed out in the comment by QiaochuYuan, note that: \begin{alignat}{1} &M^T\Omega M=\Omega \iff \\ &(M^T)^{-1}\Omega M^{-1}=\Omega \iff \\ &(M^{-1})^T\Omega M^{-1}=\Omega \end{alignat} and hence: $$M\in\operatorname{Sp}(n,{\bf R})\iff M^{-1}\in\operatorname{Sp}(n,\bf R)$$