= Solution
The identity is symplectic because $I^TJI=J$. For two <symplectic matrices> $M,N$,
$$
(MN)^TJ(MN)=N^T(M^TJM)N=N^TJN=J.
$$
Thus the set is closed under matrix multiplication. Every member is invertible by part (i). Multiplying $M^TJM=J$ on the left by $M^{-T}$ and on the right by $M^{-1}$ gives
$$
(M^{-1})^TJM^{-1}=J,
$$
so it is also closed under inverses. Associativity is inherited from the <general linear group>. These facts prove that \b[the <symplectic group> is a <subgroup> of $\operatorname{GL}(2m,\mathbb R)$].
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