Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 2 / b

The given $J$ satisfies $J^2=-I_{2m}$ and is invertible. Taking determinants in the defining identity of a symplectic matrix givesSince $\det (M^T)=\det M$ and $\det J\ne 0$, cancellation yields $(\det M{)}^2=1$. HenceIn particular every symplectic matrix is invertible. This argument establishes the requested alternatives; the later parts explicitly supply the determinant-one hypothesis that they use.

The identity is symplectic because $I^{T}JI=J$. For two symplectic matrices $M,N$,Thus the set is closed under matrix multiplication. Every member is invertible by part (i). Multiplying $M^{T}JM=J$ on the left by $M^{-T}$ and on the right by $M^{-1}$ givesso it is also closed under inverses. Associativity is inherited from the general linear group. These facts prove that the symplectic group is a subgroup of $\mathrm{GL}(2m,\mathbb{R})$.