I have to show that the symplectic group
$$\mathrm{Sp}(2n,\mathbb{C}):=\{A\in M(2n,\mathbb{C})\mid AJA^{T}=J\},$$
where $J$ denotes the symplectic matrix, is a closed matrix Lie group, e.g. its a closed subgroup of $\mathrm{Gl}(2n,\mathbb{C})$. I have already shown that its a subgroup of $\mathrm{Gl}(2n,\mathbb{C})$.
In order to show thats its closed, I was considering the determinant:
$$\mathrm{det}:\mathrm{Gl}(2n,\mathbb{C})\to\mathbb{C},$$ which is obviously continuous (bec. polynomial). If one looks than at the restriction
$$\mathrm{det}\vert_{\mathrm{Sp}(2n,\mathbb{C})}:\mathrm{Sp}(2n,\mathbb{C})\to\mathbb{C}$$ which is also a conitnuous map, we have that $$\mathrm{det}\vert_{\mathrm{Sp}(2n,\mathbb{C})}^{-1}(\{-1,1\})=\mathrm{Sp}(2n,\mathbb{C})$$ and since $\{-1,1\}$ is closed in $\mathbb{R}$, we know by continuity that $\mathrm{Sp}(2n,\mathbb{C})$ is closed.....
Is this right and/or enough to show closedness?