Consider following Lie Group: $$ \text{Sp}(2n,\mathbb{C})=\{g\in\text{Mat}_{2n}(\mathbb{C})\mid J=g^TJg\}\quad\ where\quad J=\begin{pmatrix} 0 & 1_n \\ -1_n & 0 \end{pmatrix} $$ And the corresponding Lie Algebra: $$ \text{sp}(2n,\mathbb{C})=\{g\in\text{Mat}_{2n}(\mathbb{C})\mid g^TJ+Jg=0\} $$
Are there any basic proofs that $\text{Sp}(2n,\mathbb{C})$ is a Lie Group and that $\text{sp}(2n,\mathbb{C})$ is the corresponding Lie Algebra without using submersions (seen here: Why is $Sp(2m)$ as regular set of $f(A)=A^tJA-J$, and, hence a Lie group.)? Can I proove the first statement by showing that $\text{Sp}(2n,\mathbb{C})$ is a subgroup of $\text{GL}(n,\mathbb{C})$?
Thanks!
On the Lie algebra part. By definition, $sp(2n,\mathbb C)$ is the real vector space of complex matrices
$$sp(2n,\mathbb C)=\{ A\in Mat_{2n}(\mathbb C): \exp(tA)\in Sp(2n,\mathbb C), \forall t\in \mathbb R \}.$$
$\exp(tA)$ denotes the exponential of the matrix $tA$. By definition of $Sp(2n,\mathbb C), $ we have that
$$J=\exp(tA)^{t}J\exp(tA)=\exp(tA^{t})J\exp(tA), (*)$$
where the second equality can be proved using the definition of the exponential of a matrix.
Using ($\exp(0)=1$)
$$\frac{d\exp(tA)}{dt}|_{t=0}:=\lim_{t\rightarrow 0}\frac{\exp(tA)-1}{t}= \lim_{t\rightarrow 0}\frac{(1+tA+\frac{1}{2!}t^2A^2+\dots)-1}{t}=A$$
we arrive at
$$0=(\text{using (*)})=\frac{d}{dt}(J-\exp(tA)^{t}J\exp(tA))|_{t=0}=0-A^{t}J+JA,$$
i.e. $A^{t}J+JA=0$, as $J$ is independent of $t$.