Let $e_1,\ldots,e_{2n}$ a canonical basis for $\mathbb{R}^{2n}$ and consider $E=Span(e_1,\ldots,e_{n})$. $O(2n)$ is the orthogonal group with Lie algebra $\mathfrak{o}(2n)$. Define $$C=\{A\in GL(2n;\mathbb{R})| A(E)\subset E\}.$$
(A) Prove that $C$ is a closed subgroup of $GL(2n;\mathbb{R})$.
(B) Let $\mathfrak{c}=Lie(C)$. Show that $\mathfrak{c}=\{X\in M(2n;\mathbb{R})|X(E)\subset E\}$.
(C) Show that $\mathfrak{o}(2n)+\mathfrak{c}=M(2n;\mathbb{R})$.
(D) Deduz that the map $$O(2n)\times C \to GL(2n;\mathbb{R}), \quad(B,A)\mapsto BA^{-1}$$ has surjective derivative at $(I,I)$.
Here is what I've done:
(B) $\mathfrak{c}=\{X\in \mathfrak{gl}(2n;\mathbb{R})|\forall t\in \mathbb{R},\exp(tX)\in C\}$ and let $$V=\{X\in M(2n;\mathbb{R})|X(E)\subset E\}$$ We have to prove that $V=\mathfrak{c}$.
1) $V\subset \mathfrak{c}$ Let $X\in V$, i.e. $X(E)\subset E$, let us prove that $\exp(tX)in C$, for all $t\in \mathbb{R}$
$$\exp(tX)=I+tX+\dfrac{t^2}{2}X^2+\dfrac{t^3}{6}X^3+\cdots$$ then
$$\exp(tX)(E)=I(E)+(tX)(E)+\left(\dfrac{t^2}{2}X^2\right)(E)+\left(\dfrac{t^3}{6}X^3\right)(E)+\cdots$$
$$\exp(tX)(E)=E+t(X(E))+\left(\dfrac{t^2}{2}\right)X^2(E)+\left(\dfrac{t^3}{6}\right)X^3(E)+\cdots.........(*)$$
Since $X(E)\subset E$, then $X^n(E)\subset E$ for all $n\in\mathbb N$, then in (*) we get $$\exp(tX)(E)\subset E,\forall t\in\mathbb{R}\Rightarrow \exp(tX)\in C, \forall t\in\mathbb{R}.$$
Therefore $X\in \mathfrak{c}$, i.e. $V\subset \mathfrak{c}$.
2) $\mathfrak{c}\subset V$ Let $X\in \mathfrak{c}$, then $\exp(tX)\in C, \forall t\in\mathbb{R}$, then $\exp(tX)(E)\subset E$, then $\gamma_X(t)(E)\subset E$, for all $t\in\mathbb{R}$.
$$\gamma_X(t)(e_k)=a_1(t)e_1+...+a_n(t)e_n$$ then $$\gamma'_X(0)(e_k)=a_1'(0)e_1+...+a_n'(0)e_n\Rightarrow X(e_k)a_1'(0)e_1+...+a_n'(0)e_n\in E$$ which gives $$X(E)\subset E,$$ then $X\in V$, i.e, $\mathfrak{c}\subset V$.
From 1) e 2) we get $\mathfrak{c}=V$
(C) If $X\in\mathfrak{c}$, then $X(E)\subset E$, it means, that the subspace $E$ is $X-$invariant, and I think that I have to use some property of invariant subspaces, is it right?
Thank you
(A) $C$ is closed because the elements of $C$ are those invertible matrices $(a_{ij})_{1\leqslant i,j\leqslant2n}$ such that $a_{ij}=0$ when $i>n$ and $j\leqslant n$.