Let $E$ be a linear space and $V$ a linear subspace. No assumption is made on the dimension of $E$.
We write $G = \{g \in GL(E) \mid g(V) = V\}$ where $GL(E)$ is the group of invertible linear maps of $E$. For $x \in E$ we denote $[x] = x + V$ the class of $x$ in the quotient space $E/V$.
We have a group homomorphism $$\begin{array}{l|rcl} \pi : & G & \longrightarrow & GL(E/V) \\ & g & \longmapsto & \pi(g) \end{array}$$ defined by $\pi(g)([x]) = [g(x)]$
I proved that this is indeed a group homomorphism.
We then define the group homomorphism $$\begin{array}{l|rcl} p : & G & \longrightarrow & GL(V) \times GL(E/V) \\ & g & \longmapsto & (\left.g\right|_V, \pi(g)) \end{array}$$
And for $f \in \mathcal L(E/V,V)$ (a linear map from $E/V$ to $V$), the map $\phi_f : E \to E$ is defined by $\phi_f(x) = x + f([x])$. I proved that $\phi_f$ is linear, invertible (of inverse $\phi_{-f}$) and that $\phi_f$ belongs to $\ker(p)$.
I now have to prove that the map $\phi : \mathcal L(E/V, V) \to \ker(p)$ is a group isomorphism. I prove that this is indeed a homomorphism, that it is injective... but I'm not able to prove that it is onto. Any idea for that?
The ultimate goal of the problem is to write $G$ as a semi-direct product $$\mathcal L(E/V, V) \rtimes_\rho (GL(V) \times GL(E/V)) \simeq G$$
To show that $\phi$ is onto, we take some arbitrary $g \in \ker(p)$, and we look for some $f \in \mathcal{L}(E/V,V)$ such that $\phi_f = g$. This means that for any $x \in E$, we must have $\phi_f(x) = g(x)$, i.e. $x + f([x]) = g(x)$. The only choice we have is then to take $f$ such that $f([x]) = g(x)-x$. It remains to show that this is well-defined, as the valuation of $f$ may not depend on the representative $x \in [x]$, and its image must lie in $V$.
To this end, take firstly some arbitrary $v \in V$ such that $x+v$ is another representative of $[x]$, and note that $g(x+v)-(x+v) = (g(x)-x)+(g(v)-v)$. Now, since $g \in \ker(p)$, we must have $(g|_V,\pi(g)) = (\text{Id}_V,\text{Id}_{E/V}))$, and in particular $g(v)=v$. Hence, $g(x+v)-(x+v) = g(x)-x$, which shows that $f([x])$ does not depend on the representative $x \in [x]$, and hence $f$ is well-defined as a linear map from $E/V$ to $E$.
To show that $f$ maps $E/V$ into $V$, we use the fact that $\pi(g) = \text{Id}_{E/V}$, meaning $[g(x)] = [x]$ for any $x \in E$. From this it follows that $[g(x)-x] = [0] = V$, so that indeed $f$ maps into $V$.