It is well known that given $G$ a group, $N$ a normal subgroup of $G$ and $H$ a subgroup of $G$, then the following are equivalent:
- There exists a group homomorphism $\phi: G \rightarrow H$ such that $\phi|_H=id$ and $\mbox{Ker}(\phi)=N.$
- It is satisfied that $G=NH$ and $H\cap N=\{e\}.$
When any of those requirements is satisfied we write $$ G=N\rtimes H $$ and we say that $G$ is the semidirect product of $N$ and $H$.
Example:
If we know consider a subgroup $H$ of $GL(n)$ we can consider the subgroup $G$ of $GL(n+1)$
$$
G=\left\{\begin{pmatrix}
B & v\\
0 &1\\
\end{pmatrix}:B\in H, v\in \mathbb{R}^n \right\}
$$
It is easy to show that $G$ is isomorphic to $\mathbb{R}^n \rtimes H$
$\blacksquare$
My question is: is the converse true? That is, given a group $G$ which is the semidirect product of $\mathbb{R}^n$ (additive group) and a subgroup $H$ of $GL(n)$, is $G$ isomorphic to a subgroup of $GL(n+1)$ in the form $$ G=\left\{\begin{pmatrix} B & v\\ 0 &1\\ \end{pmatrix}:B\in H, v\in \mathbb{R}^n \right\}? $$
Given any group $G=N\rtimes H$, with $N\approx \mathbb{R}^n$ and $H\approx \tilde{H}\leq GL(n)$, it can be shown that $G$ is isomorphic to a subgroup of $GL(n+1)$ of this form. Consider $$ \bar{N}=\left\{\begin{pmatrix} I & v\\ 0 &1\\ \end{pmatrix}:v\in \mathbb{R}^n \right\} $$ and $$ \bar{H}=\left\{\begin{pmatrix} B & 0\\ 0 &1\\ \end{pmatrix}:B\in \tilde{H}\right\} $$ Now it is easy to see that $G$ is isomorphic to a subgroup of $GL(n+1)$ with the form above by means of the map: $$ \begin{array}{rccccc} \varphi: & G=N\rtimes H& \to &\bar{N}\times\bar{H} &\to& GL(n+1)\\ & (n,h)&\mapsto& \left(\begin{pmatrix} I & v\\ 0 &1\\ \end{pmatrix},\begin{pmatrix} B & 0\\ 0 &1\\ \end{pmatrix}\right)& \mapsto & \begin{pmatrix} B & v\\ 0 &1\\ \end{pmatrix}\\ \end{array} $$
This map is a homomorphism. The product in $G$ takes the form $$ (n,h)(n',h')=(nhn'h^{-1},hh') $$ and with a standard computation can be checked that $$ \varphi((n,h))\cdot \varphi((n',h'))=\varphi((nhn'h^{-1},hh')) $$ Moreover, the map is also injective, as it is easy to check.