Semidirect product and matrix groups 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
$$
G=\left\{\begin{pmatrix}
B & v\\
0 &1\\
\end{pmatrix}:B\in H, v\in \mathbb{R}^n \right\}?
$$
 A: 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.
