Can this matrix group be obtained from $(\mathbb R,+)$ and $(\mathbb R^*,\cdot)$? I have stumbled upon this group of matrices in an old midterm:
$$G=\{
\begin{pmatrix}
  a & b \\
  0 & \frac1a \\
\end{pmatrix}; a,b\in\mathbb R, a\ne0
\}.$$
The students were asked to show that this is a group (under matrix multiplication).
It is easy to notice that this group contains two subgroups
\begin{align*}
H_1&=\{
\begin{pmatrix}
  a & 0 \\
  0 & \frac1a \\
\end{pmatrix}; a\in\mathbb R^*
\}\\
H_2&=\{
\begin{pmatrix}
  1 & b \\
  0 & 1 \\
\end{pmatrix}; b\in\mathbb R
\}
\end{align*}
such that $H_1 \cong (\mathbb R^*, \cdot)$ and $H_2\cong (\mathbb R,+)$.

I wonder whether there is some group theoretic construction using which we can obtain $G$ from $H_1$ and $H_2$. (I.e., some kind of construction which, given the two groups $(\mathbb R^*,\cdot)$ and $(\mathbb R,+)$ returns a group isomorphic to $G$.)

It is not very difficult to see that:


*

*Every element $g\in G$ can be expressed in exactly one way as $g=h_1h_2$ where $h_1\in H_1$ and $h_2\in H_2$.
$$
\begin{pmatrix}
  a & 0 \\
  0 & \frac1a
\end{pmatrix}
\begin{pmatrix}
  1 & b \\
  0 & 1
\end{pmatrix}=
\begin{pmatrix}
  a & ab \\
  0 & \frac1a
\end{pmatrix}
$$

*Of course, the same is true for expression of the form $g=h_2h_1$ with $h_i\in H_i$. (We can simply take such expression form $g^{-1}$ and invert it.)

*$H_2$ is a normal subgroup of $G$, it is kernel of the homomorphism $\begin{pmatrix}
  a & b \\
  0 & \frac1a
\end{pmatrix}\mapsto a$ from $G$ to $\mathbb R^*$.


Other than that, I did not notice anything special about $H_1$ and $H_2$; so it is quite possible that the answer is no. But since here we see two well-known groups as a subgroup of a relatively simple matrix group, I still thought that this is worth asking.

We have


*

*$H_2$ normal

*$H_1\cap H_2 = \{I_2\}$, where $I_2$ is the $2\times 2$ identity matrix

*Every $g \in H$ is of the form $h_1h_2$ with $h_1\in H_1, h_2\in H_2$


which is the definition of the (inner) semidirect product $H_2\rtimes H_1$.
