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.

• What you write just means that the group is the semidirect product of those subgroups. Or did I miss something? – Tobias Kildetoft Nov 3 '16 at 18:18
• @TobiasKildetoft he needs that there is a section of the map, not just that there is a s.e.s. for it, but that's the right conclusion. – Adam Hughes Nov 3 '16 at 18:21
• @AdamHughes No, he needs precisely what he has, since he has two subgroups generating the group, intersecting trivially and one of them normal. No need for any sections. – Tobias Kildetoft Nov 3 '16 at 18:22
• @TobiasKildetoft Technically, the OP hasn't pointed out that they intersect trivially, but it's not very difficult to show. – Arthur Nov 3 '16 at 18:23
• @TobiasKildetoft ah, I thought you were focusing on the last point in particular since there was no intersection bit. – Adam Hughes Nov 3 '16 at 18:24

We have

1. $H_2$ normal
2. $H_1\cap H_2 = \{I_2\}$, where $I_2$ is the $2\times 2$ identity matrix
3. 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 $H_2\rtimes H_1$.

Since there is a section of the short exact sequence

$$0\to(\Bbb R,+)\to (G,*)\to (\Bbb R^*,\cdot)\to 0$$

this is a semi-direct product of the groups $(\Bbb R, +)$ and $(\Bbb R^*, \cdot )$. The bottom of the section on outer semi-direct products I linked discusses the construction. (This is not an outer semi-direct product since they are subgroups of the big group, but that's where the section statement is)