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.

  • 1
    $\begingroup$ What you write just means that the group is the semidirect product of those subgroups. Or did I miss something? $\endgroup$ – Tobias Kildetoft Nov 3 '16 at 18:18
  • $\begingroup$ @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. $\endgroup$ – Adam Hughes Nov 3 '16 at 18:21
  • $\begingroup$ @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. $\endgroup$ – Tobias Kildetoft Nov 3 '16 at 18:22
  • $\begingroup$ @TobiasKildetoft Technically, the OP hasn't pointed out that they intersect trivially, but it's not very difficult to show. $\endgroup$ – Arthur Nov 3 '16 at 18:23
  • $\begingroup$ @TobiasKildetoft ah, I thought you were focusing on the last point in particular since there was no intersection bit. $\endgroup$ – 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)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.