IMC / 1996 / Problems / Day 2, P9

IMC 1996 · Day 2 · P9

Let $G$ be the subgroup of $GL_2(\mathbb{R})$ , generated by $A$ and $B$ , where $A = \begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.$ Let $H$ consist of those matrices $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$ in $G$ for which $a_{11} = a_{22} = 1$ .

(a) Show that $H$ is an abelian subgroup of $G$ .

(b) Show that $H$ is not finitely generated.

Remarks. $GL_2(\mathbb{R})$ denotes, as usual, the group (under matrix multiplication) of all $2 \times 2$ invertible matrices with real entries (elements). Abelian means commutative. A group is finitely generated if there are a finite number of elements of the group such that every other element of the group can be obtained from these elements using the group operation.

Solution (official)

(a) All of the matrices in $G$ are of the form $\begin{bmatrix} * & * \\ 0 & * \end{bmatrix}.$ So all of the matrices in $H$ are of the form $M(x) = \begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix},$ so they commute. Since $M(x)^{-1} = M(-x)$ , $H$ is a subgroup of $G$ .

(b) A generator of $H$ can only be of the form $M(x)$ , where $x$ is a binary rational, i.e., $x = \dfrac{p}{2^n}$ with integer $p$ and non-negative integer $n$ . In $H$ it holds $\begin{align*} M(x) M(y) &= M(x + y) \\ M(x) M(y)^{-1} &= M(x - y). \end{align*}$ The matrices of the form $M \left( \dfrac{1}{2^n} \right)$ are in $H$ for all $n \in \mathbb{N}$ . With only finite number of generators all of them cannot be achieved.

IMC 1996 · Day 2 · P9

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