# multiple choice question on group of matrices

Consider the set of matrices $$G=\left\{ \left( \begin{array}{ll}s&b\\0&1 \end{array}\right) b \in \mathbb{Z}, s \in \{1,-1\} \right\}.$$Then which of the following are true

1. G forms a group under addition
2. G forms an abelian group under multiplication
3. Every element of G is diagonolizable over $$\mathbb{C}$$
4. G is finitely generated group under multiplication

I am getting 1) is false since not closed under addition 2)Forms a group under multiplication ( abelian or not i don't know) 3)Not true if $$a=1$$ 4) dont know please help me to complete

• A few examples with $s=1$ in one matrix, $s=-1$ in the other, should convince you that $G$ isn't abelian. – Robert Shore Sep 21 '19 at 16:32
• So answer will be finitely generate right? – sabeelmsk Sep 21 '19 at 16:34

• $$1$$ is false:

Your approach is correct. Since, for example, $$\begin{pmatrix}1&*\\0&1 \end{pmatrix}+\begin{pmatrix}1&*\\0&1 \end{pmatrix}=\begin{pmatrix}2&*\\*&* \end{pmatrix} \notin G$$

• $$2$$ is false:

Take $$b \neq 0$$.$$\begin{pmatrix}1&b\\0&1 \end{pmatrix}\begin{pmatrix}-1&b\\0&1 \end{pmatrix}=\begin{pmatrix}-1&2b\\0&1 \end{pmatrix}$$ whereas $$\begin{pmatrix}-1&b\\0&1 \end{pmatrix}\begin{pmatrix}1&b\\0&1 \end{pmatrix}=\begin{pmatrix}-1&\color{red}{0}\\0&1 \end{pmatrix}$$

• $$3$$ is false too:

Since, for example, $$\begin{pmatrix}1&b\\0&1 \end{pmatrix}$$ is not diagonalizable when $$b \neq 0$$

• $$4$$ is true

The finite set $$\left\{\begin{pmatrix}1&1\\0&1 \end{pmatrix},\begin{pmatrix}1&-1\\0&1 \end{pmatrix},\begin{pmatrix}-1&0\\0&1 \end{pmatrix}\right\}$$ generates $$G$$(verify!)

• Is it generated by 2 elements ? I am not clear about generating set – sabeelmsk Sep 21 '19 at 16:36
• @sabeelmsk: See my edit – Chinnapparaj R Sep 21 '19 at 17:01