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
- G forms a group under addition
- G forms an abelian group under multiplication
- Every element of G is diagonolizable over $\mathbb{C}$
- 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
Best Answer
$1$ 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!)