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

  • 2
    $\begingroup$ A few examples with $s=1$ in one matrix, $s=-1$ in the other, should convince you that $G$ isn't abelian. $\endgroup$ – Robert Shore Sep 21 '19 at 16:32
  • $\begingroup$ So answer will be finitely generate right? $\endgroup$ – 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!)

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  • $\begingroup$ Is it generated by 2 elements ? I am not clear about generating set $\endgroup$ – sabeelmsk Sep 21 '19 at 16:36
  • 1
    $\begingroup$ @sabeelmsk: See my edit $\endgroup$ – Chinnapparaj R Sep 21 '19 at 17:01

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