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Consider the group $$G=\begin{Bmatrix} \begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix} : a,b \in \mathbb{R}, \: a>0 \end{Bmatrix}$$ with usual matrix multiplication. Let $$ N = \left\{ \begin{pmatrix}1 & b \\ 0 & 1 \end{pmatrix} : b \in \mathbb{R} \right\}.$$ Then, 

  1. $N$ is not a subgroup of $G$
  2. $N$ is a subgroup of $G$ but not normal subgroup
  3. $N$ is a normal subgroup and the quotient group $G/N$ is of finite order
  4. $N$ is a normal subgroup and the quotient group is isomorphic to $\mathbb{R}^+$ (the group of positive reals with multiplication).
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Normal subgroup is where

left coset = right coset

Now we need to verify, option $C)$ and $D)$

Say the coset is$\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$

So, according to normal subgroup

$\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$.$\begin{bmatrix} 1 &b \\ 0 &1 \end{bmatrix}$=$\begin{bmatrix} 1 &b \\ 0 &1 \end{bmatrix}$.$\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$

which satisfies.

So, it is normal subgroup


$\begin{bmatrix} 1 &b\\ 0 &1 \end{bmatrix}$ it is also a quotient group of $\begin{bmatrix} a &b\\ 0 &a^{-1} \end{bmatrix}$

where quotient value is only 1


and quotient group and normal subgroup are isomorphic and quotient group is subset of normal subgroup

Answer will be both C) and D) 

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N can be the subgroup of A as its mentioned that a,b belongs to R and a>0 

Hence option A is incorrect.

Option B is correct. As for the subgroup N  to be normal  of the the group A it should satisfy-  (A*N*A^-1) belong to N 

and on putting the matrices and finding the inverse, it doesn’t belong to N. 

Other options also doesn’t satisfy the condition.

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N can be the subgroup of A as its mentioned that a,b belongs to R and a>0 

Hence option A is incorrect.

Option B is correct. As for the subgroup N  to be normal  of the the group A it should satisfy-  (A*N*A^-1) belong to N 

and on putting the matrices and finding the inverse, it doesn’t belong to N. 

Other options also doesn’t satisfy the condition.

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