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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

                                                         $N \;=\; \begin{Bmatrix} \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}\;: b \in \mathbb{R} \end{Bmatrix}.$


(A) $N$ is not a subgroup of G

(B) $N$ is a subgroup of G but not a normal subgroup

(C) $N$ is a normal subgroup and the quotient group $G/N$ is of finite order

(D) $N$ is a normal subgroup and the quotient group is isomorphic to $\mathbb{R}^{+}$(the group of positive reals with multiplication).
in Set Theory & Algebra by Boss (17k points) | 140 views
Ans will be A)
Answer is not given. Please explain your approach.
Actually 1 is not element of G and by taking any common from G also we cannot get 1 as an element of the group. So, I told it is not a subgroup of G
$N$ is not a subgroup of $G$, this is because $N$ does not satisfy "Inverse property".

@srestha mam, elements of G is in matrix form.  @Subhanshu why N does not satisfy inverse property ?

definition of subgroup says

elements of subgroup, must be element of group, which satisfies group property

which violates here
mam, which property violating here. since a and b are real numbers so, consider a=1 and b can be any real no. in G which also present in N.
yes u r right

b can be 0,

So, associativity also maintaining

inverse !=0

what will be identity element? 1 rt?
identity element will be identity matrix
I am getting (D) if anyone has done it please confirm. I have done like this :- Since, G is a group and when a=1 then it becomes N . So, N is the part of G. So, N should definitely be subgroup. N should also be a normal subgroup because for each matrix '$x$' of $G$ , $xhx^{-1} \in N$ for each $h \in N$ for matrix multiplication operation. Since Quotient Group is a set of right cosets and it is a set of sets and here it will be infinite set , So, order will be infinite not finite. So, options (A),(B) and (C) are eliminated. So, I am getting answer as (D).


for quotient group value of a must be 1

So, I think C) also correct


mam, why value of a=1. value of 'a' can be anything. Quotient Group is a set of right cosets and coset is also a set. So, Quotient Group is a set of sets. Here it must be infinite. and order is defined as no. of elements in a group. So, since size of quotient group is infinite. So order must be infinite.

check here

quotient group means divisible group.

other than 1 what else where $\frac{G}{N}$ will satisfy?
mam, not getting what you are saying. $G/N$ is a set of left/right cosets and it also have a group properties. Considering right cosets, $G/N = \{Na_1,Na_2,....\}$, where $a_1,a_2,....$ are the elements of the group G which are matrices here and in every matrix, a,b are real numbers. So, a,b can be any real no. $Na_1,Na_2,...$ are the right cosets where N is a set of matrices of type $\begin{bmatrix} 1 &b \\ 0&1 \end{bmatrix}$ here b can take any real value.
I am not thinking about $b$

b is in both matrix $G$ and $N$

but a is in G , but not in $N$ other than $a=1$

I mean this

$\frac{\begin{pmatrix} a &b \\ 0 & a^{-1} \end{pmatrix}}{\begin{pmatrix} 1 &b \\ 0 & 1 \end{pmatrix}}$

Now quotient group only satisfies when $a=1$

So, it has finite element, hence finite order
mam, $ \frac{G}{N}$ does not mean we are dividing all the matrices of G by all the matrices of N. Here group operation is matrix multiplication and also we can't divide matrices.

Quotient Group is defined as :-

$\frac{G}{N} = \{Na_{1},Na_{2},Na_{3,...}\}$

where $a_1,a_2,a_3,...$ are the elements of group G and  N is subgroup.

Suppose, $a_{1} = \begin{bmatrix} 3 &5 \\ 0&\frac{1}{3} \end{bmatrix}$

  $a_{2} = \begin{bmatrix} 1 &3 \\ 0&\frac{1}{1} \end{bmatrix}$

$a_{3} = \begin{bmatrix} 7 &1 \\ 0&\frac{1}{7} \end{bmatrix}$

Then $Na_1$ = $\begin{bmatrix} 1 &b \\ 0&1 \end{bmatrix}$* $\begin{bmatrix} 3 &5 \\ 0&\frac{1}{3} \end{bmatrix}$

$Na_2$ = $\begin{bmatrix} 1 &b \\ 0&1 \end{bmatrix}$* $\begin{bmatrix} 1 &3 \\ 0&\frac{1}{1} \end{bmatrix}$

$Na_3$ = $\begin{bmatrix} 1 &b \\ 0&1 \end{bmatrix}$* $\begin{bmatrix} 7 &1 \\ 0&\frac{1}{7} \end{bmatrix}$

here , b can be any real number. Now do all the above matrix multiplication and make the set $\{Na_{1},Na_{2},Na_{3,...}\}$ which will be called Quotient Group. This set has infinite no. of elements . So, order should be infinite.

1 Answer

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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) 

by Veteran (119k points)

Normal subgroup is where

left coset = right coset


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

coset is a set , not an element.

So, it is normal subgroup

Yes, Correct.

where quotient value is only 1

I am not getting mam.

Answer will be both C) and D) 

Answer will be only (D). You can check here



quotient group $\frac{G}{N}=G.N^{-1}=\begin{pmatrix} a &b \\ 0 & a^{-1} \end{pmatrix}.\begin{pmatrix} 1 &b \\ 0& 1 \end{pmatrix}$

$=\begin{pmatrix} a+b^{2} & b\\ a^{-1}b & a^{-1} \end{pmatrix}$

Now normal subgroup to be isomorphic with quotient group

b need to be 0 



mam, $\frac{G}{N}\neq GN^{-1}$ . check here

mam, in the given link, nothing is given about quotient group

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