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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

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

Then,

(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).
asked in Set Theory & Algebra by Boss (11.4k points) | 106 views
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Ans will be A)
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Answer is not given. Please explain your approach.
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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
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$N$ is not a subgroup of $G$, this is because $N$ does not satisfy "Inverse property".
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@srestha mam, elements of G is in matrix form.  @Subhanshu why N does not satisfy inverse property ?

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definition of subgroup says

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

which violates here
+1
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.
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yes u r right

b can be 0,

So, associativity also maintaining

inverse !=0

what will be identity element? 1 rt?
+1
identity element will be identity matrix
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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).
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@ankitgupta.1729

for quotient group value of a must be 1

So, I think C) also correct

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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

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quotient group means divisible group.

other than 1 what else where $\frac{G}{N}$ will satisfy?
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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.
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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
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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

0 votes

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) 

answered by Veteran (114k points)
0

Normal subgroup is where

left coset = right coset

correct.

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 https://math.stackexchange.com/questions/1131458/quotient-group-g-n-order-and-isomorphic-group

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@ankitgupta.1729

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 

right?

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

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