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Solution No 1

The document presents solutions to various problems in group theory, including the structure of subgroups and their intersections. It demonstrates the verification of subgroup criteria and the homomorphism property of a logarithmic function. Additionally, it proves that the determinant of the product of two special linear matrices is equal to one.

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abdul haq
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12 views3 pages

Solution No 1

The document presents solutions to various problems in group theory, including the structure of subgroups and their intersections. It demonstrates the verification of subgroup criteria and the homomorphism property of a logarithmic function. Additionally, it proves that the determinant of the product of two special linear matrices is equal to one.

Uploaded by

abdul haq
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Solution No.

Structre eof (C8 elements

H={e,b,b2,b3,b4,b5,b6,b7}

Where ba= e

Order 2

H2= {e,b4}

Order 4

H4= {e,b,b2,b4,,b6}

Proper cycling subgroups: {e}, {e,b4} , {e,b,b2,b4,,b6}

Solution 2

H1 and H2 are sub groups of a group G is:

H1 ∩ H 2 = {× Є G1 × Є H1 and x Є H2}

Example

Let G =Z

H1 = 2Z= {0, ±2, ±4…………….}

H2=3Z ={0, ±3, ±6………………}

Their intersection:

H1 ∩ H 2= { xЄZ x Є 2Z XЄ 3Z}

Since the least common multiple of 2 and 3 is 6

H1 ∩ H 2 =6Z={ 0, ± 6, ± 12,…………}

Verify Subgroup Criteria

Closure

If ( a, b Є H1 ∩ H 2),then (a,b) are multiple of 6

Hence (a+b) is also a multiple of 6


Inverses

If (aЄ H1 ∩ H 2), then (-a) is a multiple of 6, so (-a Є H1 ∩ H 2)

Hence proved

Solution 3

To show that: K → G, (h(x) = log x), is homomorphism:

Groups Defined

(k = (R+,)) : G= (R, +)):

ISM Property:

(h) is a homomorphism if :

H(x,y Є K). – by definition of (h)

H(x,y) = log (x, y)

=log x + log y

Substituting: (x,y)= h(x) +h(y)

The homomorphism property holds: h(x,y) = H(x) +h(y).

Therefore, (h) is a homomorphism.

Solution 4

Given

AЄ SL (n, R) ↔ det (A) =1

Det (AB) = det (A). det (B).

Proof

Let A,B Є SL (n, R), so det(A) = 1 and det (B) =1

The product AB

Det (AB) = det (A). det(B).

Substituting: det (AB) =1.1=1 hence proved

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