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uksV% ;g iz'u i=k pkyhl (40) vadksa dk gSA tks nks (02) [k.Mksa d rFkk [k esa foHkkftr gSA izR;sd [k.M esa fn, x, foLr`r funsZ'kksa ds vuqlkj gh iz'uksa dks gy dhft,A

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mt-07 algebra

chtxf.kr

Bachelor of Science (BA/BSc.-12/16) Third Year, Examination-2020

Time Allowed : 2 Hours Maximum Marks : 40 Note: This paper is of Forty (40) marks divided

into Two (02) sections A and B. Attempt the question contained in these sections according to the detailed instructions given therein.

uksV% ;g iz'u i=k pkyhl (40) vadksa dk gSA tks nks (02) [k.Mksa d rFkk [k esa foHkkftr gSA izR;sd [k.M esa fn, x, foLr`r funsZ'kksa ds vuqlkj gh iz'uksa dks gy dhft,A

Section-a/

[k.M&^d*

(Long Answer type Questions/

nh?kZ mÙkjh; iz'u)

Note: Section-'A' contains Five (05) long answer type questions of Ten (10) marks each. Learners are required to answer any two (02) questions

only. (2×10=20)

(2)

uksV% [k.M&^d* esa ik¡p (05) nh?kZ mÙkjh; iz'u fn, x, gSa] izR;sd iz'u ds fy, nl (10) vad fu/kZfjr gSaA f'k{kkfFkZ;ksa dks buesa ls dsoy nks (02) iz'uksa ds mÙkj nsus gSaA

1. If H and K are two subgroups of a group G then HK will be a subgroup of G iff HK = KH.

;fn

H

rFkk

K

fdlh lewg

G

ds dksbZ nks milewg gks rks

HK, G

dk milewg gksxk ;fn vkSj dsoy ;fn

HK=KH.

2. Define order of an element of a group. If a and b are any two elements of a group G, then prove that o (b–1ab) = o (a).

lewg ds fdlh vo;o dh dksfV dh ifjHkk"kk nhft,A

;fn

a

rFkk

b

lewg

G

ds dksbZ nks LosPN vo;o gSa rks fl¼ dhft, fd

o (b–1ab) = o (a).

3. If I1 and I2 are ideals of R then I I1+ =2

{

x x x1+ 2: 1∈I x1, 2∈I2

}

is also an ideal of R containing I1 and I2 .

(3)

ekuk

I1

,oa

I2

oy;

R

dh xq.ktko fy;k gS rc

I I1+ =2

{

x x x1+ 2: 1∈I x1, 2∈I2

}

oy;

R

dh xq.ktkoyh gksrh gS rFkk

I1 + I2

esa

I1

,oa

I2

nksuksa lekfgr gksrs gSaA

4. Show that a finite dimensional vector space has a finite basis.

fn[kkb, fd fdlh fu;r vkdkj dh lfn'k lef"V dk ,d fu;r vk/kj gksrk gSA

5. If

A = (

1 2 3 4 52 3 1 5 4

)

then find the following :

(a) A–1 (b) A2

(c) A3 (d) Order of A

;fn A = (

1 2 3 4 52 3 1 5 4

) rks fuEu dks Kkr dhft;sA

(a) A–1 (b) A2

(c) A3 (d) A

dh dksfV

Section-b/

[k.M&[k

(Short answer type a question /

y?kq mÙkjh; iz'u)

Note: Section-B Contains Eight (08) short answer type questions of Five (05) marks each. Learners are required to answer any four (04) questions

only. (4×5=20)

(4)

uksV% [k.M&^[k* esa vkB (08) y?kq mÙkjh; iz'u fn, x, gSa] izR;sd iz'u ds fy, ik¡p (05) vad fu/kZfjr gSaA f'k{kkfFkZ;ksa dks buesa ls dsoy pkj (04) iz'uksa ds mÙkj nsus gSaA

1. Show that the set G={1, 2, ____ P-1} for Xp (multiplication modulo P), is an abelian group when P is prime number.

fl¼ dhft;s fd leqPp;

G={1, 2, ____ P-1} for Xp

(ekM~;wyks xq.kk

P

)lafØ;k ds fy;s vkcsyh lewg gksxk tgk¡

P

vHkkT; la[;k gSA

2. Show that in a finite cyclic group order of the group is equal to the order of its generator.

fl¼ dhft;s fd ,d ifjfer pØh; lewg dh dksfV mlds tud dh dksfV ds cjkcj gksrh gSA

3. If H is a subgroup of G, and g

G, then prove that o(H) = o (gHg-1).

;fn

H

fdlh lewg

G

dk ,d milewg gks rFkk

g

G

rks fl¼ dhft, fd

o(H) = o (gHg-1).

(5)

4. Show that center Z of a group G is normal subgroup of G.

lewg

G

dk dsUnz

Z, G

dk izlkekU; milewg gksxkA

5. If (R, + , ) is a ring and a

R then S = {x

R : ax = 0} is subring of R.

;fn

(R, + , )

,d oy; gS rFkk

a

R

rc fl¼ dhft, fd

S, R

dk mioy; gS tgk¡

S = {x

R : ax = 0}.

6. Show that the intersection of subspaces of a vector space is also a subspace.

fdlh lfn'k lef"V ds fdUgha nks milef"V;ksa dk loZfu"B Hkh ml lfn'k lef"V dh milef"V gksrh gSA

7. Show that the set P3 of all permutation on three symbols 1, 2, 3 is a finite non-abelian group of order 6 with respect to permutation multiplication as composition.

fn[kkb, fd 1] 2] 3 ds Øep; lewg dk leqPp;

P3

Øep; xq.ku ds fy, 6 Øe dk fu;r vkcsyh lewg

gksxkA

(6)

8. Show that non-zero finite integral domain is field.

fn[kkb, dh v'kwU; fu;r iw.kk±dh; izkUr ,d {ks=k gksrk gSA

*****

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