dakshina murthy group theory.ppt - karnataka · group theory already we are familiar with...
TRANSCRIPT
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Group Theory
Already we are familiar with notations;
N Set of natural numbersZ S f i
Already we are familiar with notations;
Z Set of integersZ0 Set of non‐zero integersZ+ Set of positive integersZ‐ Set of negative integersg gQ Set of rational numbersQ0 Set of non‐zero rational numbersQ0 Set of non zero rational numbersQ+ Set of positive rational numbers
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Q‐ Set of negative rational numbersQ1 Set of irrational numbersQ1 Set of irrational numbersR Set of real numbersR S f l bR0 Set of non‐zero real numbersR+ Set of positive reals R‐ Set of negative realsC Set of complex numbersC0 Set of non‐zero complex numbers.
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1) Which of the following is thebinary operation.binary operation.
1) Addition on the set Z0.
2) Subtraction on the set N2) Subtraction on the set N.
3) Division on the set Q.
4)Division on the set R+.
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Sol: I (1) 2+ ( 2) 0 Additi i t th bi0 ZIn (1) 2+ (– 2) = 0, Addition is not the binary
operation.(2) 5 – 8 = -3, Subtraction is not binary
∴∉ 00 Z
∴∉− N3(2) 5 8 3, Subtraction is not binary operation.
(3) Division is not binary ∴∉∞∞==÷ Q0202
∴∉ N3
operation.(4) If we divide any two +ve real numbers then we get
only a +ve real number
0
only a +ve real number.
Division is the binary operation.∴ y pAns is (4)
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2) In the set R which of the following is not
the binary operation under * Rba ∈∀ ,
10*)2*)1 +babaabba5*)4*)310*)2*)1
22 +=+=
−+==
babababababaabba
In (1) Let a = 4 & b = -1 then 4 * -1 = ( ) i2414 ±=−=−Sol:
operationbinarynotisRiHere
∗∴∉± 2
operationbinarynotis∗∴
Ans is (1)
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) h h f h f ll d f b3) Which of the following defines a binary operation on R ?
*)2*)1 ==aba
abbaaba b
1*)4
2*)3 22 ++
==ba
abbababa
a R, b,a, 4)In 0,00,0),1(:Sol
22
00
++∈∀
∉==
realvealwaysisbRedisnotdefinthenbletaIn
R *122
∈∴+++∴
barealvealwaysisba
Ans is (4)
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4) Which of the following is a binary operation.
1) The cross product of 2‐ dimensional vectors.2) The dot product of 3‐ dimensional vectors.) p 33) The cross product of 3‐D vectors) The dot product of 3 vectors4) The dot product of 3 vectors.
( ) &1: isavectorbathenvectorsarebaIfInSolrrrr
×( )&
&1:
ofplanetheinnotisitandbabothto
isavectorbathenvectorsarebaIfInSolr
rr⊥
×
.&barr
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( ) &2
avectornotisba
scalaraisbaofproductdotTherr
rr
∴ .. avectornotisba∴
rr( )
.3&3
spaceDtobelongsitandbabothtoisba r
−⊥×
rrrr
p
( ))4
vectoranotisitanddefinednotiscba rrr( ) ... vectoranotisitanddefinednotiscba
Ans is (3)Ans is (3)
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5) In the set of irrationals which one of the f ll i i i fi l l
|Qfollowing signs satisfies closure law.
1) + 2) - 3) x 4) None of these1) 2) 3) x 4) None of these
( ) ( ) |0&0221: QInSol ∉=−+( ) ( )( ) |0&0222 QIn
Q
∉=−
∉
( )
( ) |2&2223 QIn
Q
∉=×( )Ans is (4)
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) hi h f h f ll i i fi h l6) Which of the following satisfies the closure law under given condition on *.
Nonbaba2
*)1 +=
2) a* b = a-2b on N.
3) a * b = the smaller of (a + b) and (a –b) on N.
+= Qonbaba *)4b
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( ) thenbaLetInSol =+
=∗==312121&21: ( )
failslawClosureN
thenbaLetInSol
∴∉
==∗==
322
121&21:
f∉2In 2) Let a = 2 & b= 3 then 2 * 3 =2-(2)3 =-4 and -4 is not in the set Nis not in the set N
failslawClosure∴
In (3) Let a = 2 & b = 5 then 2 + 5 =7 2 – 5 = -3In (3) Let a = 2 & b = 5 then 2 + 5 =7, 2 – 5 = -3There fore, 2 * 5 = smaller of (2 + 5) and (2 -5) = -3
∴ 5 * 2 3 & f illCl∴ 5 * 2 = -3 & N∉− 3 failslawClosure∴
In (4) For any a & ba
is always +ve rationalsIn (4) For any a & b, b is always +ve rationals Ans is (4).
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7) Which of the following is not commutativeunder * in N.
1) a * b= 3a + 5b 2) a * b = a + b – 1) a b 3a 5b ) a b a b
3) a * b = ab + 2 4) a * b = a2 + b2) )
( ) ababbabaInSol 53*,53*1: +=+=( )abba 5353 +≠+∴
Ans is (1)Ans is (1)
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8) On the set R which of the following is commutative under *
babababa *)2*)1 −=−=
baba
baba 2*)42
*)3 +==
bb2
baababbabaInSol **)1(: −=−=−=
abbabaababbabaInSol
**,)1(:
=∴
===
Ans is (1)
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9) Which of the following defines a commutative binary operation?commutative binary operation?
1) Scalar product of vectors 2) Division on the set of all square roots of unity.2) Division on the set of all square roots of unity.
1034*,In)3 +−= babaZ4) Cross product of two vectors4) Cross product of two vectors.
)2)1:
IationbinaryoperrsisnottheoftwovectodotproductInSol
11)2−÷
In
lawsecommutativanddefinitionoperation Binary
111111
−−− Ans is (2)satisfied.are
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10) On the set R+ under * which of the following are not associative.following are not associative.
1) a * b =ab 2) a * b = a + b + 3
3) 4) a * b = a+b-100 10
* abba =cb ⎞⎛( ) ( )1.......***)1: bcacbacbacbaInSol === ⎟⎠⎞⎜
⎝⎛
⎞⎛ cb( ) ( )2.......*** ⎟
⎟⎠
⎞⎜⎜⎝
⎛
==cb
acbacba⎞
⎜⎛ cb
eassociativnotis*∴≠ ⎟⎠
⎜⎜⎝
cbabcaBut
Ans. is (1)
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11) In set Q which one is associative under *
1) * b 2b 2) * b1)a * b = a – 2b 2) a * b = a
3) a * b = a –b +10 4)aba 2* =3) a b = a –b +10 4) b
ba3
( ) ( )( ) ( )( ) ( )2***
1***),2(:−−−−==
−−−−==abacba
acacbaInSol
( ) ( )eassociativis*
,,****
∴∈∀= Qcbacbacba
eassociativis∴Ans. is (2)
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⎫⎧ ⎞⎛ xx12) In the group G= Identity
⎭⎬⎫
⎩⎨⎧
∈⎟⎟⎠
⎞⎜⎜⎝
⎛0, Rx
xxxx
element (matrix) under multiplication is
⎞⎛⎞⎛⎞⎛⎞⎛ 2211012121⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛2222
)41111
)31001
)221212121
)1⎠⎝⎠⎝⎠⎝⎠⎝
⎞⎜⎜⎛ xx
XLetSol :⎞
⎜⎜⎛
=yy
Y⎞
⎜⎜⎛
=ee
E⎠
⎜⎜⎝
=xx
XLetSol :⎠
⎜⎜⎝
=yy
Y⎠
⎜⎜⎝
=ee
E
From definition of identity axiom
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matrixidentityEthenXEX =×
⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎠
⎞⎜⎜⎝
⎛⎟⎠
⎞⎜⎜⎝
⎛ xxeexx
matrixidentityEthenXEX
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛++++
⎠⎜⎝⎠
⎜⎝⎠
⎜⎝
xxxx
xexexexexexexexe
xxeexx
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎠⎝⎠⎝ ++
xxxx
xexexexe
xxxexexexe
2222
212 =∴= exexequating
⎟⎟⎠
⎞⎜⎜⎝
⎛=
21212121
ngESubstituti Ans is (1)⎠⎝ 2121
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13) In the group Q0 under * ifthen the inverse of 4 is
0,3
2* Qbaabba ∈∀=
169)4
32)3
23)2
916)1
16329
Sol: From identity axiom; a * e = a
we get23
32
=⇒= eaae23
From Inverse axiom eaa =−1* ⇒=⇒−
23
32 1aa
aa
491 =−
99( ) 16
944
944 1 === −thenaput ans is (4)
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, )14 oftionmultiplicaunderGroup In the 0⎭⎬⎫
⎩⎨⎧
∈⎟⎟⎠
⎞⎜⎜⎝
⎛Rx
xxxx
is2222ofinversematrices ⎟⎟⎠
⎞⎜⎜⎝
⎛⎭⎩
⎟⎠
⎞⎜⎜⎝
⎛2222
)1 ⎟⎠
⎞⎜⎜⎝
⎛21212121
)2⎠
⎜⎝ 22 ⎠
⎜⎝ 2121
⎞⎛ 4141 ⎞⎛ 8181⎟⎠
⎞⎜⎜⎝
⎛41414141
)3⎟⎟⎠
⎞⎜⎜⎝
⎛81818181
)4⎠⎝
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⎞⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
xxxx
XLetSol :⎟⎟⎠
⎞⎜⎜⎝
⎛=−
yyyy
X 1
⎠⎝ ⎠⎝ yy
⎟⎠
⎞⎜⎜⎝
⎛=
21212121
... EtkwAlready⎠
⎜⎝ 2121
EXXunderdefnFrom =×× −1,.
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛eeee
yyyy
xxxx
⎠⎝⎠⎝⎠⎝ eeyyxx
⎞⎜⎜⎛⎞
⎜⎜⎛ 212122 xyxy
⎠⎜⎜⎝
=⎠
⎜⎜⎝ 212122 xyxy
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xyxyequating
41,
212 ==
( ) ===112 yxPut ( )
⎞⎜⎜⎛
=−
8242
1 yyXinSub
yxPut
⎞⎜⎜⎛
⎠⎜⎜⎝
=
− 81811X
yyXinSub
⎠⎜⎜⎝
=8181
X
Ans is (4)
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15) In the group {1, 3, 5, 7} x mod 8 the value of
( ) is53 11 −−×1) 1 2) 3 3) 5 4) 7
7531⊗Sol:
7531175318⊗
551
1−
isIdentity
3175557133
( ) ( ) 775353
551111 ===
=−−−− XX
1357731755
Ans is (4)
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( ) theRAgrouptheIn ,cossinsincos
)16 θθθθθ
θ⎭⎬⎫
⎩⎨⎧
∈⎟⎟⎠
⎞⎜⎜⎝
⎛−
=
( ) isionmultilicatmatrixunderAofinverse 7
cossin
π
θθ ⎭⎩ ⎠⎝
( )( )7)1 πA ( )7)2 π−A
( )( )2)3 π−A ( )76)4 πA= 1: AtkwSol
⎞⎜⎜⎛
=⎞⎜⎛ 7sin7cos
1...:
πππA
AtkwSol
⎠⎜⎜⎝ −
=⎠
⎜⎝ 7cos7sin7 ππ
A
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( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛ −=∴ −
cossinsincos1
θθθθ
θA
⎟⎠⎞
⎜⎝⎛≠−−−⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=⎟
⎠⎞
⎜⎝⎛∴ −
7)1(
7cos7sin7sin7cos
71 π
πππππ AA
( ) ( )( ) ( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−−−−−
=⎟⎠⎞
⎜⎝⎛ −
7cos7sin7sin7cos
7&
πππππA
⎟⎠⎞
⎜⎝⎛=−−−−⎟⎟
⎠
⎞⎜⎜⎝
⎛−
=⎟⎠⎞
⎜⎝⎛ −
⎠⎝
−
7)2(
7cos7sin7sin7cos
71 π
πππππ AA
⎠⎝
A i (2)Ans is (2)
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{ }{ }( ) ==
+=−− xthen102*3* and
ab - b a b * a if ,1 group aIn 17)1x
Q( )
1) 9 2) 19 3) -19 4) -91) 9 2) 19 3) 19 4) 9
axiomIdentity From ,3 find usLet :Sol 1−
( ) 001=−+⇒=∗ aaeeaaea
( ) 001 =∴=− eae
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( )111
111 =−+⇒=∗ −−− eaaaaeaaaxiomInverseFrom( )
,31
10
1
111
=−=∴
−=−=−+
−
−−−
aputa
aa
aaaaaaa
233
23
3133
111 =∴=
−−=
−−−
a
( ) 102*3* 1 =−xNow( ) 19102*23* =∴=⇒ xgetwegSimplifyinx
Ans is(2)
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18) In the group { e, a, b} where e is an
identity then a4 b5 =identity then a b
1) a 2) b 3) a-1 4) b-1
A group with 3 elements is always abelian.
( ) bbabbaba == 44454
bbae.( )
bbebeeabHere =4
ebaabaee
bbebe === aebbAns is (2)
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19) In an abelian group G, is( ) 1112 −−− abcba
1) c-2 2) c-1 3) acb 4) c
( ) 1112 −−− abcbagroupabeliananIn
( ) ( ) ( )( ) ( ) 22211
21111111211
−−−−−
−−−−−−−−−− ==
cceecbbaaabcbaabcba
( ) ( ) === cceecbbaa
Ans is (1).
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20) Whi h f th f ll i i20) Which one of the following is a semi group under division.
1) Q 2) Z 3) C 4) {1, -1}Sol: In 1) 2) 3 ) If we take 0 in denominator
11 −÷ In 4) From the table we can
Sol: In 1) ,2), 3 ) If we take 0 in denominator we get infinity .therefore closure law fails
11111
−÷ In 4) From the table we can
observe that Closure and Associative laws are satisfied
111 −−
{ }{ } groupsemiais1,1 −∴ Ans is (4)
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21) Which of the following is false?1) Every group is a monoid 2) Every monoid is a semi group3) Every semi group is a monoid4) Every monoid need not be a group
Sol:In 3) N is a semi group under + but it is not a monoid because under + , 0 is an identity and 0 No o d because u de , 0 s a de t ty a d 0
For 4) Z is a monoid but it not a group under xEvery semi group need not be monoid
∉
Every semi group need not be monoid Ans is (3)
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22) On the set Q if a * b = a – b + 1 , Qba ∈∀ ,
which one of the axiom is satisfied under *.
1)Identity axiom 2) Inverse axiom1)Identity axiom 2) Inverse axiom
3) Closure law 4) Commutative law.
Sol: For 1) Identity element does not exist.If a * e = a = e * a then e identitya e a e a t e e de t tyi.e. a * e = a & a = e * a
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a – e + 1 = ae 1
a = e - a + 12a e + 1-e = -1
e = 12a = e + 1e = 2a - 1
Here e values are different .Therefore Identity element does not exist.
2) Identity element does not exist implies inverse axiom is not satisfied.axiom is not satisfied.
QbaQbaQthenbifaHere ∈∀∈+−∈ ,1,,)3
Ans is (3)∴
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23) Which one of the following is not a group .1) {0, 1, 2} + mod 3 2) {1, 2, 3, 4, 5} x mod 63) {0} + mod 3 4) {1, 2, 3, 4,} x mod 5) { } ) { }Sol; Construct the table for all the options then
setOIn ∉==× .0632)2(
tiItfailslawclosure∴
)(
groupanotisIt∴
Ans is (2)
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24) {-1, 0, 1} is not a multiplicative group because of failure of g p
1)Closure law 2) Associate law
3) Identity axiom 4)Inverse axiom
101−×1011
101−−
×
existnotdoesiselementIdentity
−101
10110000
−satisfiednotisaxiomInverse∴
1011Ans is (4)
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25) For the square matrices of same order which of the following is true.c o t e o o g s t ue1)Matrix addition is not a group.
2)Matrix Multiplication is not a group
3)Matrix subtraction is a group.) g p
4)Matrix division is a group
Ans is (2)Ans. is (2)
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( )10
01,
1001
1001
,1001
matricesThe26 ⎟⎟⎠
⎞⎜⎜⎝
⎛−
−⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟⎠
⎞⎜⎜⎝
⎛
form under10101010
×⎠⎝⎠⎝⎠⎝⎠⎝
1)Infinite Semi Group 2) Infinite Group
3) Finite abelian Group 4) Infinite abelian Group.
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⎞⎛⎞⎛ 0101⎟⎟⎠
⎞⎜⎜⎝
⎛ −=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
1001
,1001
BALet
⎟⎠
⎞⎜⎜⎝
⎛ −=⎟
⎠
⎞⎜⎜⎝
⎛=
1001
,10
01DC
⎠⎜⎝ −⎠
⎜⎝ − 1010
DCBADCBAADCBA×
Ans is (3)BADCCCDABB
Ans is (3).ABCDD
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i ⎫⎧ ⎞⎛ θθXundergroupaisRGLet .,
cossinsincos
)27⎭⎬⎫
⎩⎨⎧
∈⎟⎟⎠
⎞⎜⎜⎝
⎛−
= θθθθθ
isanswerincorrectThe
1) G is an infinite group itselfofinverseis⎞⎜⎜⎛− 01
)21) G is an infinite group ff⎠
⎜⎜⎝ −10
)
Gofelementanis⎞⎜⎜⎛ 11
)3 Gofelementanis⎠
⎜⎜⎝− 11
)3
⎞⎛Gofelementanis⎟⎟
⎠
⎞⎜⎜⎝
⎛ −2123
2321)4⎠⎝
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In (1), we know that set is a group.
In (2), If we find inverse of matrix we get it selfit self.
In (4), when we get this matrix.0300=θ
In (3), tlb
cannotanyfor1
sin&cos, θθθ
( ) statementincorrectistoequalbe
3.1
∴
( )3isAns∴
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28) Which one is false in a group G
1)Identity is unique
2) Inverse is unique
3) Inverse of inverse of an element is itself
4) Inverse of an element is itself4) Inverse of an element is itself.
Sol; Inverse of an element need not be itself in a group Ans is (4)
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29) Which of the following is false?
1) In a group of even order, there exists an element other than identity which is its own inverseother than identity which is its own inverse.
2)In an abelian group, every element is its own inverse.
3) In an abelian group G ( ) Gbababa ∈∀= −−− ,111
4) In an abelian group G ( ))2(
,222
AnsisGbababa ∈∀=
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30 ) Which of the following is a semi group under subtraction.
1) {0} 2) Z 3) Q0 4) R+1) {0} 2) Z 3) Q0 4) R
Sol: In 2),3),4) Associative law fails
000−
In 1) (0 – 0) – 0 = 0 – 0 = 0i il l 0 (0 0) 0 0 000 similarly 0 – (0 – 0) = 0 – 0 = 0
A.L. Satisfies
Ans is (1)
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31) In group ( G, *) for some element Ga ∈
if a2 = e then
1) 2) a e 3) 4) a 0ea −11) 2) a = e 3) 4) a = 0ea = aa =1
Sol: Given a2 = e a *a = e
a-1 * (a * a) =a-1 * e e * a = a-1
⇒⇒a (a a) a e e a a
therefore a = a-1
⇒
Ans is (3).
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32) Which one of the following is not true in
the group G Gba ∈∀ ,)2)1 bbbb
( ) ( ) 111111 )4)3
)2)1−−−−−− ==
=⇒==⇒=
baababab
cacbabcbacab
( ) ( ))4)3 == baababab
Ans is (4)
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33) Which of the following is true in thegroup G.g p
1) if a-1 = a then G is abelian)
2)
3) if 2 th 0 1
( ) Gbababa ∈∀= −−− ,222
3) if a2=a then a = 0 or a=1
4) if a2 = e2 then a-1 = a
missedisGaIn ∈∀)1possiblenotisThis∴ Ans is (4).
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34) Which of the following is a subgroup ofadditive group of integers modulo 4.
1) {0 1} d 4 2) {0 2} d 41) {0, 1} + mod 4 2) {0, 2} + mod 4
3) {0, 2, 3} + mod4 4) {0, 4} + mod4
20020
)210010
)1++
022211
30320
23330030
)410223200
)3+
2332133
Ans is (2)
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35) Which one of the following is a subgroup of multiplicative group of set of non-zero reals R0.
1) Q0 2) Q1 3) Z0 4) Z) 0 ) 1 ) 0 )
00)1 RQIn ⊂ ∴ This is a subgroup1
01 12)For QidentitybutRofsubsetaisQ ∉
groupsubanotisQ1∴
.21&2123)For 000
satisfiednotisaxiominverseZisofinverseButRZ
∴∉⊂
0 4)For RZ ⊄ 0) ⊄
groupsubanotisZAns is (1)
∴
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36) Which of the following is a sub group of
Z11 – {0} under X mod 11Z11 {0} under X mod 11.
1) {1, 3,5, 7} 2) {1, 7} 3) {1, 10} 4) {1, 5, 7}
Sol: In 1), 2) & 4) closure law is not satisfied), ) )
10111×satisfieslawClosure∴
110101011 satisfieslawClosure∴
11010Ans. is (3)
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37) If H is a subset of group G then H is a
subgroup of G if the following axiomssubgroup of G if the following axioms
are satisfied.
1) Closure and associative 2) Closure and Identity
3) Identity & Closure 3) Closure and inverse3) Identity & Closure 3) Closure and inverse
Ans is (4)Ans is (4)
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38) If H and is subset of group G then H
is a subgroup ifHba ∈∀ ,HbHb )2)1 11
HaHabHbaHab
∈∈
∈∈−
−−
)4)3)2)1
1
11
))
Ans. is (3)
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39) If H and K are subgroups of group G then
which one is a subgroup of G
KHKH U )2)1KHKHKHKH
I
U
)3)3)2)1
+−
KHKH I)3)3 +
Ans is 4)Ans. is 4)
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40) The number of subgroups of the group
{ 1 5 7 11} under X mod 12 is{ 1, 5, 7, 11} under X mod 12 is
1) 2 2) 4 3) 5 4) 1
1175111175112×
( ) ( ) ( )1111,77,55,11 1111 ==== −−−−
511177711155
117511 ( ) ( ) ( ){ }1
11,1,7,1,5,1∴areitselfGsetandthiswith
groupsubare
1571111511177 { }
523 =+∴groupssubf
Ans is (3)
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41) Set of complex numbers Z with isunderZ ×= 1
1) an abelian group 2) Not a Group1) an abelian group 2) Not a Group
3) Not a Monoid 4)Not a Semigroup
Ans. is (1)
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42 ) Which of the following is a subgroup of additive group of integersg p g
1) N 2) Z+ 3) Set of even integers 4) Z
N, Z+, are subsets of Z but they are not groups Zunder +.
They are not subgroups.∴ y g p
Ans is (3)
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43) Which one of the following is a
semi group under x.
1)Z- 2) R-
3) Q- 4) set of odd integers) ) g
Ans is (4)
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44) The set of square matrices
form an abelian group under
1) + 2) - 3) 4) None×) ) ) )
Square matrices are of order 2x2,3x3
then addition,subtraction,multiplication may not be possible.
Ans is (4)
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iselementidentitythen*,RrealsofsetOn)45 22 babaif +=
iselementidentitythen
1) 0 2) 1 3) -1 4) does not exist) ) ) )
Sol: Identity axiom is a * e =a
00,, 222222 =∴==+=+ eeaeaaea
B t 0 i t id tit l t h 1But 0 is not identity element when a = - 1
eeBecause −=+⇒−=− 111*1 2
identitybecannoteeBut ∴−≠+ 11 2 Ans is(4)
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b( )
( )∗∗
∈∀=∗
xthen
Rbaabba
4x52and
,10
* if,R group In the46) 00
( ) ==∗∗ xthen4x5 2 and
1)10 2) 20 3) 30 4) 401)10 2) 20 3) 30 4) 40
( ) ( ) [ ]..4*5*24*5*2 == satisfiedisLAxxGiven Q
( ) 4*0152
=/// x
( ) 40410
14*1 =∴== xxxA i (4)10 Ans is (4)
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( )( )
∗⎞⎜⎛ ∗∈∀
++=−
then x142&
8,ba b* a If*Z, group In the47)1
xZba ==∗⎠
⎜⎝
∗∈∀ then x142&, xZba
112)4211)323)21)1 −−− ))))In this case given ( 2-1 * x)* 4 =1We can eliminate 2-1 by inserting 2 from left sideWe can eliminate 2 by inserting 2 from left side.
( ) 1*24**2*2 1 =− xConsider ( )( ) ( ) ( )
1812841*24**1*24**2*2
124221
∴++++⇒=⇒=−
xxxex
xConsider
181284 −=∴++=++⇒ xxAns is (1)
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×⎟⎟⎠
⎞⎜⎜⎝
⎛
⎭⎬⎫
⎩⎨⎧
∈⎟⎟⎠
⎞⎜⎜⎝
⎛ under of inversetheset theIn48)1031
,10
10Rx
x
⎠⎝⎭⎩ ⎠⎝ 1010
⎠
⎞⎜⎜⎝
⎛ 31)1 ⎞
⎜⎜⎛ −11
)2 ⎞⎜⎜⎛ 31
)3 ⎞⎜⎜⎛ 01
)4⎠
⎜⎝ 10
)⎠
⎜⎜⎝ 11
)⎠
⎜⎜⎝ 32
)3⎠
⎜⎜⎝ 10
)4
⎞⎛⎞⎛⎞⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −
101
&10
1,
101
: 1 eE
yX
xXLetSol
⎠
⎞⎜⎜⎝
⎛=
⎠
⎞⎜⎜⎝
⎛
⎠
⎞⎜⎜⎝
⎛
=
111 xexXEXaxiomidentitytheFrom
⎠⎜⎜⎝⎠
⎜⎜⎝⎠
⎜⎜⎝ 101010
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=∴=+⎟⎠
⎞⎜⎜⎝
⎛=⎟
⎠
⎞⎜⎜⎝
⎛ +0
101
101
exxexxe
=⎟⎟⎠
⎞⎜⎜⎝
⎛∴
⎠⎜⎝⎠
⎜⎝
−
1001
1010
1 EXXaxiominverseFromidentityanis
⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛ +⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛⎠⎝
1001
101
1001
101
101
10
xyyx
⎞⎜⎜⎛ −
=∴
−=∴=+⎠⎝⎠⎝⎠⎝⎠⎝⎠⎝
− 10
1 xXngSubstituti
xyxy
⎟⎠
⎞⎜⎜⎝
⎛ −=∴=
⎠⎜⎜⎝
∴
−
1031
3
10
1Xxput
XngSubstituti
⎠⎜⎝ 10
p
Ans is (1)