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Test: Group Theory - 7 - Mathematics MCQ


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20 Questions MCQ Test Topic-wise Tests & Solved Examples for Mathematics - Test: Group Theory - 7

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Test: Group Theory - 7 - Question 1

If H is a subgroup of G, then number of left cosets of H in G and number of right cosets of H in G are

Test: Group Theory - 7 - Question 2

If F is homomorphism of a group G into a group G' with kernal K, then

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Test: Group Theory - 7 - Question 3

If G is a finite group and II is a normal subgroup of G, then  is

Test: Group Theory - 7 - Question 4

Let G be a group of order 15. Then the number of sylow subgroup of G of order 3 is

Detailed Solution for Test: Group Theory - 7 - Question 4

To find number of Sylow subgroup of G of order 3.
15 = 5 x 3, by a known theorem.
Number of 3-Sylow subgroup = 3K + 1
Now, (3K + 1) must divide 15, (3K + 1) being prime to 3 should divide 5. Thus, is true only for K = 0 thus, 37K+1 = 1.
Therefore, Number of 3-Sylow subgroup = 1.

Test: Group Theory - 7 - Question 5

f = (1 2 3) (1 2) is

Detailed Solution for Test: Group Theory - 7 - Question 5

we have (1 2 3) (1 2) = (1 2) (1 3) (1 2) which is a product of odd number of transpositions.

Test: Group Theory - 7 - Question 6

In the group (Z, +) the subgroup generated by 2 and 7 is

Detailed Solution for Test: Group Theory - 7 - Question 6

Indeed, for integers m,n (not both zero),
⟨m,n⟩ = gcd(m,n)Z
In this case, that means that
⟨2,7⟩ = gcd(2,7)Z
= 1Z
= Z

Test: Group Theory - 7 - Question 7

Choose the correct answer. If His a normal subgroup of G and K is a normal subgroup of H, then

Detailed Solution for Test: Group Theory - 7 - Question 7

Let G = A4, H = {(1 2)(3 4),(1 3)(2 4),(1 4)(2 3), e},

K = {e, (1 2) (3 4)} 

Then H is normal in G and K is normal in H. But K is not normal in G

Test: Group Theory - 7 - Question 8

Let G and H be two groups. The groups G x H and H x G are isomorphic

Test: Group Theory - 7 - Question 9

The number of generators in group ({1,2,3,4,5,6} x7) are

Test: Group Theory - 7 - Question 10

Let l be the additive group of integers. Let H =  {3x/x ⊂ l} normal subgroup of l, the elements of l/H are

Test: Group Theory - 7 - Question 11

If H and K are two subgroups of G of order 6 and 8, then order of HK is 16 if

Detailed Solution for Test: Group Theory - 7 - Question 11

we have o(HK) = 
Implies 
Implies o(H ∩ K) = 3

Test: Group Theory - 7 - Question 12

If G is a group of order 10 then G have a subgroup of order 

Detailed Solution for Test: Group Theory - 7 - Question 12

By the Lagrange’s theorem, the order of subgroup divides the order of group. We claim that all elements of G cannot be of order 2. Suppose it so.

Let a,b ∈ G be two different element of order 2. Let H = < a >, K = < b > be the cyclic groups generated by a and b, then
o(H) = 2, o(K) = 2
Since all the elements of group G are of order 2, it must be abelian.
∴ HK= KH ⇒ HK is subgroup of G And so

By Lagrange’s theorem, o(HK) would divide o(G).Which is not true hence our assumption is not correct.
Again, since G is finite, o(a) I o(G) for all a ∈ G
⇒ there exist at least one element a ∈ G such that  o(a) = 5 or 10.
If o(a) = 5, then H = < a > is subgroup of order 5.

If o(a) = 10, then H = < a2 > is a subgroup of order 5.

Test: Group Theory - 7 - Question 13

The order o f the element  in Zx Z6 is

Test: Group Theory - 7 - Question 14

A homomorphism from a simple group is

Detailed Solution for Test: Group Theory - 7 - Question 14

Let G be a simple group and f:G→G′ is homomorphism. So the Kernal f is a normal subgroup of G. But only normal subgroups of G are G and {e}.

∴ Either ker f = G or ker f = {e}

⇒⇒ either f is trivial or f is one - one

Test: Group Theory - 7 - Question 15

Let the set Z/ nZ denote the ring of integers modulo n under addition and multiplication modulo n then, Z/ 9Z is not a subring of Z/12Z because

Test: Group Theory - 7 - Question 16

If a and a2 are both generators of a cycle group of order m, then

Test: Group Theory - 7 - Question 17

If a and a2 both are generator of a cyclic group, then the order of the group gcd(2, n) = 1⇒ n = odd positive integer > 1

Test: Group Theory - 7 - Question 18

Let G be a cyclic group of order 6, Then, the number of element g ∈ G such that G = (g) is

Detailed Solution for Test: Group Theory - 7 - Question 18

Test: Group Theory - 7 - Question 19

Let S be a set and P(S) be the power set of S. Then {P(S), U} is not a group because

Test: Group Theory - 7 - Question 20

Let G be a group of order 30. Let A and B be normal subgroup of order 2 and 5, respectively. Then the order o f the group G / AB is

Detailed Solution for Test: Group Theory - 7 - Question 20

we have o(AB) = 
Implies 
= 30/10 = 3

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