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If N is a set of natural numbers, then under binary operation a · b = a + b, (N, ·) is
The set of all positive rational numbers forms an abelian group under the composition defined by
Set (1,2,3,4} is a finite abelian group of order... under multiplication modulo ... as composition.
Let G be a group of order 7 and φ(x) = x^{4}, x ∈ G. Then f is
A group of prime order must be cyclic and every cyclic group is abelian. Then we can show that φ: G → G s.t. φ(x) = x^{n} is an isomorphism if 0(G) and n and are coprime.
If a, b ∈ G, a group, then b is conjugate to a, if there exist c ∈ G, such that
If H_{1} and H_{2} are two subgroups of G, then following is also a subgroups of G
If (G, *) is a group and for all a, b ∈ G, b^{1 }* a^{1}* b * a = e, then G is
Number of elements of the cyclic group of order 6 can be used as generators of the group are
Here, 6 = 2 x 3
The multiplicative group {1, 1} is a subgroup of the multiplicative group
A set G with a binary composition denoted multiplicative is a group, if
In the additive group of integers, the order of every elements a ≠ 0 is
Let Z be a set of integers, then under ordinary multiplication (Z, ·) is
Set of all n, nth roots of unity from a finite abelian group of order n with respect to
The generators of a group G = {a, a^{2}, a^{3}, a^{4}, a^{5}, a^{6} = e) are
we have o(G) = 6 and prime to 6 are 1 and 5
If G is a group such that a2 = e, for all a ∈ G, then G is
Consider the system of equations x + y + z = 1, 2x + 3y + 2z = 1, 2x + 3y + (a^{2 – 1})z = a + 1 then
Given system of linear equations:
x + y + z = 1 ….(1)
2x + 3y + 2z = 1 ….(2)
2x + 3y + (a2 – 1)z = a + 1 …..(3)
Consider a2 – 1 = 2
then LHS of (2) and (3) are same but RHS are not.
Hence a^{2} = 3 => a = √3
For a = √3, system is inconsistence.
So option (b) is correct.
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