Test: Group Theory - 4 - Mathematics MCQ

H is a non-empty complex of a group G. The necessary and sufficient condition for H to be a subgroup of G is: a, b ∈ H ⇒ ab^-1 ∈ H, where b^-1 is the inverse of b in G.

The set M of square matrices (of same order) with respect to matrix multiplication is a monoid.

A monoid is a set equipped with an associative binary operation and an identity element. In the case of square matrices, the binary operation is matrix multiplication, which is associative. The identity element is the identity matrix, a square matrix with ones on the diagonal and zeros elsewhere.

It is not a group because not every matrix has an inverse (a matrix that, when multiplied by the original matrix, yields the identity matrix).

It is not a quasi-group because although every element has a left and right inverse, these inverses are not necessarily the same.

It is a semi-group because it is a set with an associative binary operation, but this term is less specific than monoid since it doesn't specify the existence of an identity element.

So, the most specific and correct answer is monoid.

G is a group under addition modulo 6

0 is the identity

For a,b∈G

If a+b=0, then a−1 = b

Now, 3+3=0 ⟹ 3−1 = 3 ..... (i)

Also, 9+3=12=0 ⟹ 9−1 = 3 ....... (ii)

Consider, (2+3⁻¹+4)⁻¹ 

= (2+3+4)⁻¹

= 9⁻¹

= 3
Hence, (2+3⁻¹+4)⁻¹=3

Le S₅ = {a₁, a₂, a₃, a₄, a₅} be a symmetric group  of order 5. the elements of S₅ which are their own inverses are of the type (a₁, a₂)  or (a₁, a₂) (a₂, a₄).

The number of elements of the type (a₁, a₂) are

The number of elements of the type (a₃, a₄) are

So the total number of elements which are their own inverses is equal to 25.

pointed that 25 elements of order 2 make 26 self-inverse elements, because the identity is also self-inverse.