Test: Group Theory - 2 - Mathematics MCQ

An algebraic structure (P,*) is called a semigroup if a*(b*c) = (a*b)*c for all a,b,c belongs to S or the elements follow associative property under “*”. (Matrix,*) and (Set of integers,+) are examples of semigroup.

A non empty set A is called an algebraic structure w.r.t binary operation “*” if (a*b) belongs to S for all (a*b) belongs to S. Therefore “*” is closure operation on ‘A’.

A monoid(B,*) is called Group if to each element there exists an element c such that (a*c)=(c*a)=e. Here e is called an identity element and c is defined as the inverse of the corresponding element.

A Semigroup (S,*) is defined as a monoid if there exists an element e in S such that (a*e) = (e*a) = a for all a in S. This element is called identity element of S w.r.t *.

The set of two M*M non-singular matrices form a group under matrix multiplication operation. Since matrix multiplication is itself associative, it holds associative property.

A group (M,*) is said to be abelian if (x*y) = (x*y) for all x, y belongs to M. Thus Commutative property should hold in a group.

A group holds five properties simultaneously –
i) Closure
ii) associative
iii) Identity element
iv) Inverse element.

A singular element can generate a cyclic group. Every element of a cyclic group is a power of some specific element which is known as a generator ‘g’.

The set of complex numbers {1, i, -i, -1} under multiplication operation is a cyclic group. Two generators i and -i will covers all the elements of this group. Hence, it is a cyclic group.

A cyclic group is always an abelian group but every abelian group is not a cyclic group. For instance, the rational numbers under addition is an abelian group but is not a cyclic one.

Concept

• According to the Lagrange theorem order of subgroups must divide the order of the group.
• Property of group says if a group has prime order then it is cyclic.

Explanation:

• Since the order of G is 6. Therefore  its subgroup may have order 1,2,3,6
• H is one of its subgroups with condition 1< |H| <6 so H may be of order 2 or 3  which is prime 
• Hence H must be cyclic
• The order of G is 6 which is not prime and hence it may or may not  be cyclic 

Therefore option 2 is correct

Concept:
Let G be a cyclic group of order 10 generated by an element a, then o(a) = o(G) = 10
Evidently G = {a, a², a³, a⁴, a⁵, a⁶, a⁷, a⁸, a⁹, a¹⁰ = e }
If the HCF of m and n is d, then we write (m, n) = d.
An element a^m ∈ G is also a generator of G if (m, 10) = 1.
Thus there are four generators of G namely, a, a³, a⁷, a⁹

Concept:

Abelian Group: Let {G=e, a, b} where e is identity. The operation 'o' is defined by the following composition table. Then(G, o) is called Abelian if it follows the following property-

• Closure Property
• Associativity
• Existence of Identity
• Existence of Inverse
• Commutativity

Cyclic Group- A group a is said to be cyclic if it contains an element 'a' such that every element of G can be represented as some integral power of 'a'. The element 'a' is then called a generator of G, and G is denoted by (or [a]). 

Theorem: 
(i) All cyclic groups are Abelian, but an Abelian group is not necessarily cyclic. 
(ii) The order of a cyclic group is the same as the order of its generator. 
Thus it is clear that A and B both are true.

A group is said to be abelian if (a*b) = (b*a) ∀a,b ∈G
Since (G, ⋅) is a group
∴ (ab)^-1 = (b^-1a^-1)             (1)
Given: (ab)^–1 = a^–1b^–1       (2)
From (1) and (2)
 (a*b) = (b*a)
Therefore it is abelian

Concept:
A non-empty subset H of a group (G, ∗) is a group of G iff,
⇒ a, b ∈ H ⇒ a ∗ b^-1 ∈ H

A group of prime order p (in this case, 31) is always cyclic. And every cyclic group is abelian is true

 While there does exist a cyclic group of order 4, namely Z₄, there also exists a non-cyclic group of order 4, the Klein-4 group. Both are abelian hence, This statement is not always true.

 The Fundamental Theorem of Finite Abelian Groups tells us that for any integer n, there exists an abelian group of order n. In this particular case, 2021 = 43 * 47, which are both primes, so you could have a direct product of two cyclic groups of orders 43 and 47 as an abelian group of order 2021 hence, This statement is not true.

 While the cyclic group of order 6, Z₆, is abelian, there is also a non-abelian group of order 6, the symmetric group S₃ hence, This statement is not true.

Given:
Number of Improper Subgroups of Z6 is 
Concept used:
Improper subgroup means The subgroup is Group itself 
Proper Subgroup of Z_n = The subgroups which are not improper
Number of subgroups of Z_n  = Number  of divisors of n 
Calculations: 
The number of divisors of 6 = 1, 2, 3, 6 
Total subgroups is 4
but One subgroup is equal to Group that is improper subgroup generated by <1> = {0, 1, 2, 3, 4, 5}
∴ Total number of proper subgroup of Z₆  = 3

Calculation:
Let a cyclic group G of order 8 generated by an element a, then 
⇒ o(a) = o(G) = 8
To determine the number of generators of G,
Evidently, G = {a, a², a³, a⁴, a⁵, a⁶, a⁷, a⁸ = e}
An element a^m ∈ G is also a generator of G is HCF of m and 8 is 1.
HCF of 1 and 8 is 1, HCF of 3 and 8 is 1, HCF of 5 and 8 is 1, HCF of 7 and 8 is 1.
Hence, a, a3, a5, a⁷ are generators of G.
Therefore, there are four generators of G.

Concept:

In group theory, improper subgroups of any given group include the group itself and the trivial group, which contains only the identity element. These are considered improper because they are trivially subgroups of every group. All other subgroups are proper subgroups.

Therefore, every group has exactly two improper subgroups:

1. The group itself.
2. The trivial group (the subgroup containing only the identity element).

So, the correct answer is:

Option A: 2

Given:
If G is a Prime order group of p
Concept used:
Lagranges theorem:
The order of subgroups divides the Order of Group but if some m divide the order of the group that doesn't mean that group has the subgroup of order m 
Proper subgroup :  H ≠ e and H ≠ G is the subgroup of G
Calculations:
G is a Prime order Group of p 
so, it has only 2 divisors 1 and itself that divides the order of Group 
⇒ Only 2 subgroups are generated by 1 and p
⇒ <1> = G and
= {e} where e is the identity 
∴ option 1 is correct, No proper Subgroup