Groups MCQ Quiz in मल्याळम - Objective Question with Answer for Groups - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given: 

g7 = e

Concept used:

order of generator = order of group

Every element in the group must divide the order of group and satisfies the property d7 = e

Explanations:

g7 = e

o(g)\7 

Two number divides 7 that is 1 or 7 

Now, 

g ≠ e

⇒ o(g) = 7 

Let us consider that the subgroup generated by g. 

As the order of g is 7

The order of the subgroup generated by g is 7 (because order of generator = order of group)

⇒ no of elements satisfies g⁷ = 7 is odd (every element in the group must divide the order of group and satisfies the property d⁷ = e}

∴ 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

Given:

abelian groups of order 108 

Concept:

the number of abelian groups of order n is p(n) ( no of partition of n )

Calculation:

108 = 2²⋅3³.

Now, p(2) = 2 because 2 = 2 and 2 = 1 + 1,

whereas p(3) = 3 because 3 = 3, 3 = 1 + 2, and 3 = 1 + 1 + 1.

Hence, the number of abelian groups of order 108 up to isomorphism is p(2) × p(3) = 2 × 3 = 6.

Hence the option (3) is correct.

Concept:

• Any group of order 3 is cyclic.
• Or Any group of three elements is an abelian group.
• The group has 3 elements: 1, a, and b. ab can’t be a or b, because then we’d have b=1 or a=1. So ab must be 1. The same argument shows ba=1. So ab=ba, and since that’s the only nontrivial case, the group is also abelian.

Additional Information

• Every group of prime order is cyclic.
• If an abelian group of order 6 contains an element of order 3, then it must be a cyclic group.
• Every subgroup of a cyclic group is itself a cyclic group.
• Every proper subgroup of an infinite cyclic group is infinite.

Lagrange's Theorem:

The order of each subgroup of a finite group is a divisor of the order of the group.

Converse of Lagrange's Theorem:

This is to say if G is a finite group of order n and m is any divisor of n, then it is not necessary that G must have a subgroup of order m.

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-

1. Closure Property
2. Associativity
3. Existence of Identity
4. Existence of Inverse
5. 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]). 

Concept:

If G is a group, then the subgroup consisting of G itself is the improper subgroup of G. All other subgroups are proper subgroups

In any group, the number of improper subgroups is 2

Concept:

A non-empty subset H of a group (G, ∗) is a group of G iff,

⇒ a, b ∈ H ⇒ a ∗ b^-1 ∈ H

Concept:

According to Lagrange’s theorem, the order or size of a group must be completely divisible by order or size of its subgroups.

Calculation:

Group G has 35 elements, i.e. its order is 35

So, possible subgroup sizes can be 1, 5, 7, 35.

Concept:

Sylow theorem: Let G be a finite group of order p^αm, where the prime number p does not divide m. Then

(i) there exists at least one Sylow p-subgroup of G.
(ii) if P is a Sylow p-subgroup of G and Q is any p-subgroup of G, then there exists g ∈ G such that Q is a subgroup of gPg⁻¹
(iii) the number n^p of Sylow p-subgroups of G is n_p ≡ 1(mod p).

Explanation:

G is a group of order 57 and G is not a cyclic group.

57 = 3 × 19

Let n₁₉ be the number of Sylow 19-subgroups of G.

then by Sylow’s theorem, n₁₉ ≡ 1(mod19) and n₁₉∣3.

So, n19 = 1

Let g ∈ G, then the order of g is 1, 3, or 19.

Since G is not a cyclic group, the order of g cannot be 57.

As there is exactly one Sylow 19-subgroup P, any element that is not in P must have order 3.

Hence the number of elements of order 3 is 57 − 19 = 38.

Option (3) is true