Class Equations, Sylow Theorems MCQ Quiz in मल्याळम - Objective Question with Answer for Class Equations, Sylow Theorems - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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നേടുക Class Equations, Sylow Theorems ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Class Equations, Sylow Theorems MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

Latest Class Equations, Sylow Theorems MCQ Objective Questions

Top Class Equations, Sylow Theorems MCQ Objective Questions

Class Equations, Sylow Theorems Question 1:

The number of 5-Sylow subgroups in the symmetric group S5 of degree 5 is __________.

Answer (Detailed Solution Below) 6

Class Equations, Sylow Theorems Question 1 Detailed Solution

Concept use:

if G be a Finite Group and pn divides Order of G but pn + 1 does not Divide order of G Then The Subgroup of pis Called Sylow p Subgroup.

if G is a Finite Group then the number of Sylow p - Subgroup or p-ssg in G is Equal to 1 + pk which divides o(G)

Explanation:

Here O(G) = 120 

By First Concept 5 divides 120 but 52 = 25 doesn't divide 120 

So, S5 containes a Sylow Subgroup of order 5

From Second Concept if  1 + pk divides O(G) 

1 + 5K Divides 120 when k = 1 

So, The Number of Sylow subgroup is 1 + pk = 6 

So, The Number of Subgroup can be 1 or 6 

So, These Results not sure about the Actual Value of Sylows Subgroup 

The Number of Cyclic Subgroup of order n in a Group / ϕ (n) = The Number of Cyclic Subgroup of order 5 in a Group / ϕ (5) = 24/4 = 6 

The number of 5-Sylow subgroups in the symmetric group S5 of degree 5 is 6

Hence, The Correct Answer is 6.

Class Equations, Sylow Theorems Question 2:

Which of the following are class equations for a finite group?

  1. 1 + 3 + 3 + 3 + 3 + 13  + 13 = 39
  2. 1 + 1 + 2 + 2 + 2 + 2 + 2  + 2 = 14
  3. 1 + 3 + 3 + 7 + 7 = 21
  4. 1 + 1 + 1 + 2 + 5 + 5 = 15 

Answer (Detailed Solution Below)

Option :

Class Equations, Sylow Theorems Question 2 Detailed Solution

Concept:

Let G be a finite gropu of order n such that n = pq where p and q are prime and p/(q-1) then class equation of the group will be

n = 1 + (p + p +...+ p)(\(q-1\over p\) time) + q + q + ...

Explanation:

(1): 39 = 3 × 13 and 3/ (13 - 1)

So, we will get non-ableian group

\(q-1\over p\) = 12/3 = 4

So class equation is

39 = 1 + 3 + 3 + 3 + 3 + 13 + 13

(1) is correct

(2): Structure of order 14 = 2 × 7

So group will be D7 and Z14

Class equation of Z14 is 1 + 1 + ... +1 (14 times)

(2) is false

(3): 21 = 3 × 7 and 3/ (7 - 1)

So, we will get non-ableian group

\(q-1\over p\) = 6/3 = 2

So class equation is

21 = 1 + 3 + 3 + 7 + 7

(3) is correct

(4): Every group of order 15 is cyclic so abelian so every class will be of cardinality 1.

(4) is false 

Class Equations, Sylow Theorems Question 3:

Let G be a group of order 5. 7 . 11 . 1, then

  1. The number sylow 5 - subgroup is 3
  2. The number sylow 7 - subgroup is 1
  3. The number sylow 11 - subgroup is 1
  4. G is simple.

Answer (Detailed Solution Below)

Option :

Class Equations, Sylow Theorems Question 3 Detailed Solution

Concept:

Sylow's Theorem: Sylow p-subgroup (for a prime p) of a group is a subgroup of the group that has order pn, where pn is the highest power of the prime p that divides the order of the group.

Explanation:

G is a group of order 5. 7 . 11 . 1 = 1925, which is the product of unique prime powers. 

(1): By Sylow's theorems, the number of Sylow 5-subgroups (which would have order 25) is congruent to 1 mod 5 and divides 1925/25 = 77.

Hence, the number of Sylow 5-subgroups can be 1 or 1 + 4k where 1 + 4k divides 77. The possible solutions are 1, 21 and 57. 

So, (1) is false

(2): Here, a Sylow 7-subgroup would have order 7.

By Sylow's theorem, the number of such subgroups (denoted n7) is congruent to 1 mod 7 and divides 1925/7 = 275.

Since the only divisor of 275 that is congruent to 1 mod 7 is 1 itself, we conclude that n7 = 1.

(2) is correct

(3): Here a Sylow 11-subgroup would have order 11.

By Sylow's theorem, the number of such subgroups (denoted n11) is congruent to 1 mod 11 and divides 1925/11 = 175.

The only divisor of 175 that is congruent to 1 mod 11 is 1 itself, we conclude that n11 = 1.

Therefore, (3) is correct

(4): A simple group is a nontrivial group whose only normal subgroups are the trivial group and the group itself.

Now, from the Sylow's theorem, we conclude there is only one Sylow 7-subgroup and one Sylow 11-subgroup.

Therefore, these subgroups must be normal. This contradicts the definition of simple group, so G cannot be simple.

(4) is false

Class Equations, Sylow Theorems Question 4:

Let G be a finite group of order n = pkm, where p is a prime and p does not divide m. A subgroup H of order pk is called______________. 

  1. P - group 
  2. sylow P - subgroup
  3. normal subgroup
  4. simple subgroup 

Answer (Detailed Solution Below)

Option 2 : sylow P - subgroup

Class Equations, Sylow Theorems Question 4 Detailed Solution

Explanation:

A subgroup H of a finite group G, having order pk, is called a Sylow p-subgroup.

Sylow theorem stated:

"If G is a finite group with |G| = pk m, where p is a prime number, p does not divide m, then G contains a subgroup of order pk, which is a Sylow p-subgroup."

So the correct answer is: Sylow p-subgroup.

Class Equations, Sylow Theorems Question 5:

Let G be a group of order 56. Then

  1. All 7-sylow subgroups of G are normal
  2. All 2-Sylow Subgroups of G are normal
  3. Either a 7-Sylow subgroup or a 2-Sylow subgroup of G is normal
  4. There is a proper normal subgroup of G

Answer (Detailed Solution Below)

Option :

Class Equations, Sylow Theorems Question 5 Detailed Solution

Explanation:

We have

o(G) = 56 = 23 × 7

∴ No. of 7-sylow subgroup

= value of k for which 1 + 7k/2 = 1

∴ 7-sylow subgroup in group by order 56 is unique Hence, normal in G.

But no. of 2-sylow subgroup

= value of k for which 1 + 2k/7 = 2

∴ 2-sylow subgroup may or may not be normal in G.

Hence, options (1) ,(3) and (4) are correct.

Class Equations, Sylow Theorems Question 6:

For integers n > 1, let G(n) denote the number of groups of order n, up to isomorphism, i.e. G(n) is the number of isomorphism classes of groups of order n. Which of the following statements is true?

  1. If G(n) = 1, then n is prime.
  2. G(8) = 2
  3. If gcd (n, ϕ(n)) > 1, then G(n) > 1. (Here φ denotes the φ Euler-function.)
  4. \(\rm \lim sup_{n \rightarrow \infty}G(n) = 2\)

Answer (Detailed Solution Below)

Option 3 : If gcd (n, ϕ(n)) > 1, then G(n) > 1. (Here φ denotes the φ Euler-function.)

Class Equations, Sylow Theorems Question 6 Detailed Solution

Concept:

Number of Groups of a Given Order:

  • The function G(n) denotes the number of non-isomorphic groups of order n.
  • This number depends on the structure theorem for finite abelian groups and classification results for small group orders.
  • Euler’s totient function φ(n) counts integers ≤ n that are coprime to n.
  • When gcd(n, φ(n)) > 1, it implies that n and the group of units mod n share a factor, which increases the chance of non-cyclic structures.

 

Calculation:

Evaluate all options:

Option 1: If G(n) = 1, then n is prime.

⇒ False. G(15) = 1, but 15 is not prime.

Option 2: G(8) = 2

⇒ False. There are 5 non-isomorphic groups of order 8.

Option 3: If gcd(n, φ(n)) > 1, then G(n) > 1

⇒ Consider n = 15 ⇒ φ(15) = 8, gcd(15, 8) = 1 ⇒ G(15) = 1

⇒ Consider n = 21 ⇒ φ(21) = 12, gcd(21, 12) = 3 > 1 ⇒ G(21) > 1

⇒ Hence, statement is generally true.

Option 4: lim sup G(n) = 2

⇒ False. G(n) is unbounded as n → ∞

∴ The correct answer is Option 3: If gcd(n, φ(n)) > 1, then G(n) > 1.

Class Equations, Sylow Theorems Question 7:

Let p and q be two distinct prime and p < q. Then which of the following is correct?

  1. Let o(G) = p2q if subgroup of order p2 is unique then G is abelian 
  2. Let o(G) = p2q, if subgroup of order p2 is unique and normal in G then G is abelian
  3. Every abelian group of order pq is unique subgroup of order p
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : Every abelian group of order pq is unique subgroup of order p

Class Equations, Sylow Theorems Question 7 Detailed Solution

Explanation:

(1): By Sylow's Theorem, the number of Sylow p-subgroups np​ is given by

np = 1 + pk, k = 0, 1, 2, ...

and (1 + pk)/q

If the subgroup of order p2 is unique, it is normal. However, this does not imply that G is abelian because there could still be interactions between the subgroup of order q and the normal subgroup of order p2.

(1) is false.

(2): If the subgroup of order p2 is normal, then its interaction with the subgroup of order q determines the structure of G.

For G to be abelian, the subgroup of order q must also commute with all elements of the normal subgroup of order p2.

While uniqueness and normality are strong conditions, they are not sufficient to conclude that G is abelian.

(2) is false.

(3): By the Fundamental Theorem of finite abelian groups, every abelian group of order pq, where p and q are distinct primes, is cyclic because gcd(p, q) = 1.

A cyclic group of order pq has exactly one subgroup of order p (as the unique subgroup generated by gq for a generator g).

(3) is true.

Class Equations, Sylow Theorems Question 8:

Let G be a group of order 15. Then which of the following is not correct?

  1. There exist exactly one 3-sylow subgroup of G
  2. There exist exactly two 3-sylow subgroup of G
  3. There exist exactly one 5-sylow subgroup of G
  4. G is cyclic

Answer (Detailed Solution Below)

Option 2 : There exist exactly two 3-sylow subgroup of G

Class Equations, Sylow Theorems Question 8 Detailed Solution

Explanation:

o(G) = 15 = 3 × 5

By first sylow theorem, G has sylow 3-subgrop and sylow 5-subgroups. 

the number of sylow 3-subgroup (n3) is given by

n3 = 1 + 3k, k = 0, 1, 2, ...

and n3/5

i.e., (1 + 3k)/5 ⇒ k = 0 only

Hence there exist only one sylow 3-subgroup A (say).

(1) is correct, (2) is not correct.

the number of sylow 5-subgroup (n5) is given by

n5 = 1 + 5k, k = 0, 1, 2, ...

and n5/3

i.e., (1 + 5k)/3 ⇒ k = 0 only

Hence there exist only one sylow 5-subgroup B (say).

(3) is correct.

o(A) = 3 and o(B) = 5

so A ⊲ G and B ⊲ G.

Also A, B both are cyclic.

Let A =  ⇒ a3 = e and B =  ⇒ b5 = e

A ∩ B = {e}

Hence xy = yx for all x ∈ A, y ∈ B

and in particular, ab = ba

Also, gcd(o(a), o(b)) = gcd(3, 5) = 1

Hence o(ab) = o(a) o(b) = 3 × 5 = 15

Therefore G = (ab)15 = e

So, G is cyclic.

(4) is corrcet.

Class Equations, Sylow Theorems Question 9:

How many 3-Sylow Subgroup are there in the group of order 15?

  1. 1
  2. 5
  3. 3
  4. 0

Answer (Detailed Solution Below)

Option 1 : 1

Class Equations, Sylow Theorems Question 9 Detailed Solution

Explanation -

The number of Sylow p-subgroups for a given prime p, say \(n_p\), always divides the order of the group.
Moreover, \(n_p\) is congruent to 1 mod p.
Consider a group G of order 15 = 3*5. We want to determine the number of 3-Sylow subgroups.

By the first theorem, the number of 3-Sylow subgroups \(n_3\) must divide the order of the group, which is 15 in this case, and by the second theorem, \(n_3\) is congruent to 1 mod 3.

The divisors of 15 are 1, 3, 5, and 15. The only ones among these that are congruent to 1 mod 3 are 1 and 4. However, 4 cannot satisfy since it's not a divisor of 15.

Therefore, there must be exactly 1 subgroup of order 3 in a group of order 15 (i.e., one 3-Sylow subgroup).

Hence option (i) is correct.

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