Let G be a finite group of order 200, then the number of subgroup of G of order 25 isa)1b)2c)5d)10Correct answer is option 'A'. Can you explain this answer?

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Let G be a finite group of order 200, then the number of subgroup of G of order 25 is

a)
1
b)
2
c)
5
d)
10

Correct answer is option 'A'. Can you explain this answer?

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Most Upvoted Answer

Let G be a finite group of order 200, then the number of subgroup of G...

Let G be a finite group of order 200. We need to find the number of subgroups of G that have an order of 25.

Step 1: Prime Factorization of 200
To begin, let's find the prime factorization of 200:
200 = 2^3 * 5^2

Step 2: Sylow's Theorem
According to Sylow's theorem, for any prime factor p of the order of a finite group G, the number of subgroups of G of order p^k, denoted as n_p, satisfies the following conditions:
1. n_p ≡ 1 (mod p) - This means that n_p leaves a remainder of 1 when divided by p.
2. n_p | (order of G)/p^k - This means that n_p divides the order of G divided by p^k.

Step 3: Finding n_5
In this case, we are interested in finding the number of subgroups of G of order 25, which means we need to find n_5. According to Sylow's theorem, n_5 ≡ 1 (mod 5) and n_5 | 200/5^2.

Since 200/5^2 = 8, we need to find a value of n_5 that leaves a remainder of 1 when divided by 5 and divides 8.

Step 4: Possible Values of n_5
The possible values of n_5 are 1 and 6. However, since we are looking for the number of subgroups of order 25, which is a specific value, the correct answer is 1.

Step 5: Conclusion
Therefore, the number of subgroups of G of order 25 is 1.

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Let G be a finite group of order 200, then the number of subgroup of G...

Group Order and Sylow Subgroups:
Given |G| = 200 = 2³ x 5², we seek the number of subgroups of order 25 = 5². These are Sylow 5-subgroups.
Sylow's Third Theorem:
The number of Sylow 5-subgroups, n_5, must satisfy:
- n₅ divides ( |G| / 5² ) = 200 / 25 = 8,
- n₅ ≡ 1 mod 5.
Possible Values for n_5:
- Divisors of 8: 1, 2, 4, 8.
- Values congruent to 1 mod 5: Only 1.
Conclusion:
- n₅ = 1, meaning there is exactly one Sylow 5-subgroup of order 25.
- This subgroup is unique and normal in G.

Final Answer:
A

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