The Number of Subgroups of Order 25 in a Group of Order 200

To find the total number of subgroups of order 25 in a finite group G of order 200, we can use the Sylow theorems. The Sylow theorems provide information about the existence and properties of subgroups of a given order in a finite group.

1. Sylow's First Theorem
Sylow's First Theorem states that if p^k divides the order of a finite group G, then G contains a subgroup of order p^k.

In our case, we are looking for subgroups of order 25, so p = 5 and k = 2.

2. Sylow's Second Theorem
Sylow's Second Theorem states that the number of subgroups of order p^k in a finite group G is congruent to 1 modulo p and divides the order of G.

Applying Sylow's Second Theorem, we can say that the number of subgroups of order 25 in G is congruent to 1 modulo 5 and divides 200.

3. Sylow's Third Theorem
Sylow's Third Theorem states that if n_p denotes the number of subgroups of order p^k in G, then n_p satisfies the following conditions:
- n_p ≡ 1 (mod p)
- n_p divides the order of G
- n_p = |G : N_G(P)|, where N_G(P) is the normalizer of the subgroup P in G.

In our case, we are looking for subgroups of order 25, so n_5 satisfies the conditions above.

4. Reasoning
Since the number of subgroups of order 25 in G divides 200 and is congruent to 1 modulo 5, the only possibility is that the number is 1. This means there is only one subgroup of order 25 in G.

Conclusion
In conclusion, the total number of subgroups of order 25 in a finite group G of order 200 is 1. This can be determined by applying Sylow's theorems and reasoning based on the properties of subgroups.