Explanation:

Given that G is an acyclic group of order 625 and H is a cyclic subgroup of G generated by the element a^5. We need to find the order of the subgroup H.

Definition of Order:
The order of a subgroup is the number of elements in that subgroup.

Step 1: Find the Order of G
The order of a group G is the number of elements in that group. In this case, G has an order of 625.

Step 2: Find the Order of H
Since H is a cyclic subgroup generated by a^5, the order of H is equal to the order of the element a^5.

Step 3: Find the Order of a^5
To find the order of a^5, we need to find the smallest positive integer n such that (a^5)^n = e, where e is the identity element in G.

Since G is an acyclic group, every element has a finite order. Therefore, the order of a^5 exists and is finite.

Step 4: Use the Properties of Cyclic Groups
In a cyclic group, the order of an element is equal to the order of the subgroup it generates.

Since a^5 generates the cyclic subgroup H, the order of a^5 is equal to the order of H.

Step 5: Apply the Properties of Exponents
Since the order of G is 625, we know that a^625 = e, where e is the identity element in G.

Now, let's consider (a^5)^n = e. We can rewrite this as a^(5n) = e.

Since the order of a^5 is the smallest positive integer n satisfying this equation, we can conclude that the order of a^5 is a factor of 625.

Step 6: Find the Prime Factorization of 625
The prime factorization of 625 is 5^4.

Therefore, the possible orders of a^5 are 5^0, 5^1, 5^2, 5^3, and 5^4.

Step 7: Determine the Order of a^5
To find the order of a^5, we need to test each of the possible orders.

If we test 5^0, we get (a^5)^1 = a^5 ≠ e.

If we test 5^1, we get (a^5)^5 = a^25 ≠ e.

If we test 5^2, we get (a^5)^25 = a^125 = e.

Therefore, the order of a^5 is 5^2 = 25.

Step 8: Conclusion
Since the order of a^5 is 25, the order of the cyclic subgroup H generated by a^5 is also 25.

Therefore, the correct answer is '25'.