Explanation:

In order to explain why the number of 5-Sylow subgroups in a group of order 4S is 1, we need to understand the concept of Sylow subgroups and their properties.

Sylow Theorems:
Sylow theorems are a set of important theorems in group theory that provide information about the existence and properties of subgroups in a finite group. They are named after the Norwegian mathematician Ludwig Sylow.

Sylow Subgroups:
A Sylow p-subgroup of a group G is a subgroup whose order is a power of a prime p that divides the order of G. In this case, we are considering 5-Sylow subgroups.

Sylow First Theorem:
The first Sylow theorem states that if p is a prime number that divides the order of a group G, then G has at least one subgroup of order p.

Sylow Second Theorem:
The second Sylow theorem states that if p^r is the highest power of a prime p that divides the order of a group G, then G has a subgroup of order p^r.

Sylow Third Theorem:
The third Sylow theorem states that the number of Sylow p-subgroups in a group G is congruent to 1 modulo p, i.e., n_p ≡ 1 (mod p), where n_p is the number of Sylow p-subgroups.

Application to the Given Problem:
In the given problem, we have a group of order 4S. We are interested in the number of 5-Sylow subgroups in this group.

By the Sylow theorems, we know that the number of 5-Sylow subgroups, n_5, is congruent to 1 modulo 5. This means that n_5 can be written as 1 + 5k, where k is a non-negative integer.

However, we also know that the number of subgroups of a group cannot exceed the order of the group. In this case, the group has order 4S. Therefore, n_5 cannot exceed 4S.

Combining these two conditions, we have the following inequality:
1 + 5k ≤ 4S

Simplifying this inequality, we get:
k ≤ (4S - 1)/5

Since k is a non-negative integer, the maximum value of k is (4S - 1)/5 rounded down to the nearest integer.

Therefore, the maximum number of 5-Sylow subgroups in the group of order 4S is (4S - 1)/5.

However, we also know that the number of subgroups of a group must be an integer. Therefore, the maximum number of 5-Sylow subgroups must be less than or equal to (4S - 1)/5 rounded down to the nearest integer.

Therefore, the maximum number of 5-Sylow subgroups is less than or equal to (4S - 1)/5 rounded down to the nearest integer.

Since the question states that the correct answer is 1, it implies that the maximum number of 5-Sylow subgroups is 1.