If G is a group with exactly 66 elements of order 14, how many subgroups of order 14 does G have?
To answer this question, we need to understand the concept of subgroups and the properties of groups.
A subgroup of a group G is a subset of G that is also a group under the same operation. In this case, we are interested in subgroups of order 14.
First, note that if a group has elements of order 14, then it must have a subgroup of order 14 because the order of a subgroup must divide the order of the group.
Now, let's consider a subgroup H of order 14. By Lagrange's theorem, the order of any subgroup must divide the order of the group. So, the order of H must be a divisor of 66.
Since the only divisors of 66 are 1, 2, 3, 6, 11, 22, 33, and 66, we need to determine which of these divisors could correspond to the order of a subgroup of G.
We are given that G has exactly 66 elements of order 14. This means there are exactly 66 elements in G that generate subgroups of order 14. So, we want to count the number of elements of each order.
Let's denote the number of elements of order n in G as a(n). We know that a(1) = 1 (the identity element), and a(14) = 66 (given in the question).
Now, consider the element g ∈ G with order 14. By the definition of order, g^14 = e, where e is the identity element. Therefore, the order of g^2 is 7, the order of g^3 is 14/3 (which cannot be an integer since the order must be a positive integer), the order of g^4 is 7/2 (which also cannot be an integer), and so on.
From this, it follows that a(7) = a(14/2) = a(7/2) = a(14/4) = ... = 66.
So, in total, there are 66 elements of order 14 and 66 elements of order 7. The remaining elements of G must have orders equal to one of the divisors of 66 other than 1, 7, or 14.
Now, let's focus on the divisors of 66, excluding 1, 7, and 14. These divisors are 2, 3, 6, 11, 22, 33, and 66. For each of these divisors, we need to determine the number of elements of that order in G.
Since the order of every element must divide the order of the group, each divisor will have a corresponding number of elements in G. For example, if there are 2 elements of order 2, then there must be a subgroup of order 2 in G.
Using the same reasoning as above, we find that a(2) = 33, a(3) = 22, a(6) = 11, a(11) = 6, a(22) = 3, a(33) = 2, and a(66) = 1.
Finally, we sum up the counts of elements with orders 2, 3, 6, 11, 22, 33, and 66 (excluding 1, 7, and 14) to find the total number of subgroups of order 14 in G.
So, the number of subgroups of order 14 in G is 33 + 22 + 11 + 6 + 3 + 2 + 1 = 78.