Given a a group of order 30 w/ a non-normal Sylow 5-subgroup, the problem is to count the number of elements of orders 5, 3, 2, and 1.
The Attempt at a Solution
I understand the Sylow theorems and can use them to show that there must be 24 elements of order 5 and 2 of order 3. Adding in the identity element accounts for 27 of the thirty elements in the group. Where I am stuck is the number of elements of order 2. There must be a least 1 Sylow 2-subgroup, which means there is at least one element of order 2. That makes for 28 elements. But what are the orders of the remaining two elements?
Now, I also know from my group theory studies that every group of order 30 has a cyclic subgroup of order 15. This means that G must have two elements of order 15: the generator of this subgroup and its inverse. Is this correct? If so, that makes for 30 elements and, voila, problem solved. However ...
There are only four groups of order 30 (Z30, D(15), Z3 x D(5), and Z5 x D(3)) and Z30 is the only one that has only one element of order 2. The group cannot be Z30 because that would make G cyclic and, hence, every subgroup would be normal (plus there would have to be at least two elements of order 30, too). However, if I say that there then must be 3 elements of order two (i.e., there are 3 Sylow 2-subgroups instead of 1), then there are no elements of order 15, which are required to form a cyclic subgroup of order 15.
Am I right in thinking that there can be only 1 Sylow 5-subgroup because of the above contradictions? Or do I have a fundamental misunderstanding of what is at work here?