Element Orders For Group Of Order 30

Homework Statement

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?

I think you are correct. A group of order 30 must always have a normal Sylow-5 subgroup.

Also note, that in a group of order 15, there are multiple elements with order 15. You said that there were only 2, but that is incorrect...

I think you are correct. A group of order 30 must always have a normal Sylow-5 subgroup.

Also note, that in a group of order 15, there are multiple elements with order 15. You said that there were only 2, but that is incorrect...

True. I should have said that there are at least two elements of order 15 (there are eight actually). Now, does that mean that every group of order 30 will have at least eight elements of order 15? I think it does, since the group of order 15 is cyclic -- not so sure if that order is preserved from group to subgroup in general though. However, now that I think about it, that cant be true either because that would put us at over 30 elements both in this case and in cases where there are 10 Sylow 3-subgroups and only 1 Sylow 5-subgroup. So I am not sure how many elements of order 15 there must be.

I do think that there are always 8 elements of order 15.
This would imply that the 3-Sylows and the 5-Sylows are always normal!

There are indeed 4 groups of order 30: Z30, D15, Z3xD5 and Z5xD3. And they all have normal 3-Sylows and 5-Sylows...

Thanks for helping me think through this!