KB, really? "Up to isomorphism, a group of order 15 is cyclic. Thus, a subgroup of order 15 in G is automatically normal." You assert that a cyclic subgroup is always normal? Or perhaps that since the subgroup is abelian, it is normal?

Utter nonsense. I'm sick of seeing these questions given short-cut answers that are essentially (or sometimes literally) "if you know semidirect products, this is obvious." And often, as is the case here, the proof is full of not just gaps, but flaws. Maybe the idea is right, but the execution is terrible, and (importantly) not helpful.

Let's try to prove this rigorously! That would be helpful, now wouldn't it?

First, consider the Sylow theorems (theorems are the things we cite in order to do proofs). The Sylow theorems guarantee that there are subgroups of order 2, 3, and 5 since 30 = 2×3×5.

If we let:

n2 = number of 2-subgroups

n3 = number of 3-subgroups

n5 = number of 5-subgroups

then we can cite the Sylow theorems to say that:

n2 = 1 (mod 2) and n2 divides 15

n3 = 1 (mod 3) and n3 divides 10

n5 = 1 (mod 5) and n5 divides 6

Examining divisors of 10 tell us n3 must be 1 (since 2, 5, and 10 are not 1 mod 3). The same logic tells us n5=1. Call the subgroup of order 3 H and the one of order 5 K.

However, the conjugates of any of these subgroups must be a subgroup also -- if there is only one subgroup of order 5 (as we know now there is), then any conjugate of this group must be itself. Why? Well, any gHg⁻¹ is a subgroup of order 3, and thus gHg⁻¹=H, and so gH=Hg. This means H is normal. Likewise for K.

Now by the second isomorphism theorem, HK is also a normal subgroup of G. Notice that any element of H∩K must have order dividing |H| and |K|. But only 1 divides 3 and 5, so H∩K = {e}.

The second isomorphism theorem tells us:

HK / K = H / H∩K = H

Then by LaGrange's theorem |HK| / |K| = |H|, so |HK| = |H| |K| = 15.

Thus HK is a normal subgroup of index 15.

NOW you are ready to proceed.

Let N = HK, for shorthand. Let P denote one of the 2-subgroups. There could be 1, 3, 5, or 15 of these. That means there are at least four cases. But we need to show there are ONLY four cases, so we have to keep going.

We will again apply the second isomorphism theorem. By the exact same arguments above (for H and K), we can also show that G = NP. Furthermore, we can examine the multiplication in G by writing any elements as np and n'p'. Because N is normal, pN = Np, so take the n'' such that pn' = n''p.

Then npn'p' = nn'' pp'.

In a sense, what you want is to take the product on G and make it look like a product on N×P. If n''=n' then it is the standard group product. However, this need not be the case.

You will instead need to construct the map φ( n' ) = n'' as above. This is an automorphism of N. This φ exists for each p. However, what p are there? The element p comes from the group P, which is cyclic of order 2. Clearly, in the case that p is the identity, n'=n'', which means φ is also the identity map.

In the other case, we have an automorphism of N.

N is the group of order 15 -- the above arguments show it is unique, since N itself contains a normal 3-Sylow and 5-Sylow subgroup, and their product is N.

So what are the automorphisms of N? Well, they must send H to H and K to K. H has two generators, and K has four. It has to rearrange the generators.

But the automorphisms we want need to be of order 2, because they need to work like "p" from our group P. The only automorphisms that qualify are the ones that act like φ(n) = n⁻¹. It can act like this, or like the identity, on either H or K.

The only possible four cases are:

φ is the identity on H and K.

φ is the identity on H and φ(n)=n⁻¹ on K.

φ is the identity on K and φ(n)=n⁻¹ on H.

φ is φ(n)=n⁻¹ on H and K.

This implies there are at most four of these groups. But (way above) we know there must be at least four.

Thus there are exactly four.

QED