Lemma: Let and be distinct primes and let and be Sylow and subgroups of a finite group . Then is trivial.
Proof: An element of the intersection has order dividing a power of and a power of , so it must have order .
Abelian Groups
Proposition: Let be a finite abelian group and let be a prime dividing the order of . Let be the -Sylow subgroup of . Then consists of the elements of whose order is a power of .
Proof: Since is abelian, it’s Sylow -subgroup is unique. Clearly if then the order of is a power of . Conversely if has order a power of , it generates a subgroup of -power order, which must be contained in .
Corollary: Let be the order of and suppose where . Let be the elements of whose order divides . Then:
is a subgroup of .
is the trivial subgroup.
The map given by is an isomorphism.
Proof:
If and have order and , both dividing , then . Let , which divides . Then so has order dividing as well.
This is the lemma above.
If is in the kernel of the map, then since we have which means both and are in the intersection of and . By (2) this means so the map is injective. Then by counting we conclude it is surjective.
Corollary: Any finite abelian group is isomorphic to the product of its Sylow -subgroups.
This reduces the classification problem for finite abelian groups to the classification of finite abelian -groups.
Groups of order .
Suppose has order and suppose . Let be a Sylow subgroup. The number of such Sylow subgroups must be one of and must be congruent to mod . The only possibility is . Thus is a normal subgroup. Now let be any Sylow -subgroup. The intersection is trivial so is a group of order and must be .
Suppose further that . Then must also be normal, because there cannot be Sylow -subgroups. acts on by conjugation, and in that action acts trivially (since is just a copy of which is abelian). Therefore acts on by group automorphisms. But which has order and is a cyclic group of order , so there is no nontrivial map . This means commutes with , so this group is abelian and in fact cyclic of order .
If , then the group need not be abelian. For example or with and odd prime are nonabelian groups of order with .
Groups of order 30
Proposition: A group of order has a normal subgroup of index (which is necessarily cyclic of order by the previous problem).
Lemma: Let be a Sylow -subgroup and a Sylow subgroup. If either of these groups is normal in , then is a subgroup of of order – which must be normal since it is of index . Now let . Then is contained in and since is the Sylow -subgroup of we have . Therefore is normal in ; and similarly is normal in . So if either of the Sylow subgroups is normal, so is the other one, and we’ve found our subgroup.
Assume therefore that neither of them is normal. Then there must be at least conjugates of the Sylow -subgroup and of the Sylow subgroup. This accounts for $64+102=44G30$.
Groups of order .
Proposition: Let be a finite group of order . Then either contains a normal subgroup of order or is isomorphic to .
Proof: Let be the number of Sylow -subgroups. If we are done. The divisors of are . If then . Therefore the group permutes the four Sylow -subgroups transitively and this gives us a map from to PPP34N_{G}(P)=P$.
The map to given by the conjugation action is injective. To see this, suppose is in the kernel of the action. This means , . But it also means for one of the other conjugates of . That means is also in , or . Since is not , the intersection of with is a proper subgroup of , and since is cyclic of order that must be the trivial subgroup.
So the image of the permutation map is a subgroup of of order . Furthermore, the image meets in at least elements (the identity plus 8 elements of order 3). So it must be all of .
A few more
Suppose has order . The divisors of are . The number of Sylow -subgroups has to be congruent to mod , so it has to be . The number of Sylow subgroups has to be one mod , so it has to be one. So there is a normal subgroup of order and another of order . A group of order must be abelian so the only possibilities are or .
The smallest possible odd order for a nonabelian group is , and there is such a nonabelian group. It has -Sylow subgroups and a normal -Sylow subgroup.