Group theory MOC

Sylow’s theorem

Let be a finite group of order where is prime and . Then group

1. , the set of [[Sylow p-subgroup|Sylow -subgroups]], is non-empty;
2. Every [[p-group|-subgroup]] of is contained in a Sylow p-subgroup;
3. All Sylow -subgroups are conjugate in ;
4. If then divides and .

Proof due to Wielandt

In this proof group actions are taken to be right actions.

Let be the set of all subsets of of size , and act on by right multiplication. Note

which is not divisible by , so there must be a -orbit on of degree not divisible by , Let be such an orbit and . Let . We claim , whence follows is a Sylow -subgroup. Since , we know . Now

and acting on is free, so each -orbit of has size . Herefore and thus .

To recap, if and is not divisible by , then , proving ^S1.

Now let be a -subgroup of , and be as above. Then acts on . The -orbits on have sizes which are powers of . Since does not divide , it follows there must exist a singleton orbit , so , the latter being a Sylow -subgroup, proving ^S2.

Let . From directly above, and for some , thus for some . It follows

proving ^S3.

acts (transitively by ^S3) on by conjugation, and where is the normalizer of by the Orbit-stabilizer theorem. Since , it follows divides , whence divides .

acts by conjugation on , and by ^S2 for some . The only orbit of size 1 is : For if then for all , so . Since , and is a -group, so .

All orbits of on have size a -power, therefore all orbits other than have size divisible by . Therefore , proving ^S4.

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