Sylow theorems, proof of
We let be a group of order where and prove Sylow’s theorems.
First, a fact which will be used several times in the proof:
If divides the size of every conjugacy class outside the center then divides the order of the center.
has a Sylow p-subgroup
Suppose , , and the holds for all groups of smaller order. Then we can consider whether divides the order of the center, .
If it does, then by Cauchy’s theorem, there is an element of of order , and therefore a cyclic subgroup generated by , , also of order . Since this is a subgroup of the center, it is normal, so is well-defined and of order . By the inductive hypothesis, this group has a subgroup of order . Then there is a corresponding subgroup of which has .
On the other hand, if then consider the conjugacy classes not in the center. By the proposition above, since is not divisible by , at least one conjugacy class can’t be. If is a representative of this class then we have , and since , . But , since , so has a subgroup of order , and this is also a subgroup of . ∎
The following construction will be used in the remainder of the proof:
Given any Sylow p-subgroup , consider the set of its conjugates . Then for some . Observe that every is a Sylow p-subgroup (and we will show that the converse holds as well). We let act on by conjugation:
This is clearly a group action, so we can consider the orbits of under it; this remains true if we only consider elements from some subset of . Of course, if all is used then there is only one orbit, so we restrict the action to a Sylow p-subgroup . the orbits , and let be representatives of the corresponding orbits. By the orbit-stabilizer theorem, the size of an orbit is the index of the stabilizer, and under this action the stabilizer of any is just , so .
There are two easy results on this construction. If then . If then , and since the index of any subgroup of divides , .
The number of conjugates of any Sylow p-subgroup of is congruent to modulo
In the construction above, let . Then and for . Since the number of conjugates of is the sum of the number in each orbit, the number of conjugates is of the form , which is obviously congruent to modulo .
Any two Sylow p-subgroups are conjugate
Given a Sylow p-subgroup and any other Sylow p-subgroup , consider again the construction given above. If is not conjugate to then for every , and therefore for every orbit. But then the number of conjugates of is divisible by , contradicting the previous result. Therefore must be conjugate to . ∎
The number of subgroups of of order is congruent to modulo and is a factor of
Since conjugates of a Sylow p-subgroup are precisely the Sylow p-subgroups, and since a Sylow p-subgroup has modulo conjugates, there are modulo Sylow p-subgroups.
Since the number of conjugates is the index of the normalizer, it must be . Since is a subgroup of its normalizer, , and therefore . ∎
|Title||Sylow theorems, proof of|
|Date of creation||2013-03-22 12:51:02|
|Last modified on||2013-03-22 12:51:02|
|Last modified by||Henry (455)|