Products of ActionsBilateral classes, symmetry classes of mappingsAction on k-subsetsSylows Theorem

Sylows Theorem

The following is a very important application of actions on k-subsets:
Example: The regular representation of G yields, in accordance with formula, the G-sets [G choose k], for 1 <= k <= | G | . If G is finite and p a prime dividing | G | , say | G | = pr ·q, r ³1, q=pst, where p does not divide t, then we can put k:=pr and consider the particular G-set [G choose pr], as H. Wielandt did in his famous proof of Sylow's Theorem in order to show that G possesses subgroups of order pr. His argument runs as follows: ps is the exact power of p dividing the order of [G choose pr]. This is clear from
|[G choose pr] | = (prq)/(pr) ·(prq-1)/(1) ...(prq-(pr-1))/(pr-1),
as each power of p contained in the denominator cancels. Thus pr-subsets M exist, the orbit length of which is not divisible by ps+1. We consider such an M and show that its stabilizer GM is of order pr by proving that pr is both an upper and lower bound: For each m ÎM and g ÎGM we have that gm ÎM, hence
| GM | <= | M | = pr.
On the other hand, the fact that ps+1 does not divide the orbit length | G(M) | = | G | / | GM | yields
| GM | >= pr.
This proves the first item of
Sylow's Theorem Assume G to be a finite group and p to be a prime divisor of its order. Then
  • G contains subgroups of order pr, for each power pr dividing its order | G | .
The subgroups S £G of the maximal p-power order are called the Sylow p-subgroups of G. They have the following properties:
  • Each p-subgroup U of G is contained in a suitable Sylow p-subgroup S.
  • Any two Sylow p-subgroups of G are conjugate subgroups.

The proof of the second and third item follows from a consideration of double cosets. Assume a p-subgroup U of G and a Sylow p-subgroup S. Then we derive from the corollary that

( | G | )/( | S | )= åg Î D( | U | )/( | U ÇgSg-1 | ),
where D denotes a transversal of U \G/S. If all the intersections in the denominator on the right hand side were proper subgroups of U, then the right hand side were divisible by p, which contradicts the left hand side. Hence there must exist a g ÎD, such that U £gSg-1. Since gSg-1 is a Sylow p-subgroup, too, U is contained in a Sylow subgroup, which proves the second item.

The third item follows by taking for U a Sylow p-subgroup S':

( | G | )/( | S | )= åg Î D'( | S' | )/( | S' ÇgSg-1 | )
shows that S'=gSg-1, for a suitable g Î D', where D' denotes a transversal of S' \G/S.
This example shows clearly that the consideration of suitable group actions can be very helpful, at least in group theory. Applications to other fields of mathematics will follow soon.,
last changed: August 28, 2001

Products of ActionsBilateral classes, symmetry classes of mappingsAction on k-subsetsSylows Theorem