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 fromThis 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.|[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 ofSylow's Theorem Assume G to be a finite group and p to be a prime divisor of its order. Then
The subgroups S £G of the maximal p-power order are called the Sylow p-subgroups of G. They have the following properties:
- G contains subgroups of order pr, for each power pr dividing its order | G | .
- 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.