Sylows Theorem

Sylows Theorem

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 | = p^r ·q, r ³1, q=p^st, where p does not divide t, then we can put k:=p^r and consider the particular G-set [G choose p^r], as H. Wielandt did in his famous proof of Sylow's Theorem in order to show that G possesses subgroups of order p^r. His argument runs as follows: p^s is the exact power of p dividing the order of [G choose p^r]. This is clear from

|[G choose p^r] | = (p^rq)/(p^r) ·(p^rq-1)/(1) ...(p^rq-(p^r-1))/(p^r-1),

as each power of p contained in the denominator cancels. Thus p^r-subsets M exist, the orbit length of which is not divisible by p^s+1. We consider such an M and show that its stabilizer G_M is of order p^r by proving that p^r is both an upper and lower bound: For each m ÎM and g ÎG_M we have that gm ÎM, hence

| G_M | <= | M | = p^r.

On the other hand, the fact that p^s+1 does not divide the orbit length | G(M) | = | G | / | G_M | yields

| G_M | >= p^r.

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 p^r, for each power p^r 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 | )= å_{g 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 | )= å_{g 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.

Sylows Theorem