Generalization of Third Sylow Theorem

[glacier302]

Homework Statement

(a) Let G be a finite group that is divisible by by p^k, and suppose that H is a subgroup of G with order p^j, where j is less than or equal to k. Show that the number of subgroups of G of order p^k that contain H is congruent to 1 modulo p.

(b) Find an example of a finite group that has exactly p+1 Sylow p-subgroups.

Homework Equations

Theorem: Every p-group is contained in a Sylow p-subgroup.

Third Sylow Theorem: Let |G| = p^e*m where p does not divide m. Then the number of Sylow p-subgroups divides m and is congruent to 1 modulo p.

The Attempt at a Solution

I think that I should somehow be using the 3rd Sylow Theorem to prove (a). Also maybe the fact that since H is a p-group, it is contained in a Sylow p-subgroup, and any larger p-group containing H is also contained in a Sylow p-subgroup.

Any help would be much appreciated : )

 

[ystael]

Rather than trying to apply Sylow's theorem, you should try to apply (with appropriate modifications) its proof.