A divisibility question.

## Primary tabs

# A divisibility question.

Submitted by estefan34 on Sun, 04/22/2012 - 19:16

Forums:

Notation:

A int B means the intersection between A and B.

Let N be a normal subgroup of G and P a Sylow p-subgroup of G. In the chain of subgroups 1 <= P int N <= N <= PN <= G we get

|P int N||PN/N| = |P| (1)

since PN/N isomorphic to P/(P int N). And here I want to prove that |N:P int N| and |G:PN| are not divisible by p.

Suppose p | |N|/|P int N|. Then |N| = tP|P int N| and |N||PN/N| = tp|P int N||PN/N|. By (1) then, |N||PN/N| = tp|P|, which gives |PN| = tp|P|. Hence |P||N|/|P int N| = tp|P| and K:= |N|/|P int N| = tp where (t,p) = 1. That is, p^1 | K but p^2 does not divide K. And I need p^0 | K but p^1 does not divede K. Quite a pity.

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Re: A divisibility question.

I found a proof. Let |G| = sp^e, (p,e) = 1, |P| = p^e. Then p^e | |PN| | sp^e. So

(1) p^e | |PN|

(2) p^(e+1) does not divide |PN|.

From here every thing else follows. For instance |P| = |P int N||PN/N|. Hence |PN| = |P||N:P int N|. If p | |N:P int N|, then p^(e+1) | |PN|, which contradicts (2).