Normal subgroup with prime index

Click For Summary

Homework Help Overview

The discussion revolves around a group theory problem concerning normal subgroups and their properties within p-groups. The original poster seeks to prove that every subgroup of index p in a group of order p^a is normal.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster begins by assuming a subgroup H of index p and questions the implications of H not being normal. They seek guidance on how to proceed from this assumption. Other participants inquire about relevant theorems related to normalizers in p-groups and discuss theorems that may support the original poster's claim.

Discussion Status

The discussion is exploring various theorems related to normalizers in p-groups, with some participants referencing specific results from literature that may lead to the desired conclusion. There is an ongoing examination of the implications of these theorems without reaching a consensus yet.

Contextual Notes

Participants note the importance of the order of the group and the properties of maximal subgroups in the context of the problem. There is an acknowledgment of the need for clarity on theorems being referenced and their applicability to the original question.

oyolasigmaz
Messages
32
Reaction score
0

Homework Statement


Prove that if p is a prime and G is a group of order p^a for some a in Z+, then every subgroup of index p is normal in G.


Homework Equations


We know the order of H is p^(a-1). H is a maximal subgroup, if that matters.


The Attempt at a Solution


Suppose H≤G and (G:H)=p but H is not a normal subgroup of G. So for some g in G Hg≠gH. I know I didn't do much, but is this the correct way to start? What to do now?
 
Physics news on Phys.org
Do you know any general theorems about normalizers in p-groups?
 
I am not sure about which theorem you are talking about, but I just found a theorem giving me the result I want in Dummit and Foote stating that if n is the order of the group and p the largest prime dividing n, then I have the result I wanted.
 
The theorem I was talking about is that "normalizers grow" in p-groups. This means that if G is a p-group and H < G is a proper subgroup, then H &lt; N_G(H), i.e. H is a proper subgroup of its normalizer. (This is in fact true if G is any finite nilpotent group.)

Therefore if H < G and H has index p, N_G(H) must be all of G.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K