Question regarding sylow subgroups

  • Thread starter geor
  • Start date
  • #1
35
0

Main Question or Discussion Point

Hello everybody,

I I have an exercise here that I'm really stuck with..

Let P be in Syl_p(G) and assume N is a normal subgroup of G. Use the conjugacy part of Sylow's theorem to prove that P intesect N is a Sylow p-subgroup of N.

The "conjugacy part" in this book ('Algebra' by Dummit, Foote) is about
p-subgroups of G being subgroups of a conjugate of a Sylow p-subgroup
in G.

I tried some different approaches but can't get nowhere..
Using the "conjugacy part of the thm" for P intersect N has no
use since already P intersect N is a subgroup of P itself, so
we have to use it for another subgroup.

I also considered Q a sylow p-subgroup of N. Eventually I would
like to show that Q and (P intersect N) have the same cardinality.
Well, (P intersect N) is a p-subgroup of N and Q is a sylow p-subgrp of N
so that from "the conjugacy part of Sylow's thm" we have that
(P intersect N) is a subgroup of a conjugate of Q in G, that is,
(P intersect N) <= gQg^-1, some g in G.
That seems to be something but can't go more far..


Finally, the fact that N is normal smells like 2nd iso theorem..
From this, we deduce that (P intersect N) is normal sbgrp of P
and N is a normal sbgrp of NP.. Also
P / (P intersect N) is isomorphic to NP / N.
So we can play with the cardinalities.
But can't think of anything else :(

Any ideas or hints highly appreciated!
Thanks in advance!

PS. I wasn't sure if I should post this in the 'homework' section. This is an exercise
given in a first year graduate course in Algebra and I think that it shouldn't be put
with "calculus and beyond" questions..


EDIT:
Actually the exercise has another part: "Deduce that PN/N is a sylow p-sbgrp of G/N".
Well, there is a chance that we don't need the normality of N to show the first part of the question (the thing that I asked)
 
Last edited:

Answers and Replies

  • #2
mathwonk
Science Advisor
Homework Helper
10,947
1,099
im a little sleepy but it seems kind of obvious. take any p subgroup of N and enlarge it to a sylow subgroup of G which is conjugate to your sylow subgroup, so your p group is conjugate to a subgroup of your p group intersected with N, so that bthing is maximal, hence sylow. ???
 
  • #3
35
0
Oh well..
'm sleepy too, maybe that's why I didn't understand... :(

Could you please use some letters?

Let's take an p-subgroup H<N.
When you say 'enlarge it' you mean 'see it as' right (forgive my english)?
So, H<K where K = g P g^-1.

So H is conjugate to a subgroup of H intersected with N?
I don't understand that..

Sorry for the lack of understanding, maybe it will be more clear
tomorrow when I wake up :)
 

Related Threads on Question regarding sylow subgroups

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
6K
  • Last Post
Replies
10
Views
4K
Replies
3
Views
4K
  • Last Post
Replies
2
Views
2K
Top