Is H a Subgroup of N if |H| and |G:N| are Relatively Prime?

  • Thread starter Thread starter symbol0
  • Start date Start date
  • Tags Tags
    Subgroup
symbol0
Messages
77
Reaction score
0
Let G be a finite group, let H be a subgroup of G and let N be a normal subgroup of G. Show that if |H| and |G:N| are relatively prime then H is a subgroup of N.

I have tried using the fact that since N is normal, HN is a subgroup of G.
Suposing that H is not contained in N, I tried finding a common factor for |H| and |G:N|.
Numbers that divide |H| are |HnN|, |H:HnN| and |H|.
Numbers that divide |G:N| are |G:HN| and |HN:N|.
I'm stuck.

I would appreciate any suggestions.
Thanks
 
Physics news on Phys.org
Take a h\in H. You want to show that h\in N. Now, what can you say about h+N\in G/N?? What is its order??
 
The order of hN in G/N is the smallest integer k such that h^k = n for some n in N.
We know that k divides |G/N|. I don't know if k divides |H|. We know that h^k is in the intersection of H and N.
 
What I meant was: let k be the order of h. Then we know that h^k=e. And also (hN)^k=N. So the order of hN divides k. What can you conclude?
 
Thank you micromass,
I got it.
 
The world of 2\times 2 complex matrices is very colorful. They form a Banach-algebra, they act on spinors, they contain the quaternions, SU(2), su(2), SL(2,\mathbb C), sl(2,\mathbb C). Furthermore, with the determinant as Euclidean or pseudo-Euclidean norm, isu(2) is a 3-dimensional Euclidean space, \mathbb RI\oplus isu(2) is a Minkowski space with signature (1,3), i\mathbb RI\oplus su(2) is a Minkowski space with signature (3,1), SU(2) is the double cover of SO(3), sl(2,\mathbb C) is the...
Back
Top