# Show H is a subgroup of N

1. Sep 17, 2011

### symbol0

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|, |HnN| and |H|.
Numbers that divide |G:N| are |GN| and |HN:N|.
I'm stuck.

I would appreciate any suggestions.
Thanks

2. Sep 17, 2011

### micromass

Staff Emeritus
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??

3. Sep 17, 2011

### symbol0

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.

4. Sep 17, 2011

### micromass

Staff Emeritus
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?

5. Sep 17, 2011

### symbol0

Thank you micromass,
I got it.