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

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The discussion centers on the relationship between a subgroup H of a finite group G and a normal subgroup N of G, specifically addressing the condition where the orders |H| and |G:N| are relatively prime. It is established that under these conditions, H must be a subgroup of N. The participants utilized concepts such as the normality of N and the orders of elements in the quotient group G/N to derive their conclusions. The key insight is that if |H| and |G:N| share no common factors, then every element of H must reside within N.

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This discussion is beneficial for mathematicians, particularly those specializing in abstract algebra, group theorists, and students seeking to deepen their understanding of subgroup relationships and normal subgroups in finite groups.

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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
 
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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.
 

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