Prove G=HK When G, H, K are Finite Subgroups of G

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The discussion centers on proving that if G is a finite group and H and K are finite subgroups of G with relatively prime indices |G:H| and |G:K|, then G equals the product of H and K, denoted as G=HK. Participants highlight the necessity of showing that |G:H∩K| is divisible by both |G:H| and |G:K|, leveraging the counting principle. The conversation also emphasizes the importance of understanding cosets and their properties, clarifying that cosets are sets rather than equivalence classes.

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Homework Statement


Let G be a group and let H,K be subgroups of G.
Assume that G is finite and that the indices |G:H| and |G:K| are relatively prime. Show that G=HK.

Hint: Show that |G:H(intersect)K| is divisible by both |G:H| and |G:K| and then use the counting principle for |HK|.

The Attempt at a Solution



First off, why do the indices have to be relatively prime?
I don't know how to show that |G:H(intersect)K| is divisible by both |G:H| and |G:K|, but I do know that if I assume those, I know how to use the counting principle because ultimately it will come down to saying that |HK|=c|G| for some multiple c, and c must = 1 otherwise it says that for c>1, |HK|>|G| and that is not possible.

EDIT:
Is the intersection of the left coset of H and the left coset of K disjoint? Since they are both equivalence classes they would have to either be disjoint or equal, no? So then |G:H(intersect)K| would consist of both xH and xK for some x in G...?
 
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If the indices were not prime then it's easy to come up with examples where G does not equal HK.

As to showing that |G:H\capK| is divisible by both |G:H| and |G:K|, here's a hint: H\capK is a subgroup of H, K and G.
 
I understand that H(union)K is a subgroup of H, K and G. But I don't understand how the numbers would work. How do we know that |G:H(union)K| is definitely a multiple of both |G:H| and |G:K|?
 
I just added this "edit" into my original question:

Is the intersection of the left coset of H and the left coset of K disjoint? Since they are both equivalence classes they would have to either be disjoint or equal, no? So then |G:H(intersect)K| would consist of both xH and xK for some x in G...?
 
Just write down what |G:H|, |G:K| and |G:H\capK| are. It will also help to think about what |H:H\capK| and |K:H\capK| are.

fk378 said:
Is the intersection of the left coset of H and the left coset of K disjoint? Since they are both equivalence classes they would have to either be disjoint or equal, no? So then |G:H(intersect)K| would consist of both xH and xK for some x in G...?
"The" left coset of H? I think you need to review your definitions. A coset is not an equivalence class; it's a set.
 
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