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

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In summary: The left coset of H is the set of all x in H such that xH is in H. The left coset of K is the set of all x in K such that xK is in K. The intersection of two cosets is the set that contains both of them, and it's always disjoint because two cosets can't both exist in the same place.
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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[itex]\cap[/itex]K| is divisible by both |G:H| and |G:K|, here's a hint: H[itex]\cap[/itex]K is a subgroup of H, K and G.
 
  • #3
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|?
 
  • #4
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...?
 
  • #5
Just write down what |G:H|, |G:K| and |G:H[itex]\cap[/itex]K| are. It will also help to think about what |H:H[itex]\cap[/itex]K| and |K:H[itex]\cap[/itex]K| 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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1. What does it mean for G to be a finite subgroup?

For a group G to be a finite subgroup, it means that G has a finite number of elements, or in other words, the group's elements can be counted and the count is finite.

2. How are H and K related to G in this context?

H and K are both subgroups of the group G, meaning that they are smaller groups that are contained within G. They are related to G in that they both contribute to the elements of G and are necessary for proving that G=HK.

3. What does it mean for G=HK to be "proved"?

In this context, proving that G=HK means showing that the group G can be written as a combination of the subgroups H and K. This can be done by showing that every element in G can be expressed as the product of an element in H and an element in K, and vice versa.

4. How is the proof of G=HK useful in mathematics?

The proof of G=HK is useful in mathematics because it helps us understand the structure and relationships between different groups. It also allows us to simplify complex groups by breaking them down into smaller, more manageable subgroups.

5. Are there any specific conditions that must be met for G=HK to be true?

Yes, in order for G=HK to be true, certain conditions must be met. These include G, H, and K all being finite subgroups of G, and H and K being normal subgroups of G (meaning that they are invariant under conjugation by elements of G).

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