Proving G=HK: A Finite Group Theorem

In summary, the conversation discusses the theorem that states if the indices of subgroups H and K in a finite group G are coprime, then the index of their intersection is equal to the product of their indices. The discussion also considers the possibility of G being equal to HK and the importance of determining the index of G over HK to prove this. There is also a clarification on the assumption that G and H being of coprime index automatically means their intersection is equal to 1, with a counterexample given.
  • #1
happyg1
308
0

Homework Statement



Let H and K be subgroups of a finite group G with coprime indices. Prove that G=HK

Homework Equations



From a theorem we have, If |G:H| and |G:K| are finite and coprime, we have:
|G:H intersect K|=|G:H|*|G:K|

|G:H| indicates the index of G over H, not the order here...a notational point that hurts my head.

The Attempt at a Solution


I used the thoerem and I got
|G:H intersect K|=|G:H|*|G:K|

but since G and H are of coprime index, (H intersect K=1),
So that I get

|G|=|G:H|*|G:K|

if I let |G:H|=p and |G:K|=q then |G|=pq

That's where I am and I don't think I'm headed in the right direction.

pointers and clarification will be greatly appreciated.
CC
 
Physics news on Phys.org
  • #2
Well, if you determine that |G:(HK)| =1 you're golden, right?

BTW:
but since G and H are of coprime index, (H intersect K=1),
Isn't generally true:
[tex]G = \mathbb{Z}_{12}[/tex]
[tex]H=\{0,3,6,9\}[/tex]
[tex]K=\{0,2,4,6,8,10\}[/tex]
then
[tex]|G:H|=3[/tex]
and
[tex]|G:K|=2[/tex]
are coprime, but
[tex]|G \cap K|\neq 1[/tex]
and
[tex]|G:H| \times |G:K| \neq |G|[/tex]
 
Last edited:

Related to Proving G=HK: A Finite Group Theorem

1. What is the "Proving G=HK: A Finite Group Theorem"?

The "Proving G=HK: A Finite Group Theorem" is a theorem in group theory that states that if G is a finite group and H and K are subgroups of G such that HK=G and H∩K={e}, then G is isomorphic to the direct product of H and K.

2. What is the significance of this theorem?

This theorem is significant because it provides a way to prove the structure of a finite group by breaking it down into two smaller subgroups. It also helps in understanding the structure of finite groups and their relationships with their subgroups.

3. How is this theorem proven?

This theorem is proven using the Lagrange's Theorem, which states that the order of a subgroup must divide the order of the group. By applying this theorem to both H and K, it can be shown that the intersection of H and K is the identity element, and that the product of H and K generates the entire group G.

4. Can this theorem be applied to infinite groups?

No, this theorem only applies to finite groups. In infinite groups, the order of a subgroup may not divide the order of the group, and therefore the Lagrange's Theorem cannot be applied.

5. Are there any other theorems related to "Proving G=HK: A Finite Group Theorem"?

Yes, there are other theorems related to this theorem, such as the Schur-Zassenhaus Theorem and the Kurosh Subgroup Theorem. These theorems also deal with the structure of finite groups and their subgroups.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
913
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
5K
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
5K
  • Calculus and Beyond Homework Help
Replies
7
Views
9K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
Back
Top