# Poincare's theorem about subgroups of finite index

• September 26th 2009, 03:23 AM
ENRIQUESTEFANINI
Poincare's theorem about subgroups of finite index
Hi:
I'm given this problem: let G be a group and A, B subgroups of finite order in G. Let $D=A\cap B$. Prove that $u,v \epsilon At\cap Bs \Rightarrow Du=Dv$. (1) Using this, prove that $[G:D]\leq [G:A][G:B]$ (Poincare's theorem).

I can prove the theorem by defining $f: G_D \rightarrow G_A \times G_B, f(Dd)=((Ad, Bd), where G_D$ is the set of right
cosets of D in G and so on for $G_A and G_B$. f is one-to-one and therefor $G_D$ is a finite set. Furthermore, $|G_D|\leq |G_A||G_B|$.

My question is: can (1) make the proof simpler or more straightforward than that given by me? (assuming it is correct). Because I've tried to use (1) to prove the theorem but I've failed. Any suggestion would be welcome. Thanks for reading.
• September 26th 2009, 11:19 AM
Swlabr
Quote:

Originally Posted by ENRIQUESTEFANINI
Hi:
I'm given this problem: let G be a group and A, B subgroups of finite order in G. Let $D=A\cap B$. Prove that $u,v \epsilon At\cap Bs \Rightarrow Du=Dv$. (1) Using this, prove that $[G:D]\leq [G:A][G:B]$ (Poincare's theorem).

I can prove the theorem by defining $f: G_D \rightarrow G_A \times G_B, f(Dd)=((Ad, Bd), where G_D$ is the set of right
cosets of D in G and so on for $G_A and G_B$. f is one-to-one and therefor $G_D$ is a finite set. Furthermore, $|G_D|\leq |G_A||G_B|$.

My question is: can (1) make the proof simpler or more straightforward than that given by me? (assuming it is correct). Because I've tried to use (1) to prove the theorem but I've failed. Any suggestion would be welcome. Thanks for reading.

Poincare's Theorem states that the intersection of finitely many subgroups of finite index is a subgroup of finite index. So it is sufficient to prove the result for two subgroups, which is what we are doing here.

Let [tex]D= A \cap B[/MATH and $g \in Dh$ where $Dh$ is some coset of $D$, $g,h \in G$.

$\Rightarrow gh^{-1} \in D \Rightarrow g \in Ah \cap Bh$ $\Rightarrow Dh \leq Ah \cap Bh$

Clearly $Ah \cap Bh \leq Dh \Rightarrow Ah \cap Bh = Dh$.

Thus, the number of cosets of $D$ is less than or equal to $[G:A]*[G:B]$, and we are done.

Also, you can put text into your LaTeX by writing "\text{put text here}". For example, $g \text{ is an element of } G$.