Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $G$ be a finite group, let $B,N,H$ be subgroups of $G$. I believe that $$ \langle B \cap H, N \cap H \rangle = \langle B,N \rangle \cap H $$ but I do not find a satisfactory proof. I think this may be useful. I tried to use it but I did not manage to write a 100% satisfactory proof. Can someone help me with it?

Thanks in advance.

share|improve this question
add comment

2 Answers

up vote 11 down vote accepted

This is false. Let $G=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, let $B=\langle (1,0)\rangle$, $N=\langle (0,1)\rangle$, and let $H=\langle(1,1)\rangle$. Then $$\langle B\cap H,N\cap H\rangle=\langle \{(0,0)\},\{(0,0)\}\rangle=\{(0,0)\}$$ but $$\langle B,N\rangle \cap H= G\cap H=H=\{(0,0),(1,1)\}$$

share|improve this answer
3  
Took the symbols right out of my browser... –  Arturo Magidin Apr 29 '11 at 19:40
    
@Arturo: LOL! :) –  Zev Chonoles Apr 29 '11 at 20:04
add comment

This is false. Let G be the non-abelian group of order 6. Let B,N be distinct Sylow 2-subgroups of G, and let H be the Sylow 3-subgroup of G. Then B∩H=N∩H=1, but ⟨B,N⟩∩H=G∩H =H.

A related and true statement is called Dedekind's modular law.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.