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.


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 $$\langle B,N\rangle \cap H= G\cap H=H=\{(0,0),(1,1)\}$$

  • 3
    $\begingroup$ Took the symbols right out of my browser... $\endgroup$ – Arturo Magidin Apr 29 '11 at 19:40
  • $\begingroup$ @Arturo: LOL! :) $\endgroup$ – Zev Chonoles Apr 29 '11 at 20:04

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.


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