Dedekind modular law. If $A,B,C$ are subgroups of a group $G$ with $A \subseteq B$ then $A(B \cap C) = B \cap AC$.

Below is what I want to prove. Let K be a finite group with $K = LH$, where $L,H$ are subgroups of $K$ with relatively prime orders. If $U$ is a maximal subgroup of $L$ then $UH = HU$. Proof:

$HU = HU \cap LH = (HU \cap L)H = (H \cap L)UH = UH$

Is my proof true?

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    $\begingroup$ I edited your question but your should really take the time to learn how to properly write the mathematical symbols that you want to use. It's basically a matter of inserting some "$" signs here and there and it really improves the legibility of your post. $\endgroup$ – Adrián Barquero Apr 14 '12 at 18:34
  • $\begingroup$ I will try to learn. $\endgroup$ – user28083 Apr 14 '12 at 19:18
  • $\begingroup$ How did you get $HU\cap LH = (HU\cap L)H$? $\endgroup$ – anon Apr 14 '12 at 20:03
  • $\begingroup$ I put $A=H$, $B=HU$, $C=L$ and used the law in the following form $B \cap CA=(B \cap C)A$ $\endgroup$ – user28083 Apr 14 '12 at 20:10
  • $\begingroup$ Thank you very much, I did not know that I was assuming $UH$ is a group already in my proof. $\endgroup$ – user28083 Apr 18 '12 at 14:29

There is a mistake in your proof. When you state Dedekind's rule above, you assume $A, B, C$ are subgroups of a group. So, to get your second equality using Dedekind's rule you assume $B= HU$ is a group, but that's what you are trying to prove. (Recall, $UH = HU$ if and only if $UH$ is a group.)


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