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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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    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.2012-04-14
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    I will try to learn.2012-04-14
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    How did you get $HU\cap LH = (HU\cap L)H$?2012-04-14
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    I put $A=H$, $B=HU$, $C=L$ and used the law in the following form $B \cap CA=(B \cap C)A$2012-04-14
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    Thank you very much, I did not know that I was assuming $UH$ is a group already in my proof.2012-04-18

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