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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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    Thank you very much, I did not know that I was assuming $UH$ is a group already in my proof.2012-04-18

1 Answers 1

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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.)