TITLE: Dedekind modular law
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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.)