If $H, K, L$ are subgroups of a group $G$ and $LH = HL$ and $LK = KL$, is it true that $L(H \cap K) = (H \cap K) L$? Title says it all: if $H, K, L$ are subgroups of a group $G$ and $LH = HL$ and $LK = KL$, is it true that $L(H \cap K) = (H \cap K) L$?
This seems similar in flavor to this problem, but without any constraints on containment.  I'd thought one might be able to use the fact that $LHK = HKL$ together with some form of second-isomorphism-theorem argument, but wasn't able to make it stick.
 A: No. Try $G=A_5$, $L = \langle (1,2,3,4,5) \rangle$, $H=G_1 = \langle (2,3)(4,5),(3,4,5)\rangle$, $K=G_2 = \langle(1,3)(4,5),(3,4,5)\rangle$, so $H \cap K = G_{12}=\langle(3,4,5)\rangle$.
So, $|L|=5$, $|H|=|K|=12$, and $HL=LH=KL=LK=G$.
But you can check that $L(H \cap K)$ and $(H \cap K)L$ are distinct subsets of order $15$.
A: Consider the $QR$ decomposition of $G=GL(n,\mathbb{R})$.
Every element of $G$ can be written (uniquely) as a product
$$g = q \cdot r$$
where $q$ is orthogonal and $r$ is upper triangular with all positive diagonal elements, so we have the decomposition
$$G = Q \cdot R$$
Taking inverses we get $QR = RQ = G$.
Taking transposes we also get $QL= LQ=G$, where $L$ consists of lower triangular matrices with positive diagonal entries. Now $R\cap L=D$ consists of diagonal matrices with positive entries. Note that the set $DK$ consist of matrices with orthogonal rows, while $KD$ consists of matrices with orthogonal columns, so they are different.
$\bf{Added:}$ We can get a similar counterexample with finite groups. For instance, consider $G = GL(2, \mathbb{F}_p)$, where $\mathbb{F}_p$ is the field with $p$ elements, and $p \equiv 3 \pmod 4$. The condition on $p$ is such that the equation $a^2 + b^2 = 0$ has only trivial solution in $\mathbb{F}_p$. We now have a similar decomposition
$$G = Q R= RQ = QL=LQ$$
where $Q$ consists of matrices $\left( \begin{matrix} a & - b \\b & a \end{matrix}\right)$, $R$ consists of upper triangular matrices, and $L$ of lower triangular matrices. It is easy to see that $Q(L\cap R) \ne (L\cap R) Q$.
Note: This Gram-Schmidt procedure only works in dimension $2$, since a quadratic form in $n>2$ variables over a finite field has an isotropic vector, cf Chevalley-Waring theorem.
