I am trying to prove that the set $N(G)$ of normal subgroups of a group $G$ is a sublattice of the lattice of subgroups of $G$, $Sub(G)$ where the meet operation ($\lor$) is defined as usual intersection of sets and the join operation ($\land$) of two normal subgroups is given by the subgroup generated by their union.
Let $L,M,N\in Sub(G)$. To show $N(G)$ is modular, I need to show that, if $L\subseteq M$, then $(L\lor N)\land M=L\lor (N\land M)$.
I have already shown that $L\lor (N\land M)\subseteq (L\lor N)\land M$, I now need to show the reverse inequality. I tried to do this with plain algebraic methods but I didn't succeed. There is a proof on planetmath.org, but I can't follow the reasoning of the proof that "$L\land H=LH$ for normal subgroups $L,H$ of $G$" (see here).
An arbitrary element of $L\land H$ is of the form $x_1 x_2\ldots x_n$ for some $n \in\mathbb{N}$, where $x_i\in L$ or $H$ for every $1\leq i \leq n$. How can we show that we can reduce this product so that no two adjacent elements are from the same subgroup? I don't understand this part of the proof.
Any help is much appreciated. Thanks!