Help with proof that sublattice of normal subgroups of group $G$ is modular. 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!
 A: First, consider two normal subgroups $L, H$. Let $U = \{\ell \cdot h \mid \ell \in L, h \in H\}$. I claim that $L \lor H = U$.
We shall show that $U$ is a subgroup. First, we note that $e = e \cdot e \in U$. Then we claim that for all $\ell_1, \ell_2 \in L$ and $h_1, h_2 \in H$, we have $(\ell_1 h_1) (\ell_2 h_2)^{-1} \in U$. To prove that, note that $(\ell_1 h)(\ell_2 h)^{-2} = \ell_1 h_1 h_2^{-1} \ell_2^{-1} = \ell_1 \cdot h \cdot h_2^{-1} \cdot \ell_2^{-1} = (\ell_1 \cdot \ell_2^{-1}) \cdot (\ell_2 \circ (h_1 \cdot h_2^{-1}) \cdot \ell_2^{-1})$. We see that since $\ell_1 \circ \ell_2^{-1} \in \ell$, $h_1 \circ h_2^{-1} \in H$, and $H$ is normal, we do indeed have $(\ell_1 \cdot \ell_2^{-1}) \cdot (\ell_2 \circ (h_1 \cdot h_2^{-1}) \cdot \ell_2^{-1}) \in U$.
Thus, $U$ is a subgroup. Now, let us note that for all $a$, for all $\ell \in L$ and $h \in H$, we have $a \cdot \ell \cdot h \cdot a^{-1} = (a \cdot \ell \cdot a^{-1}) \cdot (a \cdot h \cdot a^{-1}) \in U$, by the normality of $H$ and $L$. Therefore, $U$ is a normal subgroup.
Clearly, we see that $L \leq U$ (since $\ell \cdot e \in U$ for all $\ell \in L$) and $H \leq U$ (since $e \cdot h \in U$ for all $h \in H$).
Now suppose we have $U'$ such that $L, H \leq U'$. I claim that $U \leq U'$. This is immediate: for all $\ell \in L$ and $h \in H$, we see that $\ell, h \in U'$ and hence $\ell \cdot h \in U'$.
Therefore, $L \lor H = U$.
Let us go back to the problem at hand. We have normal subgroups $L, M, N$ with $L \leq M$, and we must show that $(L \lor N) \land M \leq L \lor (N \land M)$.
Suppose we had some $x \in (L \lor N) \land M$. Then $x \in L \lor N$. Write $x = \ell \cdot n$. And we know that $x \in M$.
Now since $\ell \in L$ and $L \leq M$, we see that $\ell \in M$. Therefore, since $x \in M$ and $\ell \in M$, we have $\ell^{-1} \cdot x = n \in M$.
Then $n \in N$ and $n \in M$, so $n \in N \land M$. Therefore, since $x = \ell \cdot n$, $\ell \in L$, and $n \in N \land M$, we see that $x \in L \lor (N \land M)$.
