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!

• If adjacent elements come from the same subgroup, replace them with their product, which also is in the same subgroup because it's a subgroup and therefore closed under the group operation. Commented Oct 4, 2021 at 23:18

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 $$x \in L \lor (N \land M)$$.