Let $T$ be a group and $G,A_{i} \subseteq T$ be subgroups (for $i \in I$). Denote the lattice of subgroups of $G$ as $L(G)$.
I am interested in the following sublattice of $L(G)$ defined as $M=\{H \in L(G) | H \cap A_{i} = G \cap A_{i} \text{ for all } i \in I\}$ .
Instinct tells me that $M$ should be isomorphic to $L(G')$ where $G' = G/(G \cap \langle A_{i} \rangle_{i \in I})$ (providing normality of $G \cap \langle A_{i} \rangle_{i \in I}$).
When all $A_{i}=1$ we have that $L(G')=L(G)$ which is isomorphic to $M$, when all $A_{i}=G$ then $L(G')=\{1\}$ which is isomorphic to $M$.
If $M \cong L(G')$ I'm seeking a outline/draft of a proof and else a expression for the corrected $G'$ in terms of $G$? Thanks.
By the lattice theorem $L(G/\langle G \cap A_{i} \rangle_{i \in I})$ is isomorphic to $L'= \{H | \langle G \cap A_{i} \rangle_{i \in I} \subseteq H \subseteq G \}$.
If $H \in L'$ then $G \cap A_{i} \subseteq H \subseteq G$ for each $i \in I$ so $G \cap A_{i} = H \cap A_{i}$ for each $i \in I$ and hence $H \in M$.
If $H \in M$ then $G \cap A_{i} = H \cap A_{i} $ for each $i \in I$ so $G \cap A_{i} \subseteq H$ for each $i \in I$. Hence $\langle G \cap A_{i}\rangle_{i \in I} \subseteq H$ and as $H \subseteq G$ by assumption we have that $H \in L'$.
So if $G' = \langle G \cap A_{i} \rangle_{i \in I} $ then $ L(G') \cong L' = M$.