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Let $G$ be a finite group. Consider the lattice $$L=\{ G/N:\text{$ N $ is a normal subgroup of $G $}\},$$ where $G/N \leq G/K$ if and only if $K\leq N$. The lattice operations ∧ and ∨ on quotient groups $G/N, G/K$ are just $G/(NK)$ and $G/(N \cap K)$ respectively.

Is there a lattice $L^{1}$ ordered by inclusion and isomorphic to $L$ such that the cardinality of the corresponding elements of $L$ and $L^{1}$ are equal?

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  • $\begingroup$ $L$ is a lattice ordered by inclusion and isomorphic to $L$ such that the cardinality of the corresponding elements of $L$ and $L$ are equal, so you can just take $L^1$ to be $L$. But there must be something more you want from $L^1$? $\endgroup$ – Gerry Myerson May 6 '18 at 23:02
  • $\begingroup$ For any element A of $L^{1}$, I need to know what is the cardinality of the set of all elements of A that are not in any element B in $L^{1}$ and $B \subset A $. $\endgroup$ – Farid Aliniaeifard May 6 '18 at 23:20
  • $\begingroup$ Sorry, you've lost me. Do you have an example? [And is this actually a question of math research? That's what this MO website is for.] $\endgroup$ – Gerry Myerson May 6 '18 at 23:30
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    $\begingroup$ @GerryMyerson Actually, if $K<N$ then $G/K$ and $G/N$ are disjoint as sets, as a coset $aK$ has cardinality $|K|$ whereas a coset $aN$ has cardinality $|N|$. So $L$ is not itself ordered by inclusion. $\endgroup$ – Bjørn Kjos-Hanssen May 7 '18 at 6:26
  • $\begingroup$ @Bjørn, so what's happening is the $\le$ in $G/N\le G/K$ isn't inclusion but is defined by $K\le N$ where the $\le$ is inclusion. OK, now I think I understand. $\endgroup$ – Gerry Myerson May 7 '18 at 12:39

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