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Let $G$ be a group and $W(G)$ the set of all subnormal subgroups of $G$, partially ordered by inclusion. My question is if $W(G)$ always forms a lattice (but not necessarily a sublattice of $L(G)$). I've heard it holds when $W(G)$ satisfies the maximal condition.

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For finite groups $G$ the set $W(G)$ is a sublattice of $L(G)$, since the intersection $A\cap B$ of two subnormal subgroups is again subnormal in $G$, and the join $\langle A,B\rangle$ is again subnormal. For infinite groups, the first part remains true, but the last part need not be true. In fact, Zassenhaus gave an example where the join is not subnormal again. It was recognized that here chain conditions play an important role.

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  • $\begingroup$ So does $W(G)$ always form a lattice? $\endgroup$
    – Censi LI
    Jun 10, 2015 at 8:37
  • $\begingroup$ @CensiLI, since it is a sublattice... $\endgroup$ Jun 10, 2015 at 8:41
  • $\begingroup$ @MarianoSuárez-Alvarez But it's not always a sublattice of $L(G)$, if no constraint imposed on $G$. $\endgroup$
    – Censi LI
    Jun 10, 2015 at 8:42
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    $\begingroup$ @DietrichBurde In Zassenhaus example, I think that the join is not subnormal again only violates $W(G)$ to be the sublattice of $L(G)$, but does not violate $W(G)$ to be a lattice. I don't know how to show that $W(G)$ in Zassenhaus example is not a lattice. Can you give some hint? $\endgroup$
    – Censi LI
    Jun 10, 2015 at 12:17
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    $\begingroup$ @CensiLI I thought that every lattice is a sublattice of the subgroup lattice of some group, and that somehow it is the same to say that it is a sublattice of this, or a lattice in itself. But I have to admit that I don't know this really. All results with maximum conditions etc. refer to a sublattice of $L(G)$, so this seems to be more natural to consider. $\endgroup$ Jun 10, 2015 at 14:13

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