Given a lattice L with n elements, are there finite groups H < G such that L $\cong$ the lattice of subgroups between H and G? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:49:54Z http://mathoverflow.net/feeds/question/85724 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85724/given-a-lattice-l-with-n-elements-are-there-finite-groups-h-g-such-that-l-co Given a lattice L with n elements, are there finite groups H < G such that L $\cong$ the lattice of subgroups between H and G? William DeMeo 2012-01-15T09:32:23Z 2012-02-22T23:21:25Z <p>If there is no restriction on $n$, this is a famous <a href="http://garden.irmacs.sfu.ca/?q=op/finite_congruence_lattice_problem" rel="nofollow">open problem</a>. I'm wondering if any recent work has been done for small $n>6$. I believe the question is answered (positively) for $n=6$ by Watatani (1996) <a href="http://www.ams.org/mathscinet-getitem?mr=1409040" rel="nofollow">MR1409040</a> and Aschbacher (2008) <a href="http://www.ams.org/mathscinet-getitem?mr=2393428" rel="nofollow">MR2393428</a>. I also believe we can answer it for $n=7$, with one possible exception. The exceptional case is shown below.</p> <p><img src="http://math.hawaii.edu/~williamdemeo/winged2by3.png" alt="alt text"></p> <p>So my two questions are these: </p> <p>1) Does anyone know of recent work on this special case of the problem (specifically for $n=7$ or $n=8$)? </p> <p>2) Has anyone found a finite group $G$ with a subgroup $H$ such that the interval</p> <p><code>$[H, G] = \{K : H \leq K \leq G \}$</code></p> <p>is the lattice shown above? </p> <p>If not, I'd be happy to hear if anyone has ideas about how we might prove that such a group exists (without necessarily producing the group). Or, if you can put restrictions on such a group, that might be useful. For example, I believe I can show that there is no finite <em>solvable</em> group with this property.</p> <p>(By the way, I have searched many of the groups in GAP's small groups library, and this was helpful for finding some of the other difficult 7-element cases.)</p> <p><strong>Correction:</strong> We don't really have an answer to this question for $n=7$. Rather, we can show that every lattice with at most 7 elements (except possibly the one above) is the congruence lattice of a finite algebra. Pálfy and Pudlák <a href="http://www.ams.org/mathscinet-getitem?mr=593011" rel="nofollow">MR593011</a> showed the two questions are equivalent <em>when $n$ is not restricted</em>, but their proof doesn't go through for fixed $n$. Nonetheless, we can prove that a (minimal) finite algebra with the lattice above as congruence lattice must be a transitive G-set (modulo constant operations), so the existence of such a finite algebra is equivalent to the existence of finite groups H &lt; G with [H,G] isomorphic to the lattice above. (This is the reason I phrased my question in terms of subgroup lattices.)</p> <p><strong>Update:</strong> (2012/02/22) If the interval $[H,G]$ is the lattice above, with $H$ core-free, I believe I can prove that a minimal normal subgroup of $G$ must be non-abelian, which isn't much, but it's a start! In particular, I think it rules out the affine case of the Aschbacher-O'Nan-Scott Theorem.</p>