Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:29:26Z http://mathoverflow.net/feeds/question/62495 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62495/which-finite-nonabelian-groups-have-long-chains-of-subgroups-as-intervals-in-thei Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? William DeMeo 2011-04-21T05:35:58Z 2011-04-22T16:57:56Z <p>Given N, what is a finite non-abelian (and preferably non-solvable) group G in whose subgroup lattice Sub[G] there is an interval that is a chain of length at least N?</p> <p>Since N can be arbitrarily large (but fixed), perhaps there is no easy answer. In that case, can someone suggest which sorts of groups to look at to find intervals that are chains (say, on the order of 10 subgroups in the chain)?</p> <p>Thanks in advance!</p> <p>Edit: Thanks to Carnahan's answer, I see that I should have ruled out direct products of cyclic groups with nonsolvable groups. What I'm interested in are intervals in the subgroup lattice of the form:</p> <p>${ K : H \leq K \leq G }$</p> <p>where $H$ is a <em>corefree</em> subgroup of $G$.</p> http://mathoverflow.net/questions/62495/which-finite-nonabelian-groups-have-long-chains-of-subgroups-as-intervals-in-thei/62503#62503 Answer by Yiftach Barnea for Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? Yiftach Barnea 2011-04-21T07:24:55Z 2011-04-21T07:24:55Z <p>I am not completely sure what an interval means in this case. But I think that an interesting example to look at would be $p$-groups of maximal class.</p> http://mathoverflow.net/questions/62495/which-finite-nonabelian-groups-have-long-chains-of-subgroups-as-intervals-in-thei/62505#62505 Answer by S. Carnahan for Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? S. Carnahan 2011-04-21T07:45:54Z 2011-04-22T07:45:35Z <p>I think in the subgroup lattice of the non-solvable group $A_5 \times \mathbb{Z}/2^N\mathbb{Z}$, the interval between <code>$\{1\} \times \{1\}$</code> and <code>$\{ 1 \} \times \mathbb{Z}/2^N\mathbb{Z}$</code> is a chain of length $N$.</p> <p><strong>Edit:</strong> For any $N$, you can also find primes $p$ and $q$ such that $q \equiv \pm 1$ mod $p^N$, so there are cyclic groups of order $p^N$ inside any group of Lie type over <code>$\mathbb{F}_q$</code> (where you use $+1$ in the split case and $-1$ in the non-split case). These will yield intervals that are chains of length $N$. Similarly, the alternating group <code>$A_{p^N}$</code> contains a cyclic group of order $p^N$. </p> http://mathoverflow.net/questions/62495/which-finite-nonabelian-groups-have-long-chains-of-subgroups-as-intervals-in-thei/62511#62511 Answer by Geoff Robinson for Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? Geoff Robinson 2011-04-21T09:32:39Z 2011-04-21T09:32:39Z <p>It might be worthwhile to check out the paper below and its predecessors (and one successor by Alladi and Turull): Solomon, Ron; Turull, Alexandre Chains of subgroups in groups of Lie type. III. J. London Math. Soc. (2) 44 (1991), no. 3, 437–444. </p> http://mathoverflow.net/questions/62495/which-finite-nonabelian-groups-have-long-chains-of-subgroups-as-intervals-in-thei/62538#62538 Answer by Jack Schmidt for Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? Jack Schmidt 2011-04-21T13:49:40Z 2011-04-21T13:49:40Z <p>Intervals [A,B] in a subgroup lattice are the lattice formed from the subgroups C with A ≤ C ≤ B. An interval is a chain if it is totally ordered by inclusion. When discussing intervals in a subgroup lattice, it is often a good idea to assume we have chosen the group with this lattice minimally. In particular, we can assume G=B so that the interval goes all the way to the top, and we can assume A is core-free so that we cannot quotient out by any normal subgroup contained in every subgroup in the interval.</p> <p>Intervals that are chains are ubiquitous if we do not require the chain to be very long: if M is any maximal subgroup of G, then the interval [M,G] is a chain of length 1.</p> <p>Intervals that are very long chains are also easy to find: if G is a cyclic group of prime power order, p<sup>n</sup>, then the interval [1,G] is a chain of length n. However, these G are the only examples where the interval [1,G] is a chain.</p> <p>Thus the question arises if there are long chains [A,G] where 1≠A is core-free.</p> <p><strong>An example for arbitrary n</strong> is the dihedral group of order 2<sup>n+1</sup> with A any non-central subgroup of order 2. The interval [A,G] consists of the dihedral groups of order 2<sup>k</sup> for 1 ≤ k ≤ n+1, and so has length n.</p> http://mathoverflow.net/questions/62495/which-finite-nonabelian-groups-have-long-chains-of-subgroups-as-intervals-in-thei/62546#62546 Answer by F. Ladisch for Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? F. Ladisch 2011-04-21T16:33:29Z 2011-04-22T16:57:56Z <p>You can construct groups with chains of arbitrary length with the help of wreath products. Since you want to assume that the subgroup $H$ is corefree, you get a <em>faithful</em> action of $G$ on the cosets of $H$, so one may identify $G$ with a permutation group. Now assume that $G$ is a transitive permutation group on the set $\Gamma$ and $X$ is a transitive permutation group on the set $\Omega$. The wreath product of $G$ with $X$, that is the semi-direct product $W=X \ltimes G^{\Omega}$, acts on $\Omega\times \Gamma$ by $$(\omega, \gamma) (x, f) = (\omega x, \gamma ((\omega x) f ) ) \quad \text{(where f\colon \Omega\to G and thus (\omega x) f\in G.)}$$ The stabilizer of $(\omega, \gamma)$ is then $$W_{(\omega, \gamma)} = X_{\omega} \ltimes (G_{\gamma}\times G \times \dotsm \times G)$$ (where the component $G_{\gamma}$ occurs, strictly speaking, at position $\omega$). Now it is not difficult to see, that if a subgroup $K$ with $W_{(\omega, \gamma)}\leq K$ contains an element $(x, f)$ with $x\notin X_{\omega}$, then $K$ contains $G\times G \times \dotsm \times G$. So either $K$ has the form $Y \ltimes (G^{\Omega})$ with $X_{\omega} &lt; Y \leq X$, or it has the form $X_{\omega} \ltimes (I \times G \times \dotsm \times G)$ with $G_{\gamma}\leq I \leq G$.<br> So the interval $[W_{(\omega, \gamma)}, W]$ is lattice isomorphic to the lattice obtained by putting $[X_{\omega}, X]$ on top of $[G_{\gamma}, G]$. If the latter are chains, then you get a chain, where the lengths add. Starting with a primitive permutation group (non-solvable, if you wish) and repeating this contruction, you get arbitrarily large chains. Even if you are interested in non-solvable groups, I mention that the Sylow $p$-subgroup of $S_{p^n}$ is a special case of this contruction.</p> http://mathoverflow.net/questions/62495/which-finite-nonabelian-groups-have-long-chains-of-subgroups-as-intervals-in-thei/62565#62565 Answer by Derek Holt for Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? Derek Holt 2011-04-21T19:14:06Z 2011-04-21T19:14:06Z <p>I think you can get arbitrarily long chains of this type in the simple groups ${\rm PSL}(2,p)$ for $p$ prime. </p> <p>We make use of the maximal dihedral subgroups of order $p-1$. For given $N$, choose $p$ such that $(p-1)/2 = q^Nr$ with $q$ an odd prime and $r>2$. Then there is a chain of subgroups</p> <p>$H_0 &lt; H_1 &lt; \cdots H_N &lt; G$</p> <p>where $H_i$ is dihedral of order $2q^ir$. </p>