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?

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)?

Thanks in advance!

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:

$\{ K : H \leq K \leq G \}$

where $H$ is a corefree subgroup of $G$.

  • $\begingroup$ Have you considered already Heisenberg groups over a finite field? Or similiarly the upper triangular matrices in the $N \times N$ matrixes over some finite unital ring. Depending on $N$ you get many subgroups. Of course they are solvable, but if you consider them sitting inside $GL_N$? $\endgroup$ – Marc Palm Apr 21 '11 at 5:48
  • $\begingroup$ I have not looked at those yet, but I will do so now. In particular, I'll check whether the chains you get in those groups are actually intervals. Thanks. $\endgroup$ – William DeMeo Apr 21 '11 at 6:16

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 faithful 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 $X_{\omega} < Y \leq X$, or it has the form $X_{\omega} \ltimes (I \times G \times \dotsm \times G)$ with $G_{\gamma}\leq I \leq G$.
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.

  • $\begingroup$ This is interesting. Thank you for suggesting this nice, general construction. $\endgroup$ – William DeMeo Apr 21 '11 at 23:33
  • $\begingroup$ I've decided to accept this answer because, while Dr. Holt's answer is also great, your answer came first, and the construction you describe seems like it might answer another question I've been thinking about, which I'll describe in a later comment, once I've verified that it works. Thanks! $\endgroup$ – William DeMeo May 5 '11 at 9:53

I think you can get arbitrarily long chains of this type in the simple groups ${\rm PSL}(2,p)$ for $p$ prime.

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

$H_0 < H_1 < \cdots H_N < G$

where $H_i$ is dihedral of order $2q^ir$.

  • $\begingroup$ Thanks, that's really neat. The "r" makes sure anything containing H0 lands in HN. Why does q have to be odd? If we require H0 to be order not dividing 120 (so make r larger and avoid the A4,A5,S4 subgroups), can we take q=2 anyways? $\endgroup$ – Jack Schmidt Apr 21 '11 at 19:55
  • $\begingroup$ Yes you are right, $q$ need not be odd. $\endgroup$ – Derek Holt Apr 21 '11 at 21:47
  • $\begingroup$ Wow, that's pretty cool. I wouldn't have noticed these long chains without your nice use of maximal subgroups. Probably these aren't examples that are easily verified in GAP, but I'll take your word for it for now, and then convince myself that it works. Thank you!! ..and thank you to everyone else who gave helpful suggestions. $\endgroup$ – William DeMeo Apr 21 '11 at 23:39

I think in the subgroup lattice of the non-solvable group $A_5 \times \mathbb{Z}/2^N\mathbb{Z}$, the interval between $\{1\} \times \{1\}$ and $\{ 1 \} \times \mathbb{Z}/2^N\mathbb{Z}$ is a chain of length $N$.

Edit: 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 $\mathbb{F}_q$ (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 $A_{p^N}$ contains a cyclic group of order $p^N$.

  • $\begingroup$ Ha! Of course, take any cyclic group and take the direct product with any nonsolvable group! Ok, you got me there. When I ruled out cyclic groups, I should have ruled out this construction too. I will revise my question. To be more specific, for my application I need intervals above subgroups that are core-free. (So you can't mod out the non-solvable part and end up with a cyclic group.) ...but thanks for your answer, which will help me improve my question. $\endgroup$ – William DeMeo Apr 21 '11 at 8:58

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.

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.

Intervals that are very long chains are also easy to find: if G is a cyclic group of prime power order, pn, 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.

Thus the question arises if there are long chains [A,G] where 1≠A is core-free.

An example for arbitrary n is the dihedral group of order 2n+1 with A any non-central subgroup of order 2. The interval [A,G] consists of the dihedral groups of order 2k for 1 ≤ k ≤ n+1, and so has length n.

  • $\begingroup$ @Jack: He wants a non-solvable group. $\endgroup$ – Mark Sapir Apr 21 '11 at 14:10
  • $\begingroup$ Thank you! These are perfectly good solvable examples. I verified in GAP for a few k, if G:=DihedralGroup(2^k), and if H is a representative of the 3rd (GAP's labelling) conjugacy class of subgroups, then [H,G] is a chain of length k. Nice! $\endgroup$ – William DeMeo Apr 21 '11 at 23:16

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.

  • $\begingroup$ Thanks, I will look at those. (The interval between two subgroups $A$ and $B$ in the subgroup lattice of a group simply means all subgroups $C$ such that $A \leq C \leq B$.) $\endgroup$ – William DeMeo Apr 21 '11 at 9:10

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.


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