Let $X$ be a finite set and $G$ be a transitive subgroup of the symmetric group on $X.$ Recall that a (complete) block system for this action is a partition of $X = B_1 \cup \cdots \cup B_k$ into equally-sized blocks $B_1 , \ldots , B_k,$ which is preserved in the sense that blocks are always mapped to blocks under the group action. The block systems for a particular action form a lattice whose maximum is $\{ X \}$ and whose minimum is the partition of $X$ into singletons.

In the lattice of block systems, we may associate to any chain a sequence of integers: starting from the maximum element of that chain, divide the size of a block of the given block system by the size of a block of its predecessor.

Question: Is there a name for group actions or permutation groups satisfying the condition that the aforementioned sequences associated to every maximal chain determine the same multiset?

Example 1: Let $G=X=A_4$ acting on itself by multiplication. The lattice of block systems is essentially the lattice of subgroups (blocks $=$ cosets.) There are two incomparable chains of subgroups whose associated sequences are $(3, 2, 2)$ and $(4, 3),$ so the condition is not satisfied.

Example 2: $G = A_4$ acting as usual on $ X= \{ 1,2,3,4 \}$ is primitive, so the only sequence is $(4)$ and the condition is trivially satisfied.

Example 3: For $D_8$ acting on itself by multiplication, the sequence is always $(2, 2, 2),$ so the condition is satisfied.

Example 4: For $\mathbb{Z}_6$ acting on itself by multiplication, the sequences are $(3,2)$ and $(2,3),$ so the condition is satisfied.

Example 5: For the wreath product $S_n \wr S_m$ acting as usual on $[n] \times [m],$ the lattice of block systems is a chain of length $2.$ The only sequence is $(m, n),$ and hence the condition is satisfied.

This condition seems natural to me from the point of view of Ritt's polynomial decomposition theorems and related work. In that context, $G$ is a Galois group acting on the fibers of some polynomial or rational map of finite degree, and we want to "factor" that map as a composition of "irreducible" maps. The condition should then imply that the set of degrees of the irreducible elements are independent of the decomposition. For instance, corresponding to Example 4, the decompositions $(z^3)^2 = (z^2)^3$ have the same set of associated degrees.

Rather than characterizing such actions (although that would most certainly be welcome!), I am mostly just wondering if this condition or some close cousin already has an established name in the literature.

  • $\begingroup$ Could you give a more detailed (and perhaps more formal) definition of the sequence of integers associated with a chain of blocks systems?. I cannot understand it at the moment. In examples 1 - 4, the product of the integers in the sequence is the order of the group, but that is not the case in Example 5. $\endgroup$
    – Derek Holt
    Mar 15 at 19:29
  • $\begingroup$ @DerekHolt Indeed, I made mistakes writing down the sequences for both Examples 2 and 5. In general, the product of the sequence should be the cardinality of $X.$ $\endgroup$
    – tim
    Mar 15 at 19:41
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    $\begingroup$ Note that the lattice of block systems is isomorphic to the lattice of subgroups of $G$ containing a particular vertex-stabiliser $G_x$. (This is well known and not hard to prove.) The integers you have in mind are then the successive indices of subgroups in a chain. So you're asking for group-subgroups pairs, such that, in every maximal refinement as a subgroup chain, the multiset of indices is the same. (Maybe you knew this already, but just in case.) (I'm ignoring the detail of whether the point-stabiliser is core-free, because it doesn't really matter.) $\endgroup$
    – verret
    Mar 16 at 5:31
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    $\begingroup$ I believe that I could prove without too much difficulty that the property holds for all actions of a finite group $G$ if and only if $G$ is supersolvable (in which case all numbers in the integer sequences are prime). $\endgroup$
    – Derek Holt
    Mar 16 at 7:27
  • $\begingroup$ Also a technical query about the definition: would you consider the sequences $(4,4,2)$ and $(4,2,2,2)$ to be equivalent? They have the same underlying set (so I think they are equivalent by your definition) but with different multiplicities. $\endgroup$
    – Derek Holt
    Mar 16 at 7:29

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