So awhile back I asked this question here on stack exchange:

Normal subgroup $H$ of $G$ with same orbits of action on $X$.

At the time I wasn't quite sure what I was really wanting to know about that situation. After much searching around and occasionally revisiting the problem that sparked that question, I eventually figured what the name for the thing I was noticing that peaked my interest is: It's the notion of a block system for a group action.

With that in mind, I am now prepared to ask a much more specific question:

Suppose we have a group action of a group $H$ on a set $X$, with proper nontrivial normal subgroups $M$ and $N$. What is the significance of the block systems induced by restriction of the group action to the normal subgroups $M$ and $N$ being the same (that is, the induced actions of $M$ and of $N$ on $X$ have the same block system on $X$)?

If there isn't much that can be said about this in general, then my specific case of interest is when the set is a given group finite $G$, and the acting group is its automorphism group, $\operatorname{Aut}(G)$. The normal subgroups I am specifically interested in are the inner automorphism group $\operatorname{Inn}(G)$ and the group of (conjugacy) class-preserving automorphisms, which I will denote as $\Lambda_{id}(G)$.

  • 1
    $\begingroup$ Normally you only talk about block systems of a transitive action. But ${\rm Aut}(G)$ does not act transitively on $G$. So I am not sure what you mean by block systems. Do you just mean that $M$ and $N$ have the same orbits? $\endgroup$ – Derek Holt Jul 4 '16 at 10:55
  • $\begingroup$ I meant block system as it is defined here: en.wikipedia.org/wiki/Block_(permutation_group_theory) which works whether or not the action is transitive. $\endgroup$ – Justin Benfield Jul 4 '16 at 11:14
  • $\begingroup$ It looks to me as though you mean that $M$ and $N$ have the same orbits. If you mean something different, then perhaps you could explain. $\endgroup$ – Derek Holt Jul 4 '16 at 11:32
  • $\begingroup$ For example, one set of blocks for the action of $\operatorname{Aut}(G)$ on $G$ is the blocks which are comprised of all elements of the same order. Because automorphisms are operation preserving, it follows that they must be order preserving, hence they respect these blocks. In fact, the blocks are never moved, hence the (lifted) action of $\operatorname{Aut}(G)$ on that block system is trivial. $\endgroup$ – Justin Benfield Jul 4 '16 at 11:39
  • $\begingroup$ For a nontrivial example, consider the induced action of $\operatorname{Inn}(G)$ on $G$, this is the same as the conjugation action on $G$. One can take the conjugacy classes as a block system of this action. Again the blocks are never moved by $\operatorname{Inn}(G)$, but what's really interesting is that since (general) automorphisms must respect conjugacy classes, in that, two elements that are conjugate are mapped to conjugate elements, the aforementioned blocks are a block system for the action of $\operatorname{Aut}(G)$. $\endgroup$ – Justin Benfield Jul 4 '16 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.