By definition a permutation group $G$ acting on a set $\Omega$ is called primitive if $G$ acts transitively on $\Omega$ and $G$ preserves no nontrivial blocks of $\Omega$. Otherwise, if the group does preserve a nontrivial block then $G$ is called imprimitive.

Here I am asked to find an imprimitive permutation group $\Omega$ acting on $\Omega$ with $|\Omega|=12$ such that $|G|$ be of maximum possible order.

It would be difficult and unprofessionally finding a group which has a block for example with two elements. At least I cannot do that right now. :). Clearly, our $G$ is a proper subgroup of $S_{12}$ but would not be $A_{12}$.

I am wondered how can it be shown that any group I found is of maximum order. Thanks for any help.

  • $\begingroup$ You might want to consider the size of the smallest nontrivial block - what does "nontrivial" mean, and why? $\endgroup$ – Mark Bennet Jul 2 '12 at 19:36
  • $\begingroup$ @MarkBennet: The non-trivial blocks are $\Omega$, $∅$ and the singleton sets. $\endgroup$ – mrs Jul 2 '12 at 19:40
  • $\begingroup$ I'm a bit rusty on terminology here. Are we looking for a transitive group? I mean, primitivity/imprimitivity is kinda meaningless, if the group is not transitive, right? $\endgroup$ – Jyrki Lahtonen Jul 2 '12 at 19:52
  • $\begingroup$ @JyrkiLahtonen: Yes. I am looking for a transitive group, of course. $\endgroup$ – mrs Jul 2 '12 at 19:55
  • $\begingroup$ Then go with Jack Schmidt's suggestion. $\endgroup$ – Jyrki Lahtonen Jul 2 '12 at 19:57

Hint: given a particular block $B$, consider the group $H_B$ of all permutations that preserve $B$.

| cite | improve this answer | |
  • $\begingroup$ You mean; $H_B=\{\pi\in S_{12}|B^{\pi}=B or B^{\pi}\cap B=∅ \}$ would be the group? $\endgroup$ – mrs Jul 2 '12 at 19:34
  • 1
    $\begingroup$ @BabakSorouh: yes, but you can describe the group a little more explicitly in terms of those two pieces: Sym(B) wr Sym({B^pi : pi in Sym(Omega). The first part swirls the elements of a block, the second part swirls the set of blocks, without "changing the blocks inside". It has order $(m!)^n n!$ where $m=|B|$ and $mn=12$. $\endgroup$ – Jack Schmidt Jul 2 '12 at 19:43
  • $\begingroup$ @JackSchmidt: Isn't the group in your comment $(S_m×...×S_m).S_n$? Because, I see the order above. $\endgroup$ – mrs Jul 2 '12 at 19:51
  • $\begingroup$ @JackSchmidt: Thanks for your nice illustrating comment. $\endgroup$ – mrs Jul 2 '12 at 20:07
  • $\begingroup$ @Robert: Thanks for the hint. :) $\endgroup$ – mrs Jul 2 '12 at 20:08

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.