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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.

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

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  • $\begingroup$ 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
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    $\begingroup$ @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

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