1
$\begingroup$

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.

  • 0
    Then go with Jack Schmidt's suggestion.2012-07-02

1 Answers 1

2

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

  • 0
    @Babak: e$x$actly. "wr" is just shortha$n$d for what you wrote2012-07-02