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.