Finding an imprimitive group on $12$ letters

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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You might want to consider the size of the smallest nontrivial block - what does "nontrivial" mean, and why? – Mark Bennet Jul 2 '12 at 19:36
@MarkBennet: The non-trivial blocks are $\Omega$, $∅$ and the singleton sets. – Babak S. Jul 2 '12 at 19:40
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? – Jyrki Lahtonen Jul 2 '12 at 19:52
@JyrkiLahtonen: Yes. I am looking for a transitive group, of course. – Babak S. Jul 2 '12 at 19:55
Then go with Jack Schmidt's suggestion. – Jyrki Lahtonen Jul 2 '12 at 19:57

Hint: given a particular block $B$, consider the group $H_B$ of all permutations that preserve $B$.
You mean; $H_B=\{\pi\in S_{12}|B^{\pi}=B or B^{\pi}\cap B=∅ \}$ would be the group? – Babak S. Jul 2 '12 at 19:34
@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 $mn=12$. – Jack Schmidt Jul 2 '12 at 19:43
@JackSchmidt: Isn't the group in your comment $(S_m×...×S_m).S_n$? Because, I see the order above. – Babak S. Jul 2 '12 at 19:51