# minimal overgroups of a permutation group [closed]

Given a group $$G < S_N$$ is there an "efficient" way to identify (or construct) the "minimal" permutation groups $$H_i \leq S_N$$ such that $$G < H_i$$? $$H_i$$'s are minimal in the sense that there $$\not \exists H' \quad s.t.\quad G.

## closed as off-topic by Shaun, The Count, José Carlos Santos, Shailesh, 0XLRAug 11 at 3:09

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• What do you mean by "permutation group"? I've seen that mean "subgroup of a symmetric group", but with that meaning $G$ itself is the minimal such group. Unless you want $G$ to be strictly included in $H$ (which your notation does suggest). – Chris Eagle Aug 9 at 22:13
• $S_{18}$ has 598,016,157,515,302,757 subgroups (if i counted right). so if you organize them as a partially ordered set (by set inclusion), you can then read off the minimal overgroups of any particular subgroup. you might find homepages.warwick.ac.uk/~mareg/download/papers/symsubs/… of interest – David Holden Aug 9 at 22:49
• Thanks, I was hoping for a non brute force approach. – self-educator Aug 10 at 0:19
• You talk about “the” minimal subgroup, but there is no good reason to assume that such a thing is unique. For a trivial example, if $G$ is the trivial subgroup, then every subgroup of prime order of $S_n$ will work, and none of them should be the minimal subgroup of $S_n$. – Arturo Magidin Aug 10 at 21:46
• I really cannot understand why people vote to close questions like this. Can anyone explain? It is an interesting question, and I don't know the answer, although I don't believe that there is any currently known efficient way to do this. – Derek Holt Aug 12 at 7:43