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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 $\not \exists H' \quad s.t.\quad G<H'<H_i$.

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

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  • $\begingroup$ 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). $\endgroup$ – Chris Eagle Aug 9 at 22:13
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    $\begingroup$ $S_{18}$ 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 $\endgroup$ – David Holden Aug 9 at 22:49
  • $\begingroup$ Thanks, I was hoping for a non brute force approach. $\endgroup$ – self-educator Aug 10 at 0:19
  • $\begingroup$ 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 $S_n$. $\endgroup$ – Arturo Magidin Aug 10 at 21:46
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Aug 12 at 7:43