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I am trying to find a counterexample to a claim about groups.

I need a random generator (a program in any programming language, or an algorithm) which will generate a random group $G$ of any order such that:

  • $G$ has a p-subgroup $P$ which is not a Sylow p-subgroup
  • $G$ has a normal subgroup $N$ whose order is coprime with $|P|$

Is there any known work on generating random groups in general? The only way I see now is to create a multiplication table, check group axioms, check the properties I need. But this is inefficient. I do not think I can every generate a group satisfying the above properties by simply trying all tables. Any thoughts?

Evgenii.Balai
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  • Any finite groups is a subgroup of some symmetric group. If you take a language that can generate subgroups of the symmetric group (e.g. magma or GAP) you might be able to generate some examples (it is still inefficient though). – Clément Guérin Nov 09 '15 at 11:27
  • If you want to choose a group of order $n$ whose isomorphism class is uniformly distributed over all isomorphism types of groups of order $n$, then I think that is only even approximately possible if a complete list of isomorphism types is available. So there would be no problem for orders up to $2000$. – Derek Holt Nov 09 '15 at 12:53
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    I am minded of Gordon Royle's quote: sometimes a couple of days enthusiatic programming can save you an hour's careful thought. – Chris Godsil Nov 09 '15 at 13:04

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It is easy to write down a group with the desired properties; and, you should be able to see how to generalise this to provide infinitely many more examples. Take $G = C_2^2\times C_3$, and $P = C_2$ and $N = C_3$. The Sylow $2$-subgroup of $G$ is $C_2^2$, $P$ is a non-Sylow $2$-subgroup, and $N$ - with order coprime to $2$ - is certainly normal in $G$ because $G$ is abelian.

James
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    Thanks. This approach helps to generate some subgroups. We can more elements to the direct product with orders coprime to 2 to get more groups. – Evgenii.Balai Nov 12 '15 at 04:52