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As we read from wiki, informally, the reconstruction conjecture in graph theory says that graphs are determined uniquely by their subgraphs.

  1. Is there a group-theoretic formulation of this conjecture?
  2. Has an analogous conjecture been made in group theory(in any sensible way)?

References such as books will do, thanks.

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  • $\begingroup$ Re: 1. Do you have a reason for thinking that there might be a group-theoretic formulation? $\endgroup$
    – HJRW
    Commented Aug 8, 2010 at 11:46
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    $\begingroup$ Maybe, he is asking whether there is an algebraic attack as there has been many in algebraic graph theory using group theory. And as his previous question on intersections suggests:mathoverflow.net/questions/33366/… , he has a bias towards an object-subobject relation. $\endgroup$
    – Unknown
    Commented Aug 8, 2010 at 12:02
  • $\begingroup$ @Elohemahab, as I understand, you are seeking a way by which the following can be made precise "groups are determined uniquely by their subgroups...." $\endgroup$
    – Unknown
    Commented Aug 8, 2010 at 12:07
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    $\begingroup$ You can find two groups of order 16 whose lattices of subgroups are isomorphic. This suggests you may have to look at finer structure. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 8, 2010 at 15:00
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    $\begingroup$ I don't think anything analogous can hold for groups, because graphs are far more general. There's no notion of index of a subgraph in a graph, for example. Intuitively, the set of all subgraphs of a graph tells you much more than the set of all subgroups of a group (as Scott suggested). $\endgroup$
    – Eric Tressler
    Commented Aug 8, 2010 at 18:27

1 Answer 1

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There are continuum 2-generated groups where all proper subgroups are cyclic of order $p$ (for the same prime $p\sim 10^{70}$), the Tarski monsters. All these groups have the same lists of proper subgroups, and the same lattice of subgroups. Also all cyclic groups of prime order have the same proper subgroups: the trivial group. So in general the answer is "no". But every finite non-cyclic abelian group is determined by the list of its proper subgroups, which follows easily from the description of finite abelian groups (note that I and, I think, the question, are not talking about the lattice of subgroups, but about the list of all proper subgroups). I suspect that at least for a large class of finite solvable groups, the list of proper subgroups determines the group. I would look first at A-groups, finite groups with all Sylow subgroups abelian (see the book of Huppert, "Endliche Gruppen") because the structure of subgroups of A-groups is more known, and one can use induction on the order of the group.

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  • $\begingroup$ Nice explanation. That makes the formulation of such conjectures for groups as remote as can be. I think question (1) remains. $\endgroup$
    – Unknown
    Commented Feb 6, 2011 at 0:36
  • $\begingroup$ Am I right that you want a group theoretic statement that is equivalent to the statement about graphs? With every graph $\Gamma$, associate a group (called Right Angled Artin group) $R(\Gamma)$ it is generated by the vertices of $\Gamma$ subject to the relations $ab=ba$ iff $a$ is adjacent to $b$. Every induced subgraph of $\Gamma$ gives a subgroup of $R(\Gamma)$. If $\Gamma$ is finite, one can use a finite nilpotent quotient of $R(\Gamma)$. $\endgroup$
    – user6976
    Commented Feb 6, 2011 at 1:54
  • $\begingroup$ @Mark, yes. I want a group theoretic statement that is equivalent to the statement about graphs. $\endgroup$
    – Unknown
    Commented Feb 27, 2011 at 22:50
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    $\begingroup$ @Elohemahab: The following is probably true: the group $R(\Gamma)$ is defined uniquely by its subgroups iff $\Gamma$ is defined by its subgraphs. $\endgroup$
    – user6976
    Commented Feb 27, 2011 at 23:19

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