Does the hypergraph of subgroups determine a group?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite












A hypergraph is a pair $H=(V,E)$ where $Vneq emptyset$ is a set and $Esubseteqcal P(V)$ is a collection of subsets of $V$. We say two hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic if there is a bijection $f:V_1to V_2$ such that $f(e_1) in E_2$ for all $e_1in E_1$, and $f^-1(e_2) in E_1$ for all $e_2in E_2$.



If $G$ is a group, denote by $textSub(G)$ the collection of the subgroups of $G$.



Are there non-isomorphic groups $G,H$ such that the hypergraphs $(G, textSub(G))$ and $(H, textSub(H))$ are isomorphic?










share|cite|improve this question



















  • 2




    See mathoverflow.net/questions/37738/…
    – Derek Holt
    4 hours ago










  • What is the subgroup hypergraph of a cyclic group of prime order?
    – Gerald Edgar
    3 hours ago










  • So there is the important vertex which is characterized by being on every hyperedge.
    – AHusain
    1 hour ago










  • @GeraldEdgar If I understand you correctly, the answer is $(mathbbZ_p, big0,mathbbZ_pbig)$ for $p$ prime?
    – Dominic van der Zypen
    1 hour ago















up vote
3
down vote

favorite












A hypergraph is a pair $H=(V,E)$ where $Vneq emptyset$ is a set and $Esubseteqcal P(V)$ is a collection of subsets of $V$. We say two hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic if there is a bijection $f:V_1to V_2$ such that $f(e_1) in E_2$ for all $e_1in E_1$, and $f^-1(e_2) in E_1$ for all $e_2in E_2$.



If $G$ is a group, denote by $textSub(G)$ the collection of the subgroups of $G$.



Are there non-isomorphic groups $G,H$ such that the hypergraphs $(G, textSub(G))$ and $(H, textSub(H))$ are isomorphic?










share|cite|improve this question



















  • 2




    See mathoverflow.net/questions/37738/…
    – Derek Holt
    4 hours ago










  • What is the subgroup hypergraph of a cyclic group of prime order?
    – Gerald Edgar
    3 hours ago










  • So there is the important vertex which is characterized by being on every hyperedge.
    – AHusain
    1 hour ago










  • @GeraldEdgar If I understand you correctly, the answer is $(mathbbZ_p, big0,mathbbZ_pbig)$ for $p$ prime?
    – Dominic van der Zypen
    1 hour ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











A hypergraph is a pair $H=(V,E)$ where $Vneq emptyset$ is a set and $Esubseteqcal P(V)$ is a collection of subsets of $V$. We say two hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic if there is a bijection $f:V_1to V_2$ such that $f(e_1) in E_2$ for all $e_1in E_1$, and $f^-1(e_2) in E_1$ for all $e_2in E_2$.



If $G$ is a group, denote by $textSub(G)$ the collection of the subgroups of $G$.



Are there non-isomorphic groups $G,H$ such that the hypergraphs $(G, textSub(G))$ and $(H, textSub(H))$ are isomorphic?










share|cite|improve this question















A hypergraph is a pair $H=(V,E)$ where $Vneq emptyset$ is a set and $Esubseteqcal P(V)$ is a collection of subsets of $V$. We say two hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic if there is a bijection $f:V_1to V_2$ such that $f(e_1) in E_2$ for all $e_1in E_1$, and $f^-1(e_2) in E_1$ for all $e_2in E_2$.



If $G$ is a group, denote by $textSub(G)$ the collection of the subgroups of $G$.



Are there non-isomorphic groups $G,H$ such that the hypergraphs $(G, textSub(G))$ and $(H, textSub(H))$ are isomorphic?







co.combinatorics gr.group-theory graph-theory hypergraph






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago

























asked 4 hours ago









Dominic van der Zypen

13.1k43171




13.1k43171







  • 2




    See mathoverflow.net/questions/37738/…
    – Derek Holt
    4 hours ago










  • What is the subgroup hypergraph of a cyclic group of prime order?
    – Gerald Edgar
    3 hours ago










  • So there is the important vertex which is characterized by being on every hyperedge.
    – AHusain
    1 hour ago










  • @GeraldEdgar If I understand you correctly, the answer is $(mathbbZ_p, big0,mathbbZ_pbig)$ for $p$ prime?
    – Dominic van der Zypen
    1 hour ago













  • 2




    See mathoverflow.net/questions/37738/…
    – Derek Holt
    4 hours ago










  • What is the subgroup hypergraph of a cyclic group of prime order?
    – Gerald Edgar
    3 hours ago










  • So there is the important vertex which is characterized by being on every hyperedge.
    – AHusain
    1 hour ago










  • @GeraldEdgar If I understand you correctly, the answer is $(mathbbZ_p, big0,mathbbZ_pbig)$ for $p$ prime?
    – Dominic van der Zypen
    1 hour ago








2




2




See mathoverflow.net/questions/37738/…
– Derek Holt
4 hours ago




See mathoverflow.net/questions/37738/…
– Derek Holt
4 hours ago












What is the subgroup hypergraph of a cyclic group of prime order?
– Gerald Edgar
3 hours ago




What is the subgroup hypergraph of a cyclic group of prime order?
– Gerald Edgar
3 hours ago












So there is the important vertex which is characterized by being on every hyperedge.
– AHusain
1 hour ago




So there is the important vertex which is characterized by being on every hyperedge.
– AHusain
1 hour ago












@GeraldEdgar If I understand you correctly, the answer is $(mathbbZ_p, big0,mathbbZ_pbig)$ for $p$ prime?
– Dominic van der Zypen
1 hour ago





@GeraldEdgar If I understand you correctly, the answer is $(mathbbZ_p, big0,mathbbZ_pbig)$ for $p$ prime?
– Dominic van der Zypen
1 hour ago











1 Answer
1






active

oldest

votes

















up vote
4
down vote



accepted










In the comments to the question, I notice something which might be an error, or at least is an incomplete response. It is pointed out in the comments that there exist nonisomorphic groups with isomorphic subgroup lattices. While true, that fact doesn't answer this question, since it is possible to have isomorphic subgroup lattices and nonisomorphic subgroup hypergraphs. Even if you assume that the groups have the same order and isomorphic subgroup lattices, it does not immediately follow that they have isomorphic subgroup hypergraphs.




I am sure the negative answer to this question can be found in Roland Schmidt's book. But I would like to point to a theorem I coauthored after Schmidt's book was published, which applies to this question. Namely:



Thm. For any finite $N$, there is a finite set $X=X_N$ and $N$ binary operations $circ_i$ defined on $X$ such that



(1) $G_i = (X,circ_i)$ is a group for all $i$,

(2) $G_inotcong G_j$ when $ineq j$, and

(3) for all $i, j$, the groups
$G_i^kappa$ and $G_j^kappa$ have exactly the same subgroups (as sets) for all cardinals $kappa$.



The last item means that the subgroup hypergraphs of $G_i^kappa$ and $G_j^kappa$ are equal for all $i, j, kappa$.




The paper is



Keith A. Kearnes and Agnes Szendrei,

Groups with identical subgroup lattices in all powers.

J. Group Theory 7 (2004), no. 3, 385--402.






share|cite|improve this answer






















    Your Answer




    StackExchange.ifUsing("editor", function ()
    return StackExchange.using("mathjaxEditing", function ()
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    );
    );
    , "mathjax-editing");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "504"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f312177%2fdoes-the-hypergraph-of-subgroups-determine-a-group%23new-answer', 'question_page');

    );

    Post as a guest






























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    In the comments to the question, I notice something which might be an error, or at least is an incomplete response. It is pointed out in the comments that there exist nonisomorphic groups with isomorphic subgroup lattices. While true, that fact doesn't answer this question, since it is possible to have isomorphic subgroup lattices and nonisomorphic subgroup hypergraphs. Even if you assume that the groups have the same order and isomorphic subgroup lattices, it does not immediately follow that they have isomorphic subgroup hypergraphs.




    I am sure the negative answer to this question can be found in Roland Schmidt's book. But I would like to point to a theorem I coauthored after Schmidt's book was published, which applies to this question. Namely:



    Thm. For any finite $N$, there is a finite set $X=X_N$ and $N$ binary operations $circ_i$ defined on $X$ such that



    (1) $G_i = (X,circ_i)$ is a group for all $i$,

    (2) $G_inotcong G_j$ when $ineq j$, and

    (3) for all $i, j$, the groups
    $G_i^kappa$ and $G_j^kappa$ have exactly the same subgroups (as sets) for all cardinals $kappa$.



    The last item means that the subgroup hypergraphs of $G_i^kappa$ and $G_j^kappa$ are equal for all $i, j, kappa$.




    The paper is



    Keith A. Kearnes and Agnes Szendrei,

    Groups with identical subgroup lattices in all powers.

    J. Group Theory 7 (2004), no. 3, 385--402.






    share|cite|improve this answer


























      up vote
      4
      down vote



      accepted










      In the comments to the question, I notice something which might be an error, or at least is an incomplete response. It is pointed out in the comments that there exist nonisomorphic groups with isomorphic subgroup lattices. While true, that fact doesn't answer this question, since it is possible to have isomorphic subgroup lattices and nonisomorphic subgroup hypergraphs. Even if you assume that the groups have the same order and isomorphic subgroup lattices, it does not immediately follow that they have isomorphic subgroup hypergraphs.




      I am sure the negative answer to this question can be found in Roland Schmidt's book. But I would like to point to a theorem I coauthored after Schmidt's book was published, which applies to this question. Namely:



      Thm. For any finite $N$, there is a finite set $X=X_N$ and $N$ binary operations $circ_i$ defined on $X$ such that



      (1) $G_i = (X,circ_i)$ is a group for all $i$,

      (2) $G_inotcong G_j$ when $ineq j$, and

      (3) for all $i, j$, the groups
      $G_i^kappa$ and $G_j^kappa$ have exactly the same subgroups (as sets) for all cardinals $kappa$.



      The last item means that the subgroup hypergraphs of $G_i^kappa$ and $G_j^kappa$ are equal for all $i, j, kappa$.




      The paper is



      Keith A. Kearnes and Agnes Szendrei,

      Groups with identical subgroup lattices in all powers.

      J. Group Theory 7 (2004), no. 3, 385--402.






      share|cite|improve this answer
























        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        In the comments to the question, I notice something which might be an error, or at least is an incomplete response. It is pointed out in the comments that there exist nonisomorphic groups with isomorphic subgroup lattices. While true, that fact doesn't answer this question, since it is possible to have isomorphic subgroup lattices and nonisomorphic subgroup hypergraphs. Even if you assume that the groups have the same order and isomorphic subgroup lattices, it does not immediately follow that they have isomorphic subgroup hypergraphs.




        I am sure the negative answer to this question can be found in Roland Schmidt's book. But I would like to point to a theorem I coauthored after Schmidt's book was published, which applies to this question. Namely:



        Thm. For any finite $N$, there is a finite set $X=X_N$ and $N$ binary operations $circ_i$ defined on $X$ such that



        (1) $G_i = (X,circ_i)$ is a group for all $i$,

        (2) $G_inotcong G_j$ when $ineq j$, and

        (3) for all $i, j$, the groups
        $G_i^kappa$ and $G_j^kappa$ have exactly the same subgroups (as sets) for all cardinals $kappa$.



        The last item means that the subgroup hypergraphs of $G_i^kappa$ and $G_j^kappa$ are equal for all $i, j, kappa$.




        The paper is



        Keith A. Kearnes and Agnes Szendrei,

        Groups with identical subgroup lattices in all powers.

        J. Group Theory 7 (2004), no. 3, 385--402.






        share|cite|improve this answer














        In the comments to the question, I notice something which might be an error, or at least is an incomplete response. It is pointed out in the comments that there exist nonisomorphic groups with isomorphic subgroup lattices. While true, that fact doesn't answer this question, since it is possible to have isomorphic subgroup lattices and nonisomorphic subgroup hypergraphs. Even if you assume that the groups have the same order and isomorphic subgroup lattices, it does not immediately follow that they have isomorphic subgroup hypergraphs.




        I am sure the negative answer to this question can be found in Roland Schmidt's book. But I would like to point to a theorem I coauthored after Schmidt's book was published, which applies to this question. Namely:



        Thm. For any finite $N$, there is a finite set $X=X_N$ and $N$ binary operations $circ_i$ defined on $X$ such that



        (1) $G_i = (X,circ_i)$ is a group for all $i$,

        (2) $G_inotcong G_j$ when $ineq j$, and

        (3) for all $i, j$, the groups
        $G_i^kappa$ and $G_j^kappa$ have exactly the same subgroups (as sets) for all cardinals $kappa$.



        The last item means that the subgroup hypergraphs of $G_i^kappa$ and $G_j^kappa$ are equal for all $i, j, kappa$.




        The paper is



        Keith A. Kearnes and Agnes Szendrei,

        Groups with identical subgroup lattices in all powers.

        J. Group Theory 7 (2004), no. 3, 385--402.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 37 mins ago

























        answered 45 mins ago









        Keith Kearnes

        5,31412438




        5,31412438



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f312177%2fdoes-the-hypergraph-of-subgroups-determine-a-group%23new-answer', 'question_page');

            );

            Post as a guest













































































            Comments

            Popular posts from this blog

            What does second last employer means? [closed]

            Installing NextGIS Connect into QGIS 3?

            One-line joke