Are there any computational problems in groups that are harder than P?












22












$begingroup$


There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



$$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?










share|cite|improve this question











$endgroup$

















    22












    $begingroup$


    There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



    Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



    Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



    On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



    $$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



    in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



    So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?










    share|cite|improve this question











    $endgroup$















      22












      22








      22


      4



      $begingroup$


      There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



      Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



      Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



      On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



      $$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



      in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



      So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?










      share|cite|improve this question











      $endgroup$




      There are several well known classes of groups for which the word problem, conjugacy etc. are solvable in polynomial time (hyperbolic, automatic).



      Then there are several classes of groups like asynchronously automatic groups for which it is known that there is an exponential time algorithm to solve the word problem (and whether this can be improved to polynomial is open and conjectured as far as I'm aware).



      Going several steps further, there is an algorithm to solve the word problem in one-relator groups in time not bounded by any finite tower of exponentials (and again it is open and conjectured whether this can be improved to P).



      On the other, there are algorithms to solve word problems in pathological groups like the Baumslag-Gersten group:



      $$G_{(1,2)} = langle a, b | a^{a^b}= a^2 rangle$$



      in polynomial time. So even though general algorithms can be very bad, it is unknown whether there are group-specific algorithms for every group in a given class that solve the word problem quickly.



      So my question is, are there any groups in which it is known that the word problem (or any other computational problem) is at least exponential, but still solvable? Maybe exp-complete? What are the approaches to proving such a thing?







      gr.group-theory computational-complexity geometric-group-theory computational-group-theory word-problem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 17:58







      MSL

















      asked Jan 23 at 17:45









      MSLMSL

      11618




      11618






















          5 Answers
          5






          active

          oldest

          votes


















          24












          $begingroup$

          An earlier reference for groups with this property is



          J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



          There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



          The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks Derek, this seems to be exactly what I was looking for!
            $endgroup$
            – MSL
            Jan 27 at 19:39



















          19












          $begingroup$

          As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



          Isoperimetric and Isodiametric Functions of Groups,
          Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
          Annals of Mathematics
          Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



          and



          Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
          Isoperimetric functions of groups and computational complexity of the word problem.
          Ann. of Math. (2) 156 (2002), no. 2, 467–518.



          groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



          To add more information, Corollary 1.1 says:




          There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




          So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






          share|cite|improve this answer











          $endgroup$





















            11












            $begingroup$

            There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
              $endgroup$
              – MSL
              Jan 23 at 17:57






            • 7




              $begingroup$
              @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
              $endgroup$
              – Andreas Blass
              Jan 23 at 18:06



















            6












            $begingroup$

            Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



            Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



            There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              Checking if a cyclic group is simple would seem to be primality testing which is in P.
              $endgroup$
              – Benjamin Steinberg
              Jan 24 at 0:58










            • $begingroup$
              You are right; I have corrected my answer to reflect this.
              $endgroup$
              – Dima Pasechnik
              Jan 24 at 1:09






            • 1




              $begingroup$
              As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
              $endgroup$
              – Yakk
              Jan 24 at 16:58










            • $begingroup$
              “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
              $endgroup$
              – Dima Pasechnik
              Jan 24 at 17:09










            • $begingroup$
              Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
              $endgroup$
              – Pål GD
              Jan 24 at 19:00



















            1












            $begingroup$

            While most other answers have mentioned computational problems related to finitely presented (but generally infinite) groups, there are many problems in finite group theory which are either conjectured or known to be NP-complete. As an example, we have the following result from A. Lubiw, "Some NP-complete problems similar to graph isomorphism", SIAM J. Comp. 10(1981) 11-21.




            The problem of determining whether $G$ has a fixed point free element is NP-complete, even for elementary abelian $2$-groups.




            The classic problem of determining the number of subgroups of a group of order $n$ is also, as far as I can tell, conjectured to be very difficult, but I am not aware of any definite statement in either direction.



            Moving back to the world of finitely presented groups, the subset sum problem is defined as follows. Given $g_1, dots, g_k, g in G$, decide whether $$ g = g_1^{varepsilon_1} cdots g_k^{varepsilon_k}$$ for some $varepsilon_1, dots varepsilon_k in { 0, 1}$. This problem is known to be NP-complete for a vast range of different classes of groups, including, but not limited to, free metabelian non-abelian groups of finite rank, and the wreath product of two finitely generated infinite abelian groups. It is also known to be NP-complete in some particular cases, such as for $BS(1, 2)$ and Thompson's group $F$. In hyperbolic groups, however, the problem is P-time decidable, so the problem certainly is an interesting one.



            The subset sum problem and many other related problems for finitely presented groups are studied in detail in A. Miasnikov, A. Nikolaev, A. Ushakov, "Knapsack Problems in Groups", Math. of Comp. 84(2013).






            share|cite|improve this answer









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              24












              $begingroup$

              An earlier reference for groups with this property is



              J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



              There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



              The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thanks Derek, this seems to be exactly what I was looking for!
                $endgroup$
                – MSL
                Jan 27 at 19:39
















              24












              $begingroup$

              An earlier reference for groups with this property is



              J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



              There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



              The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Thanks Derek, this seems to be exactly what I was looking for!
                $endgroup$
                – MSL
                Jan 27 at 19:39














              24












              24








              24





              $begingroup$

              An earlier reference for groups with this property is



              J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



              There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



              The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,






              share|cite|improve this answer









              $endgroup$



              An earlier reference for groups with this property is



              J. Avenhaus and K. Madlener. Subrekursive Komplexität der Gruppen. I. Gruppen mit vorgeschriebenen Komplexität. Acta Infomat., 9 (1): 87-104, 1977/78.



              There is a hierarchy of the recursive functions known as the (difficult to pronounce) Grzegorczyk Hierarchy $E_0 subset E_1 subset E_2 subset cdots$, where (roughly) $E_1$ contains the linearly bounded functions, $E_2$ polynomially bounded functions, and $E_3$ those functions that are bounded by iterated exponentials.



              The above paper describes constructions of finitely presented groups $G_n$ for $n ge 3$, in which solving the word problem has time complexity bounded by a function $E_n$ but not by any function in $E_{n-1}$,







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 23 at 22:16









              Derek HoltDerek Holt

              27.3k464112




              27.3k464112












              • $begingroup$
                Thanks Derek, this seems to be exactly what I was looking for!
                $endgroup$
                – MSL
                Jan 27 at 19:39


















              • $begingroup$
                Thanks Derek, this seems to be exactly what I was looking for!
                $endgroup$
                – MSL
                Jan 27 at 19:39
















              $begingroup$
              Thanks Derek, this seems to be exactly what I was looking for!
              $endgroup$
              – MSL
              Jan 27 at 19:39




              $begingroup$
              Thanks Derek, this seems to be exactly what I was looking for!
              $endgroup$
              – MSL
              Jan 27 at 19:39











              19












              $begingroup$

              As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



              Isoperimetric and Isodiametric Functions of Groups,
              Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
              Annals of Mathematics
              Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



              and



              Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
              Isoperimetric functions of groups and computational complexity of the word problem.
              Ann. of Math. (2) 156 (2002), no. 2, 467–518.



              groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



              To add more information, Corollary 1.1 says:




              There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




              So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






              share|cite|improve this answer











              $endgroup$


















                19












                $begingroup$

                As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



                Isoperimetric and Isodiametric Functions of Groups,
                Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                Annals of Mathematics
                Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



                and



                Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                Isoperimetric functions of groups and computational complexity of the word problem.
                Ann. of Math. (2) 156 (2002), no. 2, 467–518.



                groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



                To add more information, Corollary 1.1 says:




                There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




                So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






                share|cite|improve this answer











                $endgroup$
















                  19












                  19








                  19





                  $begingroup$

                  As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



                  Isoperimetric and Isodiametric Functions of Groups,
                  Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                  Annals of Mathematics
                  Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



                  and



                  Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                  Isoperimetric functions of groups and computational complexity of the word problem.
                  Ann. of Math. (2) 156 (2002), no. 2, 467–518.



                  groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



                  To add more information, Corollary 1.1 says:




                  There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




                  So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.






                  share|cite|improve this answer











                  $endgroup$



                  As Andreas says (in his answer and his comment to it), there are groups whose word problem is undecidable and one could similarly set up a group that encodes the halting problem of a class of Turing machines where this is decidable but difficult. However, one must be careful in the encoding. In



                  Isoperimetric and Isodiametric Functions of Groups,
                  Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                  Annals of Mathematics
                  Second Series, Vol. 156, No. 2 (Sep., 2002), pp. 345-466



                  and



                  Mark V. Sapir, Jean-Camille Birget and Eliyahu Rips
                  Isoperimetric functions of groups and computational complexity of the word problem.
                  Ann. of Math. (2) 156 (2002), no. 2, 467–518.



                  groups with NP complete word problem are constructed and other similar results. See in particular, Corollary 1.1 of the first paper listed above.



                  To add more information, Corollary 1.1 says:




                  There exists a finitely presented group with NP-complete word problem. Moreover for every language $Lsubseteq A^*$ for some finite alphabet $A$ there exists a finitely presented group $G$ such that the nondeterministic complexity of $G$ is polynomially equivalent to the nondeterministic complexity of $L$.




                  So, for instance, if $L$ is an EXP-time complete problem, then the word problem of $G$ is in NEXP-time and not in NP (hence not in P). Of course you can replace EXP by your favorite time complexity class strictly above NP.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 24 at 18:01

























                  answered Jan 23 at 18:09









                  Benjamin SteinbergBenjamin Steinberg

                  23.4k265125




                  23.4k265125























                      11












                      $begingroup$

                      There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                        $endgroup$
                        – MSL
                        Jan 23 at 17:57






                      • 7




                        $begingroup$
                        @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                        $endgroup$
                        – Andreas Blass
                        Jan 23 at 18:06
















                      11












                      $begingroup$

                      There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                        $endgroup$
                        – MSL
                        Jan 23 at 17:57






                      • 7




                        $begingroup$
                        @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                        $endgroup$
                        – Andreas Blass
                        Jan 23 at 18:06














                      11












                      11








                      11





                      $begingroup$

                      There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .






                      share|cite|improve this answer









                      $endgroup$



                      There are finitely presented groups whose word problem is undecidable. See, for example, https://en.wikipedia.org/wiki/Word_problem_for_groups .







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 23 at 17:55









                      Andreas BlassAndreas Blass

                      57.8k7137224




                      57.8k7137224












                      • $begingroup$
                        True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                        $endgroup$
                        – MSL
                        Jan 23 at 17:57






                      • 7




                        $begingroup$
                        @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                        $endgroup$
                        – Andreas Blass
                        Jan 23 at 18:06


















                      • $begingroup$
                        True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                        $endgroup$
                        – MSL
                        Jan 23 at 17:57






                      • 7




                        $begingroup$
                        @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                        $endgroup$
                        – Andreas Blass
                        Jan 23 at 18:06
















                      $begingroup$
                      True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                      $endgroup$
                      – MSL
                      Jan 23 at 17:57




                      $begingroup$
                      True, what I really meant was among the groups whose word problem is solvable. Thanks, I'll edit my question!
                      $endgroup$
                      – MSL
                      Jan 23 at 17:57




                      7




                      7




                      $begingroup$
                      @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                      $endgroup$
                      – Andreas Blass
                      Jan 23 at 18:06




                      $begingroup$
                      @MSL I'd expect that the same method can handle the decidable case. Instead of coding into the presentation a Turing machine whose halting problem is undecidable, code one whose halting problem is decidable but takes a very long time to decide.
                      $endgroup$
                      – Andreas Blass
                      Jan 23 at 18:06











                      6












                      $begingroup$

                      Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                      Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                      There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        Checking if a cyclic group is simple would seem to be primality testing which is in P.
                        $endgroup$
                        – Benjamin Steinberg
                        Jan 24 at 0:58










                      • $begingroup$
                        You are right; I have corrected my answer to reflect this.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 1:09






                      • 1




                        $begingroup$
                        As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                        $endgroup$
                        – Yakk
                        Jan 24 at 16:58










                      • $begingroup$
                        “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 17:09










                      • $begingroup$
                        Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                        $endgroup$
                        – Pål GD
                        Jan 24 at 19:00
















                      6












                      $begingroup$

                      Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                      Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                      There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        Checking if a cyclic group is simple would seem to be primality testing which is in P.
                        $endgroup$
                        – Benjamin Steinberg
                        Jan 24 at 0:58










                      • $begingroup$
                        You are right; I have corrected my answer to reflect this.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 1:09






                      • 1




                        $begingroup$
                        As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                        $endgroup$
                        – Yakk
                        Jan 24 at 16:58










                      • $begingroup$
                        “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 17:09










                      • $begingroup$
                        Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                        $endgroup$
                        – Pål GD
                        Jan 24 at 19:00














                      6












                      6








                      6





                      $begingroup$

                      Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                      Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                      There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289






                      share|cite|improve this answer











                      $endgroup$



                      Classical problem which is believed not to be in P is number factoring, which can be cast as computing a decomposition of a cyclic group into simple ones.



                      Several problems in permutation groups are known to be as hard as graph isomorphism, thus not believed to be in P, too.



                      There are also NP-hard problems known for permutation groups, see e.g. https://www.sciencedirect.com/science/article/pii/S0012365X09001289







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 24 at 1:08

























                      answered Jan 24 at 0:50









                      Dima PasechnikDima Pasechnik

                      9,15311851




                      9,15311851








                      • 1




                        $begingroup$
                        Checking if a cyclic group is simple would seem to be primality testing which is in P.
                        $endgroup$
                        – Benjamin Steinberg
                        Jan 24 at 0:58










                      • $begingroup$
                        You are right; I have corrected my answer to reflect this.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 1:09






                      • 1




                        $begingroup$
                        As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                        $endgroup$
                        – Yakk
                        Jan 24 at 16:58










                      • $begingroup$
                        “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 17:09










                      • $begingroup$
                        Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                        $endgroup$
                        – Pål GD
                        Jan 24 at 19:00














                      • 1




                        $begingroup$
                        Checking if a cyclic group is simple would seem to be primality testing which is in P.
                        $endgroup$
                        – Benjamin Steinberg
                        Jan 24 at 0:58










                      • $begingroup$
                        You are right; I have corrected my answer to reflect this.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 1:09






                      • 1




                        $begingroup$
                        As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                        $endgroup$
                        – Yakk
                        Jan 24 at 16:58










                      • $begingroup$
                        “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                        $endgroup$
                        – Dima Pasechnik
                        Jan 24 at 17:09










                      • $begingroup$
                        Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                        $endgroup$
                        – Pål GD
                        Jan 24 at 19:00








                      1




                      1




                      $begingroup$
                      Checking if a cyclic group is simple would seem to be primality testing which is in P.
                      $endgroup$
                      – Benjamin Steinberg
                      Jan 24 at 0:58




                      $begingroup$
                      Checking if a cyclic group is simple would seem to be primality testing which is in P.
                      $endgroup$
                      – Benjamin Steinberg
                      Jan 24 at 0:58












                      $begingroup$
                      You are right; I have corrected my answer to reflect this.
                      $endgroup$
                      – Dima Pasechnik
                      Jan 24 at 1:09




                      $begingroup$
                      You are right; I have corrected my answer to reflect this.
                      $endgroup$
                      – Dima Pasechnik
                      Jan 24 at 1:09




                      1




                      1




                      $begingroup$
                      As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                      $endgroup$
                      – Yakk
                      Jan 24 at 16:58




                      $begingroup$
                      As I have mentioned elsewhere, no problem in NP has ever been proven to be outside of P. So problems in NP seem like the exact opposite of what the OP asked for. You need problems complete in a complexity class proven to be separate from P, not just assumed (with good reason) to be separate from P.
                      $endgroup$
                      – Yakk
                      Jan 24 at 16:58












                      $begingroup$
                      “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                      $endgroup$
                      – Dima Pasechnik
                      Jan 24 at 17:09




                      $begingroup$
                      “NP-hard” does not merely mean “in NP”. Roughly, it means “at least as hard as an NP-complete problem”, and NP-complete problems are believed not to be in P.
                      $endgroup$
                      – Dima Pasechnik
                      Jan 24 at 17:09












                      $begingroup$
                      Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                      $endgroup$
                      – Pål GD
                      Jan 24 at 19:00




                      $begingroup$
                      Yes, it means at least as hard as an NP-complete problem, but you surely see how that doesn't answer the question. NP-hard problems are only conjecturally harder than P.
                      $endgroup$
                      – Pål GD
                      Jan 24 at 19:00











                      1












                      $begingroup$

                      While most other answers have mentioned computational problems related to finitely presented (but generally infinite) groups, there are many problems in finite group theory which are either conjectured or known to be NP-complete. As an example, we have the following result from A. Lubiw, "Some NP-complete problems similar to graph isomorphism", SIAM J. Comp. 10(1981) 11-21.




                      The problem of determining whether $G$ has a fixed point free element is NP-complete, even for elementary abelian $2$-groups.




                      The classic problem of determining the number of subgroups of a group of order $n$ is also, as far as I can tell, conjectured to be very difficult, but I am not aware of any definite statement in either direction.



                      Moving back to the world of finitely presented groups, the subset sum problem is defined as follows. Given $g_1, dots, g_k, g in G$, decide whether $$ g = g_1^{varepsilon_1} cdots g_k^{varepsilon_k}$$ for some $varepsilon_1, dots varepsilon_k in { 0, 1}$. This problem is known to be NP-complete for a vast range of different classes of groups, including, but not limited to, free metabelian non-abelian groups of finite rank, and the wreath product of two finitely generated infinite abelian groups. It is also known to be NP-complete in some particular cases, such as for $BS(1, 2)$ and Thompson's group $F$. In hyperbolic groups, however, the problem is P-time decidable, so the problem certainly is an interesting one.



                      The subset sum problem and many other related problems for finitely presented groups are studied in detail in A. Miasnikov, A. Nikolaev, A. Ushakov, "Knapsack Problems in Groups", Math. of Comp. 84(2013).






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        While most other answers have mentioned computational problems related to finitely presented (but generally infinite) groups, there are many problems in finite group theory which are either conjectured or known to be NP-complete. As an example, we have the following result from A. Lubiw, "Some NP-complete problems similar to graph isomorphism", SIAM J. Comp. 10(1981) 11-21.




                        The problem of determining whether $G$ has a fixed point free element is NP-complete, even for elementary abelian $2$-groups.




                        The classic problem of determining the number of subgroups of a group of order $n$ is also, as far as I can tell, conjectured to be very difficult, but I am not aware of any definite statement in either direction.



                        Moving back to the world of finitely presented groups, the subset sum problem is defined as follows. Given $g_1, dots, g_k, g in G$, decide whether $$ g = g_1^{varepsilon_1} cdots g_k^{varepsilon_k}$$ for some $varepsilon_1, dots varepsilon_k in { 0, 1}$. This problem is known to be NP-complete for a vast range of different classes of groups, including, but not limited to, free metabelian non-abelian groups of finite rank, and the wreath product of two finitely generated infinite abelian groups. It is also known to be NP-complete in some particular cases, such as for $BS(1, 2)$ and Thompson's group $F$. In hyperbolic groups, however, the problem is P-time decidable, so the problem certainly is an interesting one.



                        The subset sum problem and many other related problems for finitely presented groups are studied in detail in A. Miasnikov, A. Nikolaev, A. Ushakov, "Knapsack Problems in Groups", Math. of Comp. 84(2013).






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          While most other answers have mentioned computational problems related to finitely presented (but generally infinite) groups, there are many problems in finite group theory which are either conjectured or known to be NP-complete. As an example, we have the following result from A. Lubiw, "Some NP-complete problems similar to graph isomorphism", SIAM J. Comp. 10(1981) 11-21.




                          The problem of determining whether $G$ has a fixed point free element is NP-complete, even for elementary abelian $2$-groups.




                          The classic problem of determining the number of subgroups of a group of order $n$ is also, as far as I can tell, conjectured to be very difficult, but I am not aware of any definite statement in either direction.



                          Moving back to the world of finitely presented groups, the subset sum problem is defined as follows. Given $g_1, dots, g_k, g in G$, decide whether $$ g = g_1^{varepsilon_1} cdots g_k^{varepsilon_k}$$ for some $varepsilon_1, dots varepsilon_k in { 0, 1}$. This problem is known to be NP-complete for a vast range of different classes of groups, including, but not limited to, free metabelian non-abelian groups of finite rank, and the wreath product of two finitely generated infinite abelian groups. It is also known to be NP-complete in some particular cases, such as for $BS(1, 2)$ and Thompson's group $F$. In hyperbolic groups, however, the problem is P-time decidable, so the problem certainly is an interesting one.



                          The subset sum problem and many other related problems for finitely presented groups are studied in detail in A. Miasnikov, A. Nikolaev, A. Ushakov, "Knapsack Problems in Groups", Math. of Comp. 84(2013).






                          share|cite|improve this answer









                          $endgroup$



                          While most other answers have mentioned computational problems related to finitely presented (but generally infinite) groups, there are many problems in finite group theory which are either conjectured or known to be NP-complete. As an example, we have the following result from A. Lubiw, "Some NP-complete problems similar to graph isomorphism", SIAM J. Comp. 10(1981) 11-21.




                          The problem of determining whether $G$ has a fixed point free element is NP-complete, even for elementary abelian $2$-groups.




                          The classic problem of determining the number of subgroups of a group of order $n$ is also, as far as I can tell, conjectured to be very difficult, but I am not aware of any definite statement in either direction.



                          Moving back to the world of finitely presented groups, the subset sum problem is defined as follows. Given $g_1, dots, g_k, g in G$, decide whether $$ g = g_1^{varepsilon_1} cdots g_k^{varepsilon_k}$$ for some $varepsilon_1, dots varepsilon_k in { 0, 1}$. This problem is known to be NP-complete for a vast range of different classes of groups, including, but not limited to, free metabelian non-abelian groups of finite rank, and the wreath product of two finitely generated infinite abelian groups. It is also known to be NP-complete in some particular cases, such as for $BS(1, 2)$ and Thompson's group $F$. In hyperbolic groups, however, the problem is P-time decidable, so the problem certainly is an interesting one.



                          The subset sum problem and many other related problems for finitely presented groups are studied in detail in A. Miasnikov, A. Nikolaev, A. Ushakov, "Knapsack Problems in Groups", Math. of Comp. 84(2013).







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Feb 2 at 10:44









                          Carl-Fredrik Nyberg BroddaCarl-Fredrik Nyberg Brodda

                          33328




                          33328






























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