Quotient group of the free group on the alphabet of a string.












1












$begingroup$


For instance, let $Sigma = {a,b}$, and consider $F = F(Sigma)$, the free group.



Let $s$ be a string, say $s = ab$ and consider the subgroup $H leqslant F$, $H = langle t: t nleqslant srangle$ where $leqslant$ on strings indicates "is a substring of". Then is $H$ normal?



Let $h in H$, ie $h nleqslant s$, and let $f in F$. Then $fhf^{-1}$ is certainly in $H$ as you can't form a substring via concatenation if it involves any non substring, so $fhf^{-1} in {t nleqslant s}$ is in the generating set of $H$ so is in $H$. Therefore $F/H$ is a group.



What does $F/H$ look like!? It must be ${epsilon, a,b, ab} approx Bbb{Z}_4$ right?



Does this group have a name in literature?






abstract-algebra group-theory formal-languages free-groups







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    For instance, let $Sigma = {a,b}$, and consider $F = F(Sigma)$, the free group.



    Let $s$ be a string, say $s = ab$ and consider the subgroup $H leqslant F$, $H = langle t: t nleqslant srangle$ where $leqslant$ on strings indicates "is a substring of". Then is $H$ normal?



    Let $h in H$, ie $h nleqslant s$, and let $f in F$. Then $fhf^{-1}$ is certainly in $H$ as you can't form a substring via concatenation if it involves any non substring, so $fhf^{-1} in {t nleqslant s}$ is in the generating set of $H$ so is in $H$. Therefore $F/H$ is a group.



    What does $F/H$ look like!? It must be ${epsilon, a,b, ab} approx Bbb{Z}_4$ right?



    Does this group have a name in literature?






    abstract-algebra group-theory formal-languages free-groups







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      For instance, let $Sigma = {a,b}$, and consider $F = F(Sigma)$, the free group.



      Let $s$ be a string, say $s = ab$ and consider the subgroup $H leqslant F$, $H = langle t: t nleqslant srangle$ where $leqslant$ on strings indicates "is a substring of". Then is $H$ normal?



      Let $h in H$, ie $h nleqslant s$, and let $f in F$. Then $fhf^{-1}$ is certainly in $H$ as you can't form a substring via concatenation if it involves any non substring, so $fhf^{-1} in {t nleqslant s}$ is in the generating set of $H$ so is in $H$. Therefore $F/H$ is a group.



      What does $F/H$ look like!? It must be ${epsilon, a,b, ab} approx Bbb{Z}_4$ right?



      Does this group have a name in literature?






      abstract-algebra group-theory formal-languages free-groups







      share|cite|improve this question











      $endgroup$




      For instance, let $Sigma = {a,b}$, and consider $F = F(Sigma)$, the free group.



      Let $s$ be a string, say $s = ab$ and consider the subgroup $H leqslant F$, $H = langle t: t nleqslant srangle$ where $leqslant$ on strings indicates "is a substring of". Then is $H$ normal?



      Let $h in H$, ie $h nleqslant s$, and let $f in F$. Then $fhf^{-1}$ is certainly in $H$ as you can't form a substring via concatenation if it involves any non substring, so $fhf^{-1} in {t nleqslant s}$ is in the generating set of $H$ so is in $H$. Therefore $F/H$ is a group.



      What does $F/H$ look like!? It must be ${epsilon, a,b, ab} approx Bbb{Z}_4$ right?



      Does this group have a name in literature?






      abstract-algebra group-theory formal-languages free-groups




      abstract-algebra group-theory formal-languages free-groups






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 4:48









      Eric Wofsey

      184k13212338




      184k13212338










      asked Mar 18 '17 at 19:12









      Hermit with AdjointHermit with Adjoint

      9,12052458




      9,12052458






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Your subgroup $H$ is all of $F$. For instance, if $s=ab$, then $aba$ and $abab$ are both in $H$, and hence so is $b=(aba)^{-1}(abab)$. Similarly, $ain H$ as well. You can easily construct similar examples for any value of $s$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            If you pick $s = ab$ as your string, $h = b^{-1}ab$, as your not substring and $f=b$ as your other string, then
            $$fhf^{-1} = bb^{-1}abb^{-1} = a leq ab = s.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yeah, I saw that happening. I wonder if it's ever normal.
              $endgroup$
              – Hermit with Adjoint
              Mar 18 '17 at 22:11










            • $begingroup$
              @FruitfulApproach Now that I think about it, subgroups share the parent's identity, so "", the empty string should be in $H$, but I think the empty string should also be a substring of every string... If so, then H isn't even a subgroup, because the identity isn't in it. :/ The topic seems interesting enough, though!
              $endgroup$
              – Andrew
              Mar 18 '17 at 22:19










            • $begingroup$
              It's called you "artificialy add it in", as it's trivial enough to do. Actually, $epsilon in H$ automatically, by definition of generated subgroup.
              $endgroup$
              – Hermit with Adjoint
              Mar 18 '17 at 22:20












            • $begingroup$
              I'm making an answer. I think $s = ababab$ might shed some light
              $endgroup$
              – Hermit with Adjoint
              Mar 18 '17 at 22:21










            • $begingroup$
              Eh, I somehow missed the $ langle $, and its pair, don't ask me how ... And sure thing. :)
              $endgroup$
              – Andrew
              Mar 18 '17 at 22:25













            Your Answer





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            2 Answers
            2






            active

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            2 Answers
            2






            active

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            active

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            active

            oldest

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            2












            $begingroup$

            Your subgroup $H$ is all of $F$. For instance, if $s=ab$, then $aba$ and $abab$ are both in $H$, and hence so is $b=(aba)^{-1}(abab)$. Similarly, $ain H$ as well. You can easily construct similar examples for any value of $s$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Your subgroup $H$ is all of $F$. For instance, if $s=ab$, then $aba$ and $abab$ are both in $H$, and hence so is $b=(aba)^{-1}(abab)$. Similarly, $ain H$ as well. You can easily construct similar examples for any value of $s$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Your subgroup $H$ is all of $F$. For instance, if $s=ab$, then $aba$ and $abab$ are both in $H$, and hence so is $b=(aba)^{-1}(abab)$. Similarly, $ain H$ as well. You can easily construct similar examples for any value of $s$.






                share|cite|improve this answer









                $endgroup$



                Your subgroup $H$ is all of $F$. For instance, if $s=ab$, then $aba$ and $abab$ are both in $H$, and hence so is $b=(aba)^{-1}(abab)$. Similarly, $ain H$ as well. You can easily construct similar examples for any value of $s$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 19 '17 at 1:07









                Eric WofseyEric Wofsey

                184k13212338




                184k13212338























                    1












                    $begingroup$

                    If you pick $s = ab$ as your string, $h = b^{-1}ab$, as your not substring and $f=b$ as your other string, then
                    $$fhf^{-1} = bb^{-1}abb^{-1} = a leq ab = s.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Yeah, I saw that happening. I wonder if it's ever normal.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:11










                    • $begingroup$
                      @FruitfulApproach Now that I think about it, subgroups share the parent's identity, so "", the empty string should be in $H$, but I think the empty string should also be a substring of every string... If so, then H isn't even a subgroup, because the identity isn't in it. :/ The topic seems interesting enough, though!
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:19










                    • $begingroup$
                      It's called you "artificialy add it in", as it's trivial enough to do. Actually, $epsilon in H$ automatically, by definition of generated subgroup.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:20












                    • $begingroup$
                      I'm making an answer. I think $s = ababab$ might shed some light
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:21










                    • $begingroup$
                      Eh, I somehow missed the $ langle $, and its pair, don't ask me how ... And sure thing. :)
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:25


















                    1












                    $begingroup$

                    If you pick $s = ab$ as your string, $h = b^{-1}ab$, as your not substring and $f=b$ as your other string, then
                    $$fhf^{-1} = bb^{-1}abb^{-1} = a leq ab = s.$$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      Yeah, I saw that happening. I wonder if it's ever normal.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:11










                    • $begingroup$
                      @FruitfulApproach Now that I think about it, subgroups share the parent's identity, so "", the empty string should be in $H$, but I think the empty string should also be a substring of every string... If so, then H isn't even a subgroup, because the identity isn't in it. :/ The topic seems interesting enough, though!
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:19










                    • $begingroup$
                      It's called you "artificialy add it in", as it's trivial enough to do. Actually, $epsilon in H$ automatically, by definition of generated subgroup.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:20












                    • $begingroup$
                      I'm making an answer. I think $s = ababab$ might shed some light
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:21










                    • $begingroup$
                      Eh, I somehow missed the $ langle $, and its pair, don't ask me how ... And sure thing. :)
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:25
















                    1












                    1








                    1





                    $begingroup$

                    If you pick $s = ab$ as your string, $h = b^{-1}ab$, as your not substring and $f=b$ as your other string, then
                    $$fhf^{-1} = bb^{-1}abb^{-1} = a leq ab = s.$$






                    share|cite|improve this answer









                    $endgroup$



                    If you pick $s = ab$ as your string, $h = b^{-1}ab$, as your not substring and $f=b$ as your other string, then
                    $$fhf^{-1} = bb^{-1}abb^{-1} = a leq ab = s.$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 18 '17 at 19:51









                    AndrewAndrew

                    2,45021324




                    2,45021324












                    • $begingroup$
                      Yeah, I saw that happening. I wonder if it's ever normal.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:11










                    • $begingroup$
                      @FruitfulApproach Now that I think about it, subgroups share the parent's identity, so "", the empty string should be in $H$, but I think the empty string should also be a substring of every string... If so, then H isn't even a subgroup, because the identity isn't in it. :/ The topic seems interesting enough, though!
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:19










                    • $begingroup$
                      It's called you "artificialy add it in", as it's trivial enough to do. Actually, $epsilon in H$ automatically, by definition of generated subgroup.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:20












                    • $begingroup$
                      I'm making an answer. I think $s = ababab$ might shed some light
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:21










                    • $begingroup$
                      Eh, I somehow missed the $ langle $, and its pair, don't ask me how ... And sure thing. :)
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:25




















                    • $begingroup$
                      Yeah, I saw that happening. I wonder if it's ever normal.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:11










                    • $begingroup$
                      @FruitfulApproach Now that I think about it, subgroups share the parent's identity, so "", the empty string should be in $H$, but I think the empty string should also be a substring of every string... If so, then H isn't even a subgroup, because the identity isn't in it. :/ The topic seems interesting enough, though!
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:19










                    • $begingroup$
                      It's called you "artificialy add it in", as it's trivial enough to do. Actually, $epsilon in H$ automatically, by definition of generated subgroup.
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:20












                    • $begingroup$
                      I'm making an answer. I think $s = ababab$ might shed some light
                      $endgroup$
                      – Hermit with Adjoint
                      Mar 18 '17 at 22:21










                    • $begingroup$
                      Eh, I somehow missed the $ langle $, and its pair, don't ask me how ... And sure thing. :)
                      $endgroup$
                      – Andrew
                      Mar 18 '17 at 22:25


















                    $begingroup$
                    Yeah, I saw that happening. I wonder if it's ever normal.
                    $endgroup$
                    – Hermit with Adjoint
                    Mar 18 '17 at 22:11




                    $begingroup$
                    Yeah, I saw that happening. I wonder if it's ever normal.
                    $endgroup$
                    – Hermit with Adjoint
                    Mar 18 '17 at 22:11












                    $begingroup$
                    @FruitfulApproach Now that I think about it, subgroups share the parent's identity, so "", the empty string should be in $H$, but I think the empty string should also be a substring of every string... If so, then H isn't even a subgroup, because the identity isn't in it. :/ The topic seems interesting enough, though!
                    $endgroup$
                    – Andrew
                    Mar 18 '17 at 22:19




                    $begingroup$
                    @FruitfulApproach Now that I think about it, subgroups share the parent's identity, so "", the empty string should be in $H$, but I think the empty string should also be a substring of every string... If so, then H isn't even a subgroup, because the identity isn't in it. :/ The topic seems interesting enough, though!
                    $endgroup$
                    – Andrew
                    Mar 18 '17 at 22:19












                    $begingroup$
                    It's called you "artificialy add it in", as it's trivial enough to do. Actually, $epsilon in H$ automatically, by definition of generated subgroup.
                    $endgroup$
                    – Hermit with Adjoint
                    Mar 18 '17 at 22:20






                    $begingroup$
                    It's called you "artificialy add it in", as it's trivial enough to do. Actually, $epsilon in H$ automatically, by definition of generated subgroup.
                    $endgroup$
                    – Hermit with Adjoint
                    Mar 18 '17 at 22:20














                    $begingroup$
                    I'm making an answer. I think $s = ababab$ might shed some light
                    $endgroup$
                    – Hermit with Adjoint
                    Mar 18 '17 at 22:21




                    $begingroup$
                    I'm making an answer. I think $s = ababab$ might shed some light
                    $endgroup$
                    – Hermit with Adjoint
                    Mar 18 '17 at 22:21












                    $begingroup$
                    Eh, I somehow missed the $ langle $, and its pair, don't ask me how ... And sure thing. :)
                    $endgroup$
                    – Andrew
                    Mar 18 '17 at 22:25






                    $begingroup$
                    Eh, I somehow missed the $ langle $, and its pair, don't ask me how ... And sure thing. :)
                    $endgroup$
                    – Andrew
                    Mar 18 '17 at 22:25




















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