What is wrong with this fake proof that subgroup of a cyclic group is cyclic?

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Let $G$ be a cyclic group generated by $a$ and $H$ its subgroup. This is a proof by contradiction. Assume there is no $r$ in $H$ such that $langle r rangle = H$. If some $s$ in $G$ is not in $H$ then $langle s rangle neq H$. So for all $s$ in $G$, $langle
s rangle neq H$ Thus for all $s$ in $G$ there exists an $hin H$ such that $h notin langle s rangle$. But this is a contradiction with the fact that $a$ generates $G$.



I know this proof is fake, because it does not use the fact that $H$ is a subgroup but I am unable to find the mistake(s).






fake-proofs







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    Let $G$ be a cyclic group generated by $a$ and $H$ its subgroup. This is a proof by contradiction. Assume there is no $r$ in $H$ such that $langle r rangle = H$. If some $s$ in $G$ is not in $H$ then $langle s rangle neq H$. So for all $s$ in $G$, $langle
    s rangle neq H$     Thus for all $s$ in $G$ there exists an $hin H$ such that $h notin langle s rangle$. But this is a contradiction with the fact that $a$ generates $G$.



    I know this proof is fake, because it does not use the fact that $H$ is a subgroup but I am unable to find the mistake(s).






    fake-proofs







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      Let $G$ be a cyclic group generated by $a$ and $H$ its subgroup. This is a proof by contradiction. Assume there is no $r$ in $H$ such that $langle r rangle = H$. If some $s$ in $G$ is not in $H$ then $langle s rangle neq H$. So for all $s$ in $G$, $langle
      s rangle neq H$       Thus for all $s$ in $G$ there exists an $hin H$ such that $h notin langle s rangle$. But this is a contradiction with the fact that $a$ generates $G$.



      I know this proof is fake, because it does not use the fact that $H$ is a subgroup but I am unable to find the mistake(s).






      fake-proofs







      share|cite|improve this question













      Let $G$ be a cyclic group generated by $a$ and $H$ its subgroup. This is a proof by contradiction. Assume there is no $r$ in $H$ such that $langle r rangle = H$. If some $s$ in $G$ is not in $H$ then $langle s rangle neq H$. So for all $s$ in $G$, $langle
      s rangle neq H$       Thus for all $s$ in $G$ there exists an $hin H$ such that $h notin langle s rangle$. But this is a contradiction with the fact that $a$ generates $G$.



      I know this proof is fake, because it does not use the fact that $H$ is a subgroup but I am unable to find the mistake(s).






      fake-proofs




      fake-proofs






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      asked 3 hours ago









      Adam

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          3 Answers
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          You are correct in proving that, for every $xin G$, $langle xranglene H$. But this doesn't imply that, for every $xin G$, there is $hin H$ such that $hnotinlangle xrangle$.



          For instance, this is false as soon as $Hne G$ and $x$ is a generator of $G$.






          share|cite|improve this answer



























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            No $sin G$ such that $langle srangle=H$ is clear.



            But then you can't follow with for all $sin G$ there's $hin H$ such that $hnotinlangle srangle$ because $H$ could be a subset of $langle srangle$ even though $snotin G$.



            Which, incidentally, is what happens if $s$ is a generator of $G$






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              It is correct that $forall s in G, langle srangle ne H$, but this does not imply $exists h in H, h notin langle srangle$, because $langle srangle$ may be a proper superset of $H$.






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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes








                up vote
                6
                down vote



                accepted










                You are correct in proving that, for every $xin G$, $langle xranglene H$. But this doesn't imply that, for every $xin G$, there is $hin H$ such that $hnotinlangle xrangle$.



                For instance, this is false as soon as $Hne G$ and $x$ is a generator of $G$.






                share|cite|improve this answer
























                  up vote
                  6
                  down vote



                  accepted










                  You are correct in proving that, for every $xin G$, $langle xranglene H$. But this doesn't imply that, for every $xin G$, there is $hin H$ such that $hnotinlangle xrangle$.



                  For instance, this is false as soon as $Hne G$ and $x$ is a generator of $G$.






                  share|cite|improve this answer






















                    up vote
                    6
                    down vote



                    accepted







                    up vote
                    6
                    down vote



                    accepted






                    You are correct in proving that, for every $xin G$, $langle xranglene H$. But this doesn't imply that, for every $xin G$, there is $hin H$ such that $hnotinlangle xrangle$.



                    For instance, this is false as soon as $Hne G$ and $x$ is a generator of $G$.






                    share|cite|improve this answer












                    You are correct in proving that, for every $xin G$, $langle xranglene H$. But this doesn't imply that, for every $xin G$, there is $hin H$ such that $hnotinlangle xrangle$.



                    For instance, this is false as soon as $Hne G$ and $x$ is a generator of $G$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    egreg

                    170k1283192




                    170k1283192




















                        up vote
                        3
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                        No $sin G$ such that $langle srangle=H$ is clear.



                        But then you can't follow with for all $sin G$ there's $hin H$ such that $hnotinlangle srangle$ because $H$ could be a subset of $langle srangle$ even though $snotin G$.



                        Which, incidentally, is what happens if $s$ is a generator of $G$






                        share|cite|improve this answer
























                          up vote
                          3
                          down vote













                          No $sin G$ such that $langle srangle=H$ is clear.



                          But then you can't follow with for all $sin G$ there's $hin H$ such that $hnotinlangle srangle$ because $H$ could be a subset of $langle srangle$ even though $snotin G$.



                          Which, incidentally, is what happens if $s$ is a generator of $G$






                          share|cite|improve this answer






















                            up vote
                            3
                            down vote










                            up vote
                            3
                            down vote









                            No $sin G$ such that $langle srangle=H$ is clear.



                            But then you can't follow with for all $sin G$ there's $hin H$ such that $hnotinlangle srangle$ because $H$ could be a subset of $langle srangle$ even though $snotin G$.



                            Which, incidentally, is what happens if $s$ is a generator of $G$






                            share|cite|improve this answer












                            No $sin G$ such that $langle srangle=H$ is clear.



                            But then you can't follow with for all $sin G$ there's $hin H$ such that $hnotinlangle srangle$ because $H$ could be a subset of $langle srangle$ even though $snotin G$.



                            Which, incidentally, is what happens if $s$ is a generator of $G$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









                            Andrea Mori

                            19.2k13465




                            19.2k13465




















                                up vote
                                1
                                down vote













                                It is correct that $forall s in G, langle srangle ne H$, but this does not imply $exists h in H, h notin langle srangle$, because $langle srangle$ may be a proper superset of $H$.






                                share|cite|improve this answer
























                                  up vote
                                  1
                                  down vote













                                  It is correct that $forall s in G, langle srangle ne H$, but this does not imply $exists h in H, h notin langle srangle$, because $langle srangle$ may be a proper superset of $H$.






                                  share|cite|improve this answer






















                                    up vote
                                    1
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                                    up vote
                                    1
                                    down vote









                                    It is correct that $forall s in G, langle srangle ne H$, but this does not imply $exists h in H, h notin langle srangle$, because $langle srangle$ may be a proper superset of $H$.






                                    share|cite|improve this answer












                                    It is correct that $forall s in G, langle srangle ne H$, but this does not imply $exists h in H, h notin langle srangle$, because $langle srangle$ may be a proper superset of $H$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 3 hours ago









                                    fkraiem

                                    2,8361510




                                    2,8361510



























                                         

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