Abstract algebra subgroup proof verification












8












$begingroup$


This is my first attempt at a formally written proof so I would appreciate any pointers as far as proof-writing technique or the validity of the actual proof itself.



Note: I have not formally taken abstract algebra or a proof-writing course so I am sure that I am lacking in many proof-writing aspects, so I would really love a lot of constructive criticism both on the actual proof itself, and the way I wrote the proof. I also go into greater detail than might be appropriate for this type of proof because I am shaky on a lot of the math foundations so I figure that any imperfections in my knowledge will be more easily seen with a more explicit construction of this proof. Thank you all in advance.




Let $G$ be a finite group, and let $S$ be a nonempty subset of $G$. Suppose $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$. (HINT: It remains to prove that $S$ contains $e$ and is closed with respect to inverses. Let $S$ = {$a_1$ ... $a_n$}. If $a_i$ $∈$ $S$, consider the distinct elements $a_ia_1$, $a_ia_2$, $...$ $a_ia_n$




Proof:



First we will define a function $A_1 : S rightarrow S$ that maps $s mapsto a_1s$. This function is injective because $$a_1y = a_1x$$ $$a^{-1}_1a^1y = a^{-1}_1a_1x$$ $$y = x$$



The function is then surjective because $A_1(S) subseteq S$ and since $A_1$ is injective, it contains $|S|$ elements. Therefore $A_1$ maps onto every element in $S$ and is therefore surjective as well.



This means that $a_1$ is in the image of $A_1$. Therefore $$A_1(a_1) = a_1$$ $$a_1s = a_1$$ $$s = e$$



Since $S$ is closed under multiplication, $e in S$.



Next, we will define a function $A_2 : S rightarrow S$ that maps $s mapsto a^2_1s$. This function is also injective $$a^2_1x = a^2_1y$$ $$x = y$$



It follows that this function is also surjective since it too is injective and contains |S| elements.



This means that $a_1$ is in the image of $A_2$ as well. Therefore $$A_2(z) = a_1$$ $$a^2_1z = a_1$$ $$z = a^{-1}_1$$



Since $S$ is closed under multiplication $a^{-1}_1 in S$.



Therefore $e, a^{-1}_1 in S$ so $S$ is a subgroup of $G$.



Please tear this apart! Thanks in advance.










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    8












    $begingroup$


    This is my first attempt at a formally written proof so I would appreciate any pointers as far as proof-writing technique or the validity of the actual proof itself.



    Note: I have not formally taken abstract algebra or a proof-writing course so I am sure that I am lacking in many proof-writing aspects, so I would really love a lot of constructive criticism both on the actual proof itself, and the way I wrote the proof. I also go into greater detail than might be appropriate for this type of proof because I am shaky on a lot of the math foundations so I figure that any imperfections in my knowledge will be more easily seen with a more explicit construction of this proof. Thank you all in advance.




    Let $G$ be a finite group, and let $S$ be a nonempty subset of $G$. Suppose $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$. (HINT: It remains to prove that $S$ contains $e$ and is closed with respect to inverses. Let $S$ = {$a_1$ ... $a_n$}. If $a_i$ $∈$ $S$, consider the distinct elements $a_ia_1$, $a_ia_2$, $...$ $a_ia_n$




    Proof:



    First we will define a function $A_1 : S rightarrow S$ that maps $s mapsto a_1s$. This function is injective because $$a_1y = a_1x$$ $$a^{-1}_1a^1y = a^{-1}_1a_1x$$ $$y = x$$



    The function is then surjective because $A_1(S) subseteq S$ and since $A_1$ is injective, it contains $|S|$ elements. Therefore $A_1$ maps onto every element in $S$ and is therefore surjective as well.



    This means that $a_1$ is in the image of $A_1$. Therefore $$A_1(a_1) = a_1$$ $$a_1s = a_1$$ $$s = e$$



    Since $S$ is closed under multiplication, $e in S$.



    Next, we will define a function $A_2 : S rightarrow S$ that maps $s mapsto a^2_1s$. This function is also injective $$a^2_1x = a^2_1y$$ $$x = y$$



    It follows that this function is also surjective since it too is injective and contains |S| elements.



    This means that $a_1$ is in the image of $A_2$ as well. Therefore $$A_2(z) = a_1$$ $$a^2_1z = a_1$$ $$z = a^{-1}_1$$



    Since $S$ is closed under multiplication $a^{-1}_1 in S$.



    Therefore $e, a^{-1}_1 in S$ so $S$ is a subgroup of $G$.



    Please tear this apart! Thanks in advance.










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      8












      8








      8


      1



      $begingroup$


      This is my first attempt at a formally written proof so I would appreciate any pointers as far as proof-writing technique or the validity of the actual proof itself.



      Note: I have not formally taken abstract algebra or a proof-writing course so I am sure that I am lacking in many proof-writing aspects, so I would really love a lot of constructive criticism both on the actual proof itself, and the way I wrote the proof. I also go into greater detail than might be appropriate for this type of proof because I am shaky on a lot of the math foundations so I figure that any imperfections in my knowledge will be more easily seen with a more explicit construction of this proof. Thank you all in advance.




      Let $G$ be a finite group, and let $S$ be a nonempty subset of $G$. Suppose $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$. (HINT: It remains to prove that $S$ contains $e$ and is closed with respect to inverses. Let $S$ = {$a_1$ ... $a_n$}. If $a_i$ $∈$ $S$, consider the distinct elements $a_ia_1$, $a_ia_2$, $...$ $a_ia_n$




      Proof:



      First we will define a function $A_1 : S rightarrow S$ that maps $s mapsto a_1s$. This function is injective because $$a_1y = a_1x$$ $$a^{-1}_1a^1y = a^{-1}_1a_1x$$ $$y = x$$



      The function is then surjective because $A_1(S) subseteq S$ and since $A_1$ is injective, it contains $|S|$ elements. Therefore $A_1$ maps onto every element in $S$ and is therefore surjective as well.



      This means that $a_1$ is in the image of $A_1$. Therefore $$A_1(a_1) = a_1$$ $$a_1s = a_1$$ $$s = e$$



      Since $S$ is closed under multiplication, $e in S$.



      Next, we will define a function $A_2 : S rightarrow S$ that maps $s mapsto a^2_1s$. This function is also injective $$a^2_1x = a^2_1y$$ $$x = y$$



      It follows that this function is also surjective since it too is injective and contains |S| elements.



      This means that $a_1$ is in the image of $A_2$ as well. Therefore $$A_2(z) = a_1$$ $$a^2_1z = a_1$$ $$z = a^{-1}_1$$



      Since $S$ is closed under multiplication $a^{-1}_1 in S$.



      Therefore $e, a^{-1}_1 in S$ so $S$ is a subgroup of $G$.



      Please tear this apart! Thanks in advance.










      share|cite|improve this question







      New contributor




      ForIgreaterthanJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      $endgroup$




      This is my first attempt at a formally written proof so I would appreciate any pointers as far as proof-writing technique or the validity of the actual proof itself.



      Note: I have not formally taken abstract algebra or a proof-writing course so I am sure that I am lacking in many proof-writing aspects, so I would really love a lot of constructive criticism both on the actual proof itself, and the way I wrote the proof. I also go into greater detail than might be appropriate for this type of proof because I am shaky on a lot of the math foundations so I figure that any imperfections in my knowledge will be more easily seen with a more explicit construction of this proof. Thank you all in advance.




      Let $G$ be a finite group, and let $S$ be a nonempty subset of $G$. Suppose $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$. (HINT: It remains to prove that $S$ contains $e$ and is closed with respect to inverses. Let $S$ = {$a_1$ ... $a_n$}. If $a_i$ $∈$ $S$, consider the distinct elements $a_ia_1$, $a_ia_2$, $...$ $a_ia_n$




      Proof:



      First we will define a function $A_1 : S rightarrow S$ that maps $s mapsto a_1s$. This function is injective because $$a_1y = a_1x$$ $$a^{-1}_1a^1y = a^{-1}_1a_1x$$ $$y = x$$



      The function is then surjective because $A_1(S) subseteq S$ and since $A_1$ is injective, it contains $|S|$ elements. Therefore $A_1$ maps onto every element in $S$ and is therefore surjective as well.



      This means that $a_1$ is in the image of $A_1$. Therefore $$A_1(a_1) = a_1$$ $$a_1s = a_1$$ $$s = e$$



      Since $S$ is closed under multiplication, $e in S$.



      Next, we will define a function $A_2 : S rightarrow S$ that maps $s mapsto a^2_1s$. This function is also injective $$a^2_1x = a^2_1y$$ $$x = y$$



      It follows that this function is also surjective since it too is injective and contains |S| elements.



      This means that $a_1$ is in the image of $A_2$ as well. Therefore $$A_2(z) = a_1$$ $$a^2_1z = a_1$$ $$z = a^{-1}_1$$



      Since $S$ is closed under multiplication $a^{-1}_1 in S$.



      Therefore $e, a^{-1}_1 in S$ so $S$ is a subgroup of $G$.



      Please tear this apart! Thanks in advance.







      abstract-algebra group-theory proof-verification proof-writing finite-groups






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      asked 7 hours ago









      ForIgreaterthanJForIgreaterthanJ

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          2 Answers
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          $begingroup$

          You don't need to use the second map $A_2$. Once you know that $e in S$, you know by the first part of your proof that $A_1$ is surjective, so for some $s in S, a_1s=e$. Then $s=a^{-1}$.



          Another way to prove this result, by the way, is to just take all possible powers of $a_1$, which all are contained within $S$ because $S$ is multiplicatively closed. They have to repeat at some point because $S$ is finite. If $a^m=a^{m+k}=a^ma^k$, then $a^k=e in S$, and $aa^{k-1}=a^k=e Rightarrow a^{k-1}=a^{-1} in S$.



          As for proof-writing, one small nit. I don't think the line $A_1(a_1) = a_1$ conveys your intended meaning. I think you mean to say $a_1 in A_1(S)$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            The proof presented here by our OP ForIgreaterthan_I appears to be logically valid; indeed, I like it because it is clever and innovative.



            The way I would approach this, which alas is standard and not in fact overly clever, being as it is indeed rather obvious and straightforward, is as follows:



            Let



            $s in S; tag 1$



            consider the powers



            $s^i in S; tag 2$



            since



            $S subset G tag 3$



            and



            $vert G vert < infty, tag 4$



            we have



            $vert S vert < infty tag 5$



            as well; thus the sequence



            $s, s^2, s^3, ldots, s^i, s^{i + 1}, ldots tag 6$



            must repeat itself at some point; that is,



            $exists k, l in Bbb N, ; l ge k + 1, tag 7$



            with



            $s^l = s^k; tag 8$



            then



            $s^{l - k} = e; tag 9$



            thus



            $e = s^{l - k} in S; tag{10}$



            it follows then that



            $s^{l - k - 1}s = e tag{11}$



            and clearly



            $s^{l - k - 1} in S; tag{12}$



            thus the group identity $e in S$, and every $s in S$ has an inverse in $S$; $S$ is thus a subgroup of $G$.






            share|cite|improve this answer









            $endgroup$













              Your Answer





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              2 Answers
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              $begingroup$

              You don't need to use the second map $A_2$. Once you know that $e in S$, you know by the first part of your proof that $A_1$ is surjective, so for some $s in S, a_1s=e$. Then $s=a^{-1}$.



              Another way to prove this result, by the way, is to just take all possible powers of $a_1$, which all are contained within $S$ because $S$ is multiplicatively closed. They have to repeat at some point because $S$ is finite. If $a^m=a^{m+k}=a^ma^k$, then $a^k=e in S$, and $aa^{k-1}=a^k=e Rightarrow a^{k-1}=a^{-1} in S$.



              As for proof-writing, one small nit. I don't think the line $A_1(a_1) = a_1$ conveys your intended meaning. I think you mean to say $a_1 in A_1(S)$.






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                You don't need to use the second map $A_2$. Once you know that $e in S$, you know by the first part of your proof that $A_1$ is surjective, so for some $s in S, a_1s=e$. Then $s=a^{-1}$.



                Another way to prove this result, by the way, is to just take all possible powers of $a_1$, which all are contained within $S$ because $S$ is multiplicatively closed. They have to repeat at some point because $S$ is finite. If $a^m=a^{m+k}=a^ma^k$, then $a^k=e in S$, and $aa^{k-1}=a^k=e Rightarrow a^{k-1}=a^{-1} in S$.



                As for proof-writing, one small nit. I don't think the line $A_1(a_1) = a_1$ conveys your intended meaning. I think you mean to say $a_1 in A_1(S)$.






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  You don't need to use the second map $A_2$. Once you know that $e in S$, you know by the first part of your proof that $A_1$ is surjective, so for some $s in S, a_1s=e$. Then $s=a^{-1}$.



                  Another way to prove this result, by the way, is to just take all possible powers of $a_1$, which all are contained within $S$ because $S$ is multiplicatively closed. They have to repeat at some point because $S$ is finite. If $a^m=a^{m+k}=a^ma^k$, then $a^k=e in S$, and $aa^{k-1}=a^k=e Rightarrow a^{k-1}=a^{-1} in S$.



                  As for proof-writing, one small nit. I don't think the line $A_1(a_1) = a_1$ conveys your intended meaning. I think you mean to say $a_1 in A_1(S)$.






                  share|cite|improve this answer









                  $endgroup$



                  You don't need to use the second map $A_2$. Once you know that $e in S$, you know by the first part of your proof that $A_1$ is surjective, so for some $s in S, a_1s=e$. Then $s=a^{-1}$.



                  Another way to prove this result, by the way, is to just take all possible powers of $a_1$, which all are contained within $S$ because $S$ is multiplicatively closed. They have to repeat at some point because $S$ is finite. If $a^m=a^{m+k}=a^ma^k$, then $a^k=e in S$, and $aa^{k-1}=a^k=e Rightarrow a^{k-1}=a^{-1} in S$.



                  As for proof-writing, one small nit. I don't think the line $A_1(a_1) = a_1$ conveys your intended meaning. I think you mean to say $a_1 in A_1(S)$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  Robert ShoreRobert Shore

                  1,47915




                  1,47915























                      2












                      $begingroup$

                      The proof presented here by our OP ForIgreaterthan_I appears to be logically valid; indeed, I like it because it is clever and innovative.



                      The way I would approach this, which alas is standard and not in fact overly clever, being as it is indeed rather obvious and straightforward, is as follows:



                      Let



                      $s in S; tag 1$



                      consider the powers



                      $s^i in S; tag 2$



                      since



                      $S subset G tag 3$



                      and



                      $vert G vert < infty, tag 4$



                      we have



                      $vert S vert < infty tag 5$



                      as well; thus the sequence



                      $s, s^2, s^3, ldots, s^i, s^{i + 1}, ldots tag 6$



                      must repeat itself at some point; that is,



                      $exists k, l in Bbb N, ; l ge k + 1, tag 7$



                      with



                      $s^l = s^k; tag 8$



                      then



                      $s^{l - k} = e; tag 9$



                      thus



                      $e = s^{l - k} in S; tag{10}$



                      it follows then that



                      $s^{l - k - 1}s = e tag{11}$



                      and clearly



                      $s^{l - k - 1} in S; tag{12}$



                      thus the group identity $e in S$, and every $s in S$ has an inverse in $S$; $S$ is thus a subgroup of $G$.






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        The proof presented here by our OP ForIgreaterthan_I appears to be logically valid; indeed, I like it because it is clever and innovative.



                        The way I would approach this, which alas is standard and not in fact overly clever, being as it is indeed rather obvious and straightforward, is as follows:



                        Let



                        $s in S; tag 1$



                        consider the powers



                        $s^i in S; tag 2$



                        since



                        $S subset G tag 3$



                        and



                        $vert G vert < infty, tag 4$



                        we have



                        $vert S vert < infty tag 5$



                        as well; thus the sequence



                        $s, s^2, s^3, ldots, s^i, s^{i + 1}, ldots tag 6$



                        must repeat itself at some point; that is,



                        $exists k, l in Bbb N, ; l ge k + 1, tag 7$



                        with



                        $s^l = s^k; tag 8$



                        then



                        $s^{l - k} = e; tag 9$



                        thus



                        $e = s^{l - k} in S; tag{10}$



                        it follows then that



                        $s^{l - k - 1}s = e tag{11}$



                        and clearly



                        $s^{l - k - 1} in S; tag{12}$



                        thus the group identity $e in S$, and every $s in S$ has an inverse in $S$; $S$ is thus a subgroup of $G$.






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          The proof presented here by our OP ForIgreaterthan_I appears to be logically valid; indeed, I like it because it is clever and innovative.



                          The way I would approach this, which alas is standard and not in fact overly clever, being as it is indeed rather obvious and straightforward, is as follows:



                          Let



                          $s in S; tag 1$



                          consider the powers



                          $s^i in S; tag 2$



                          since



                          $S subset G tag 3$



                          and



                          $vert G vert < infty, tag 4$



                          we have



                          $vert S vert < infty tag 5$



                          as well; thus the sequence



                          $s, s^2, s^3, ldots, s^i, s^{i + 1}, ldots tag 6$



                          must repeat itself at some point; that is,



                          $exists k, l in Bbb N, ; l ge k + 1, tag 7$



                          with



                          $s^l = s^k; tag 8$



                          then



                          $s^{l - k} = e; tag 9$



                          thus



                          $e = s^{l - k} in S; tag{10}$



                          it follows then that



                          $s^{l - k - 1}s = e tag{11}$



                          and clearly



                          $s^{l - k - 1} in S; tag{12}$



                          thus the group identity $e in S$, and every $s in S$ has an inverse in $S$; $S$ is thus a subgroup of $G$.






                          share|cite|improve this answer









                          $endgroup$



                          The proof presented here by our OP ForIgreaterthan_I appears to be logically valid; indeed, I like it because it is clever and innovative.



                          The way I would approach this, which alas is standard and not in fact overly clever, being as it is indeed rather obvious and straightforward, is as follows:



                          Let



                          $s in S; tag 1$



                          consider the powers



                          $s^i in S; tag 2$



                          since



                          $S subset G tag 3$



                          and



                          $vert G vert < infty, tag 4$



                          we have



                          $vert S vert < infty tag 5$



                          as well; thus the sequence



                          $s, s^2, s^3, ldots, s^i, s^{i + 1}, ldots tag 6$



                          must repeat itself at some point; that is,



                          $exists k, l in Bbb N, ; l ge k + 1, tag 7$



                          with



                          $s^l = s^k; tag 8$



                          then



                          $s^{l - k} = e; tag 9$



                          thus



                          $e = s^{l - k} in S; tag{10}$



                          it follows then that



                          $s^{l - k - 1}s = e tag{11}$



                          and clearly



                          $s^{l - k - 1} in S; tag{12}$



                          thus the group identity $e in S$, and every $s in S$ has an inverse in $S$; $S$ is thus a subgroup of $G$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 6 hours ago









                          Robert LewisRobert Lewis

                          46.9k23067




                          46.9k23067






















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