$A_n$ is the only subgroup of $S_n$ of index $2$.












25












$begingroup$


How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?



Why isn't there other possibility?



Thanks :)






group-theory finite-groups permutations symmetric-groups







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
    $endgroup$
    – Mariano Suárez-Álvarez
    Mar 14 '11 at 21:00










  • $begingroup$
    I am terribly sorry. But I don't know how to reedit it.
    $endgroup$
    – ShinyaSakai
    Mar 14 '11 at 21:07






  • 4




    $begingroup$
    There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
    $endgroup$
    – Arturo Magidin
    Mar 14 '11 at 21:11
















25












$begingroup$


How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?



Why isn't there other possibility?



Thanks :)






group-theory finite-groups permutations symmetric-groups







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
    $endgroup$
    – Mariano Suárez-Álvarez
    Mar 14 '11 at 21:00










  • $begingroup$
    I am terribly sorry. But I don't know how to reedit it.
    $endgroup$
    – ShinyaSakai
    Mar 14 '11 at 21:07






  • 4




    $begingroup$
    There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
    $endgroup$
    – Arturo Magidin
    Mar 14 '11 at 21:11














25












25








25


17



$begingroup$


How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?



Why isn't there other possibility?



Thanks :)






group-theory finite-groups permutations symmetric-groups







share|cite|improve this question











$endgroup$




How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?



Why isn't there other possibility?



Thanks :)






group-theory finite-groups permutations symmetric-groups




group-theory finite-groups permutations symmetric-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 '16 at 10:22









mrs

1




1










asked Mar 14 '11 at 20:57









ShinyaSakaiShinyaSakai

3,01012550




3,01012550








  • 4




    $begingroup$
    Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
    $endgroup$
    – Mariano Suárez-Álvarez
    Mar 14 '11 at 21:00










  • $begingroup$
    I am terribly sorry. But I don't know how to reedit it.
    $endgroup$
    – ShinyaSakai
    Mar 14 '11 at 21:07






  • 4




    $begingroup$
    There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
    $endgroup$
    – Arturo Magidin
    Mar 14 '11 at 21:11














  • 4




    $begingroup$
    Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
    $endgroup$
    – Mariano Suárez-Álvarez
    Mar 14 '11 at 21:00










  • $begingroup$
    I am terribly sorry. But I don't know how to reedit it.
    $endgroup$
    – ShinyaSakai
    Mar 14 '11 at 21:07






  • 4




    $begingroup$
    There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
    $endgroup$
    – Arturo Magidin
    Mar 14 '11 at 21:11








4




4




$begingroup$
Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
$endgroup$
– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00




$begingroup$
Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
$endgroup$
– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00












$begingroup$
I am terribly sorry. But I don't know how to reedit it.
$endgroup$
– ShinyaSakai
Mar 14 '11 at 21:07




$begingroup$
I am terribly sorry. But I don't know how to reedit it.
$endgroup$
– ShinyaSakai
Mar 14 '11 at 21:07




4




4




$begingroup$
There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
$endgroup$
– Arturo Magidin
Mar 14 '11 at 21:11




$begingroup$
There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
$endgroup$
– Arturo Magidin
Mar 14 '11 at 21:11










4 Answers
4






active

oldest

votes


















31












$begingroup$

As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Very detailed~ thank you very much~
    $endgroup$
    – ShinyaSakai
    Mar 15 '11 at 21:28










  • $begingroup$
    $S_n$ in the last line, not $S_2$. Very nice solution.
    $endgroup$
    – ReverseFlow
    Sep 7 '14 at 23:41










  • $begingroup$
    @Genomeme: thanks, corrected
    $endgroup$
    – user8268
    Sep 8 '14 at 21:14






  • 1




    $begingroup$
    @user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
    $endgroup$
    – jstnchng
    Nov 30 '14 at 18:49





















12












$begingroup$

subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
    $endgroup$
    – ShinyaSakai
    Mar 15 '11 at 21:30






  • 6




    $begingroup$
    How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
    $endgroup$
    – Vladhagen
    Oct 30 '13 at 19:42



















3












$begingroup$

Other Way :



$A_n$ is generated by all $3$-cycles in $S_n$.



If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$



WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:




    Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.




    Keeping that in mind, we first prove the two easy facts about subgroups of index $2$




    Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.




    Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.




    Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.




    Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.



    From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How do 3-cycles have an odd order? Or were you referring to something else?
      $endgroup$
      – Tomás Palamás
      Oct 30 '18 at 20:14











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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    31












    $begingroup$

    As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Very detailed~ thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:28










    • $begingroup$
      $S_n$ in the last line, not $S_2$. Very nice solution.
      $endgroup$
      – ReverseFlow
      Sep 7 '14 at 23:41










    • $begingroup$
      @Genomeme: thanks, corrected
      $endgroup$
      – user8268
      Sep 8 '14 at 21:14






    • 1




      $begingroup$
      @user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
      $endgroup$
      – jstnchng
      Nov 30 '14 at 18:49


















    31












    $begingroup$

    As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Very detailed~ thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:28










    • $begingroup$
      $S_n$ in the last line, not $S_2$. Very nice solution.
      $endgroup$
      – ReverseFlow
      Sep 7 '14 at 23:41










    • $begingroup$
      @Genomeme: thanks, corrected
      $endgroup$
      – user8268
      Sep 8 '14 at 21:14






    • 1




      $begingroup$
      @user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
      $endgroup$
      – jstnchng
      Nov 30 '14 at 18:49
















    31












    31








    31





    $begingroup$

    As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.






    share|cite|improve this answer











    $endgroup$



    As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 8 '14 at 21:14

























    answered Mar 14 '11 at 23:01









    user8268user8268

    16.9k12747




    16.9k12747












    • $begingroup$
      Very detailed~ thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:28










    • $begingroup$
      $S_n$ in the last line, not $S_2$. Very nice solution.
      $endgroup$
      – ReverseFlow
      Sep 7 '14 at 23:41










    • $begingroup$
      @Genomeme: thanks, corrected
      $endgroup$
      – user8268
      Sep 8 '14 at 21:14






    • 1




      $begingroup$
      @user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
      $endgroup$
      – jstnchng
      Nov 30 '14 at 18:49




















    • $begingroup$
      Very detailed~ thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:28










    • $begingroup$
      $S_n$ in the last line, not $S_2$. Very nice solution.
      $endgroup$
      – ReverseFlow
      Sep 7 '14 at 23:41










    • $begingroup$
      @Genomeme: thanks, corrected
      $endgroup$
      – user8268
      Sep 8 '14 at 21:14






    • 1




      $begingroup$
      @user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
      $endgroup$
      – jstnchng
      Nov 30 '14 at 18:49


















    $begingroup$
    Very detailed~ thank you very much~
    $endgroup$
    – ShinyaSakai
    Mar 15 '11 at 21:28




    $begingroup$
    Very detailed~ thank you very much~
    $endgroup$
    – ShinyaSakai
    Mar 15 '11 at 21:28












    $begingroup$
    $S_n$ in the last line, not $S_2$. Very nice solution.
    $endgroup$
    – ReverseFlow
    Sep 7 '14 at 23:41




    $begingroup$
    $S_n$ in the last line, not $S_2$. Very nice solution.
    $endgroup$
    – ReverseFlow
    Sep 7 '14 at 23:41












    $begingroup$
    @Genomeme: thanks, corrected
    $endgroup$
    – user8268
    Sep 8 '14 at 21:14




    $begingroup$
    @Genomeme: thanks, corrected
    $endgroup$
    – user8268
    Sep 8 '14 at 21:14




    1




    1




    $begingroup$
    @user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
    $endgroup$
    – jstnchng
    Nov 30 '14 at 18:49






    $begingroup$
    @user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
    $endgroup$
    – jstnchng
    Nov 30 '14 at 18:49













    12












    $begingroup$

    subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:30






    • 6




      $begingroup$
      How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
      $endgroup$
      – Vladhagen
      Oct 30 '13 at 19:42
















    12












    $begingroup$

    subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:30






    • 6




      $begingroup$
      How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
      $endgroup$
      – Vladhagen
      Oct 30 '13 at 19:42














    12












    12








    12





    $begingroup$

    subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.






    share|cite|improve this answer









    $endgroup$



    subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 14 '11 at 21:08









    yoyoyoyo

    6,5991726




    6,5991726












    • $begingroup$
      It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:30






    • 6




      $begingroup$
      How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
      $endgroup$
      – Vladhagen
      Oct 30 '13 at 19:42


















    • $begingroup$
      It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
      $endgroup$
      – ShinyaSakai
      Mar 15 '11 at 21:30






    • 6




      $begingroup$
      How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
      $endgroup$
      – Vladhagen
      Oct 30 '13 at 19:42
















    $begingroup$
    It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
    $endgroup$
    – ShinyaSakai
    Mar 15 '11 at 21:30




    $begingroup$
    It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
    $endgroup$
    – ShinyaSakai
    Mar 15 '11 at 21:30




    6




    6




    $begingroup$
    How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
    $endgroup$
    – Vladhagen
    Oct 30 '13 at 19:42




    $begingroup$
    How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
    $endgroup$
    – Vladhagen
    Oct 30 '13 at 19:42











    3












    $begingroup$

    Other Way :



    $A_n$ is generated by all $3$-cycles in $S_n$.



    If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$



    WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Other Way :



      $A_n$ is generated by all $3$-cycles in $S_n$.



      If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$



      WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Other Way :



        $A_n$ is generated by all $3$-cycles in $S_n$.



        If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$



        WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.






        share|cite|improve this answer











        $endgroup$



        Other Way :



        $A_n$ is generated by all $3$-cycles in $S_n$.



        If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$



        WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 19:22









        Namaste

        1




        1










        answered Nov 21 '18 at 15:24









        SRJSRJ

        1,8331620




        1,8331620























            2












            $begingroup$

            I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:




            Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.




            Keeping that in mind, we first prove the two easy facts about subgroups of index $2$




            Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.




            Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.




            Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.




            Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.



            From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How do 3-cycles have an odd order? Or were you referring to something else?
              $endgroup$
              – Tomás Palamás
              Oct 30 '18 at 20:14
















            2












            $begingroup$

            I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:




            Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.




            Keeping that in mind, we first prove the two easy facts about subgroups of index $2$




            Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.




            Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.




            Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.




            Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.



            From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              How do 3-cycles have an odd order? Or were you referring to something else?
              $endgroup$
              – Tomás Palamás
              Oct 30 '18 at 20:14














            2












            2








            2





            $begingroup$

            I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:




            Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.




            Keeping that in mind, we first prove the two easy facts about subgroups of index $2$




            Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.




            Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.




            Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.




            Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.



            From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.






            share|cite|improve this answer











            $endgroup$



            I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:




            Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.




            Keeping that in mind, we first prove the two easy facts about subgroups of index $2$




            Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.




            Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.




            Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.




            Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.



            From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 3 '17 at 18:23

























            answered May 3 '17 at 18:09









            user193319user193319

            2,4362927




            2,4362927












            • $begingroup$
              How do 3-cycles have an odd order? Or were you referring to something else?
              $endgroup$
              – Tomás Palamás
              Oct 30 '18 at 20:14


















            • $begingroup$
              How do 3-cycles have an odd order? Or were you referring to something else?
              $endgroup$
              – Tomás Palamás
              Oct 30 '18 at 20:14
















            $begingroup$
            How do 3-cycles have an odd order? Or were you referring to something else?
            $endgroup$
            – Tomás Palamás
            Oct 30 '18 at 20:14




            $begingroup$
            How do 3-cycles have an odd order? Or were you referring to something else?
            $endgroup$
            – Tomás Palamás
            Oct 30 '18 at 20:14


















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