How is it possible that $textrm{HK}=textrm{G}$?












1












$begingroup$


For the following problem:




If $textrm{H}$ and $textrm{K}$ are distinct subgroups of $textrm{G}$ of index $2$, then $textrm{H}captextrm{K}$ is a normal subgroup of $textrm{G}$ of index $4$.




It is obvious that both $textrm{H}$ and $textrm{K}$ are normal subgroups since both have index $2$. Hence their intersection $textrm{H}captextrm{K}$ is also normal.



The main issue comes with proving that it has an index $4$, using 2nd Isomorphism Theorem. Clearly $displaystylefrac{textrm{H}}{textrm{H}captextrm{K}}$ is isomorphic to $displaystylefrac{textrm{HK}}{textrm{K}}$ and using this we can show that $textrm{H}captextrm{K}$ that index $4$. But on this forum, I have seen another proof of this problem where a person wrote that $textrm{H}textrm{K} = textrm{G}$.



How the heck are they equal? I'm having extreme difficulty in proving this.



Has this got something to do with both $textrm{H}$ and $textrm{K}$ being distinct ?






abstract-algebra group-theory normal-subgroups group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    $H$ and $K$ are maximal subgroups of $G$. This follows from Lagarange's theorem. Since $H subset HK$, it follows that $HK=G$.
    $endgroup$
    – Crostul
    Dec 30 '18 at 11:25










  • $begingroup$
    An addendum to Crostul's answer: $H$ and $K$ distinct implies $HK not = H$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 11:28






  • 1




    $begingroup$
    I suggest you use a less informal title, as it may scare off certain people.
    $endgroup$
    – Math_QED
    Dec 30 '18 at 11:33






  • 2




    $begingroup$
    @Crostul You're invoking Lagrange. Is $G$ assumed finite?
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 11:35








  • 1




    $begingroup$
    Thank you, @Derek Holt, for editing the title. Minto, please avoid language you wouldn't use in front of your professor.
    $endgroup$
    – Ennar
    Dec 30 '18 at 11:46
















1












$begingroup$


For the following problem:




If $textrm{H}$ and $textrm{K}$ are distinct subgroups of $textrm{G}$ of index $2$, then $textrm{H}captextrm{K}$ is a normal subgroup of $textrm{G}$ of index $4$.




It is obvious that both $textrm{H}$ and $textrm{K}$ are normal subgroups since both have index $2$. Hence their intersection $textrm{H}captextrm{K}$ is also normal.



The main issue comes with proving that it has an index $4$, using 2nd Isomorphism Theorem. Clearly $displaystylefrac{textrm{H}}{textrm{H}captextrm{K}}$ is isomorphic to $displaystylefrac{textrm{HK}}{textrm{K}}$ and using this we can show that $textrm{H}captextrm{K}$ that index $4$. But on this forum, I have seen another proof of this problem where a person wrote that $textrm{H}textrm{K} = textrm{G}$.



How the heck are they equal? I'm having extreme difficulty in proving this.



Has this got something to do with both $textrm{H}$ and $textrm{K}$ being distinct ?






abstract-algebra group-theory normal-subgroups group-homomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    $H$ and $K$ are maximal subgroups of $G$. This follows from Lagarange's theorem. Since $H subset HK$, it follows that $HK=G$.
    $endgroup$
    – Crostul
    Dec 30 '18 at 11:25










  • $begingroup$
    An addendum to Crostul's answer: $H$ and $K$ distinct implies $HK not = H$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 11:28






  • 1




    $begingroup$
    I suggest you use a less informal title, as it may scare off certain people.
    $endgroup$
    – Math_QED
    Dec 30 '18 at 11:33






  • 2




    $begingroup$
    @Crostul You're invoking Lagrange. Is $G$ assumed finite?
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 11:35








  • 1




    $begingroup$
    Thank you, @Derek Holt, for editing the title. Minto, please avoid language you wouldn't use in front of your professor.
    $endgroup$
    – Ennar
    Dec 30 '18 at 11:46














1












1








1


1



$begingroup$


For the following problem:




If $textrm{H}$ and $textrm{K}$ are distinct subgroups of $textrm{G}$ of index $2$, then $textrm{H}captextrm{K}$ is a normal subgroup of $textrm{G}$ of index $4$.




It is obvious that both $textrm{H}$ and $textrm{K}$ are normal subgroups since both have index $2$. Hence their intersection $textrm{H}captextrm{K}$ is also normal.



The main issue comes with proving that it has an index $4$, using 2nd Isomorphism Theorem. Clearly $displaystylefrac{textrm{H}}{textrm{H}captextrm{K}}$ is isomorphic to $displaystylefrac{textrm{HK}}{textrm{K}}$ and using this we can show that $textrm{H}captextrm{K}$ that index $4$. But on this forum, I have seen another proof of this problem where a person wrote that $textrm{H}textrm{K} = textrm{G}$.



How the heck are they equal? I'm having extreme difficulty in proving this.



Has this got something to do with both $textrm{H}$ and $textrm{K}$ being distinct ?






abstract-algebra group-theory normal-subgroups group-homomorphism







share|cite|improve this question











$endgroup$




For the following problem:




If $textrm{H}$ and $textrm{K}$ are distinct subgroups of $textrm{G}$ of index $2$, then $textrm{H}captextrm{K}$ is a normal subgroup of $textrm{G}$ of index $4$.




It is obvious that both $textrm{H}$ and $textrm{K}$ are normal subgroups since both have index $2$. Hence their intersection $textrm{H}captextrm{K}$ is also normal.



The main issue comes with proving that it has an index $4$, using 2nd Isomorphism Theorem. Clearly $displaystylefrac{textrm{H}}{textrm{H}captextrm{K}}$ is isomorphic to $displaystylefrac{textrm{HK}}{textrm{K}}$ and using this we can show that $textrm{H}captextrm{K}$ that index $4$. But on this forum, I have seen another proof of this problem where a person wrote that $textrm{H}textrm{K} = textrm{G}$.



How the heck are they equal? I'm having extreme difficulty in proving this.



Has this got something to do with both $textrm{H}$ and $textrm{K}$ being distinct ?






abstract-algebra group-theory normal-subgroups group-homomorphism




abstract-algebra group-theory normal-subgroups group-homomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 11:42









Derek Holt

52.8k53570




52.8k53570










asked Dec 30 '18 at 11:22









Minto PMinto P

415




415












  • $begingroup$
    $H$ and $K$ are maximal subgroups of $G$. This follows from Lagarange's theorem. Since $H subset HK$, it follows that $HK=G$.
    $endgroup$
    – Crostul
    Dec 30 '18 at 11:25










  • $begingroup$
    An addendum to Crostul's answer: $H$ and $K$ distinct implies $HK not = H$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 11:28






  • 1




    $begingroup$
    I suggest you use a less informal title, as it may scare off certain people.
    $endgroup$
    – Math_QED
    Dec 30 '18 at 11:33






  • 2




    $begingroup$
    @Crostul You're invoking Lagrange. Is $G$ assumed finite?
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 11:35








  • 1




    $begingroup$
    Thank you, @Derek Holt, for editing the title. Minto, please avoid language you wouldn't use in front of your professor.
    $endgroup$
    – Ennar
    Dec 30 '18 at 11:46


















  • $begingroup$
    $H$ and $K$ are maximal subgroups of $G$. This follows from Lagarange's theorem. Since $H subset HK$, it follows that $HK=G$.
    $endgroup$
    – Crostul
    Dec 30 '18 at 11:25










  • $begingroup$
    An addendum to Crostul's answer: $H$ and $K$ distinct implies $HK not = H$.
    $endgroup$
    – mathworker21
    Dec 30 '18 at 11:28






  • 1




    $begingroup$
    I suggest you use a less informal title, as it may scare off certain people.
    $endgroup$
    – Math_QED
    Dec 30 '18 at 11:33






  • 2




    $begingroup$
    @Crostul You're invoking Lagrange. Is $G$ assumed finite?
    $endgroup$
    – MathematicsStudent1122
    Dec 30 '18 at 11:35








  • 1




    $begingroup$
    Thank you, @Derek Holt, for editing the title. Minto, please avoid language you wouldn't use in front of your professor.
    $endgroup$
    – Ennar
    Dec 30 '18 at 11:46
















$begingroup$
$H$ and $K$ are maximal subgroups of $G$. This follows from Lagarange's theorem. Since $H subset HK$, it follows that $HK=G$.
$endgroup$
– Crostul
Dec 30 '18 at 11:25




$begingroup$
$H$ and $K$ are maximal subgroups of $G$. This follows from Lagarange's theorem. Since $H subset HK$, it follows that $HK=G$.
$endgroup$
– Crostul
Dec 30 '18 at 11:25












$begingroup$
An addendum to Crostul's answer: $H$ and $K$ distinct implies $HK not = H$.
$endgroup$
– mathworker21
Dec 30 '18 at 11:28




$begingroup$
An addendum to Crostul's answer: $H$ and $K$ distinct implies $HK not = H$.
$endgroup$
– mathworker21
Dec 30 '18 at 11:28




1




1




$begingroup$
I suggest you use a less informal title, as it may scare off certain people.
$endgroup$
– Math_QED
Dec 30 '18 at 11:33




$begingroup$
I suggest you use a less informal title, as it may scare off certain people.
$endgroup$
– Math_QED
Dec 30 '18 at 11:33




2




2




$begingroup$
@Crostul You're invoking Lagrange. Is $G$ assumed finite?
$endgroup$
– MathematicsStudent1122
Dec 30 '18 at 11:35






$begingroup$
@Crostul You're invoking Lagrange. Is $G$ assumed finite?
$endgroup$
– MathematicsStudent1122
Dec 30 '18 at 11:35






1




1




$begingroup$
Thank you, @Derek Holt, for editing the title. Minto, please avoid language you wouldn't use in front of your professor.
$endgroup$
– Ennar
Dec 30 '18 at 11:46




$begingroup$
Thank you, @Derek Holt, for editing the title. Minto, please avoid language you wouldn't use in front of your professor.
$endgroup$
– Ennar
Dec 30 '18 at 11:46










3 Answers
3






active

oldest

votes


















2












$begingroup$

We first remark, in general, if $G$ is a group (not necessarily finite) and $H$ is a subgroup of index $2$, $H$ must be normal. This is because the left and right cosets of $H$ necessarily coincide: there is one and only one partition of $G$ into $2$ parts such that one of the parts is $H$.



It follows that for this problem, $H$, $K$, and $H cap K$ are all normal. In particular, this means $HK$ is a subgroup. It suffices to show that $H cap K$ has index $2$ in $H$. This will imply that $H cap K$ has index $4$ in $G$.



Observe that $HK = G$, since $H, K$ are distinct maximal subgroups. Moreover, by one of the isomorphism theorems, we have that $H/(H cap K) cong (HK)/K cong G/K$. Hence $H cap K$ has index $2$ in $H$, as desired.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good answer! This one is clearest to me so far; @Bernard's, most elegant.
    $endgroup$
    – Shaun
    Dec 30 '18 at 12:01



















2












$begingroup$

It has to do with $H$ and $K$ being distinct and having index $2$: these hypotheses imply that none of them is contained in the other. So take an element $hin Hsmallsetminus K$. Then $;G=hKcup K$ since $K$ has index $2$. Also, since both $hK,Ksubset HK$, we have $G=HK$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $H,K$ have index $2$. This means that:




    1. $H,K$ are normal.


    2. $H,K$ are maximal.
      Indeed, let $M$ be a proper subgroup of $G$ containing $H$. Then $M/H$ is a proper subgroup of $G/H$. Since
      $G/H$ is the group with two elements, it follows that $M/H$ is trivial, i.e. $M=H$. The same holds for $K$.


    3. By maximality of $H$ and $K$, you have that either $HK=G$ or $HK=H$ or $HK=K$. But the latter two can hold only for $H=HK=K$ (and this contradicts the hypothesis).







    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I like the style & format of this answer.
      $endgroup$
      – Shaun
      Dec 30 '18 at 12:01











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    We first remark, in general, if $G$ is a group (not necessarily finite) and $H$ is a subgroup of index $2$, $H$ must be normal. This is because the left and right cosets of $H$ necessarily coincide: there is one and only one partition of $G$ into $2$ parts such that one of the parts is $H$.



    It follows that for this problem, $H$, $K$, and $H cap K$ are all normal. In particular, this means $HK$ is a subgroup. It suffices to show that $H cap K$ has index $2$ in $H$. This will imply that $H cap K$ has index $4$ in $G$.



    Observe that $HK = G$, since $H, K$ are distinct maximal subgroups. Moreover, by one of the isomorphism theorems, we have that $H/(H cap K) cong (HK)/K cong G/K$. Hence $H cap K$ has index $2$ in $H$, as desired.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Good answer! This one is clearest to me so far; @Bernard's, most elegant.
      $endgroup$
      – Shaun
      Dec 30 '18 at 12:01
















    2












    $begingroup$

    We first remark, in general, if $G$ is a group (not necessarily finite) and $H$ is a subgroup of index $2$, $H$ must be normal. This is because the left and right cosets of $H$ necessarily coincide: there is one and only one partition of $G$ into $2$ parts such that one of the parts is $H$.



    It follows that for this problem, $H$, $K$, and $H cap K$ are all normal. In particular, this means $HK$ is a subgroup. It suffices to show that $H cap K$ has index $2$ in $H$. This will imply that $H cap K$ has index $4$ in $G$.



    Observe that $HK = G$, since $H, K$ are distinct maximal subgroups. Moreover, by one of the isomorphism theorems, we have that $H/(H cap K) cong (HK)/K cong G/K$. Hence $H cap K$ has index $2$ in $H$, as desired.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Good answer! This one is clearest to me so far; @Bernard's, most elegant.
      $endgroup$
      – Shaun
      Dec 30 '18 at 12:01














    2












    2








    2





    $begingroup$

    We first remark, in general, if $G$ is a group (not necessarily finite) and $H$ is a subgroup of index $2$, $H$ must be normal. This is because the left and right cosets of $H$ necessarily coincide: there is one and only one partition of $G$ into $2$ parts such that one of the parts is $H$.



    It follows that for this problem, $H$, $K$, and $H cap K$ are all normal. In particular, this means $HK$ is a subgroup. It suffices to show that $H cap K$ has index $2$ in $H$. This will imply that $H cap K$ has index $4$ in $G$.



    Observe that $HK = G$, since $H, K$ are distinct maximal subgroups. Moreover, by one of the isomorphism theorems, we have that $H/(H cap K) cong (HK)/K cong G/K$. Hence $H cap K$ has index $2$ in $H$, as desired.






    share|cite|improve this answer









    $endgroup$



    We first remark, in general, if $G$ is a group (not necessarily finite) and $H$ is a subgroup of index $2$, $H$ must be normal. This is because the left and right cosets of $H$ necessarily coincide: there is one and only one partition of $G$ into $2$ parts such that one of the parts is $H$.



    It follows that for this problem, $H$, $K$, and $H cap K$ are all normal. In particular, this means $HK$ is a subgroup. It suffices to show that $H cap K$ has index $2$ in $H$. This will imply that $H cap K$ has index $4$ in $G$.



    Observe that $HK = G$, since $H, K$ are distinct maximal subgroups. Moreover, by one of the isomorphism theorems, we have that $H/(H cap K) cong (HK)/K cong G/K$. Hence $H cap K$ has index $2$ in $H$, as desired.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 30 '18 at 11:54









    MathematicsStudent1122MathematicsStudent1122

    8,56822466




    8,56822466












    • $begingroup$
      Good answer! This one is clearest to me so far; @Bernard's, most elegant.
      $endgroup$
      – Shaun
      Dec 30 '18 at 12:01


















    • $begingroup$
      Good answer! This one is clearest to me so far; @Bernard's, most elegant.
      $endgroup$
      – Shaun
      Dec 30 '18 at 12:01
















    $begingroup$
    Good answer! This one is clearest to me so far; @Bernard's, most elegant.
    $endgroup$
    – Shaun
    Dec 30 '18 at 12:01




    $begingroup$
    Good answer! This one is clearest to me so far; @Bernard's, most elegant.
    $endgroup$
    – Shaun
    Dec 30 '18 at 12:01











    2












    $begingroup$

    It has to do with $H$ and $K$ being distinct and having index $2$: these hypotheses imply that none of them is contained in the other. So take an element $hin Hsmallsetminus K$. Then $;G=hKcup K$ since $K$ has index $2$. Also, since both $hK,Ksubset HK$, we have $G=HK$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      It has to do with $H$ and $K$ being distinct and having index $2$: these hypotheses imply that none of them is contained in the other. So take an element $hin Hsmallsetminus K$. Then $;G=hKcup K$ since $K$ has index $2$. Also, since both $hK,Ksubset HK$, we have $G=HK$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        It has to do with $H$ and $K$ being distinct and having index $2$: these hypotheses imply that none of them is contained in the other. So take an element $hin Hsmallsetminus K$. Then $;G=hKcup K$ since $K$ has index $2$. Also, since both $hK,Ksubset HK$, we have $G=HK$.






        share|cite|improve this answer











        $endgroup$



        It has to do with $H$ and $K$ being distinct and having index $2$: these hypotheses imply that none of them is contained in the other. So take an element $hin Hsmallsetminus K$. Then $;G=hKcup K$ since $K$ has index $2$. Also, since both $hK,Ksubset HK$, we have $G=HK$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 12:00

























        answered Dec 30 '18 at 11:46









        BernardBernard

        119k639112




        119k639112























            1












            $begingroup$

            $H,K$ have index $2$. This means that:




            1. $H,K$ are normal.


            2. $H,K$ are maximal.
              Indeed, let $M$ be a proper subgroup of $G$ containing $H$. Then $M/H$ is a proper subgroup of $G/H$. Since
              $G/H$ is the group with two elements, it follows that $M/H$ is trivial, i.e. $M=H$. The same holds for $K$.


            3. By maximality of $H$ and $K$, you have that either $HK=G$ or $HK=H$ or $HK=K$. But the latter two can hold only for $H=HK=K$ (and this contradicts the hypothesis).







            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I like the style & format of this answer.
              $endgroup$
              – Shaun
              Dec 30 '18 at 12:01
















            1












            $begingroup$

            $H,K$ have index $2$. This means that:




            1. $H,K$ are normal.


            2. $H,K$ are maximal.
              Indeed, let $M$ be a proper subgroup of $G$ containing $H$. Then $M/H$ is a proper subgroup of $G/H$. Since
              $G/H$ is the group with two elements, it follows that $M/H$ is trivial, i.e. $M=H$. The same holds for $K$.


            3. By maximality of $H$ and $K$, you have that either $HK=G$ or $HK=H$ or $HK=K$. But the latter two can hold only for $H=HK=K$ (and this contradicts the hypothesis).







            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I like the style & format of this answer.
              $endgroup$
              – Shaun
              Dec 30 '18 at 12:01














            1












            1








            1





            $begingroup$

            $H,K$ have index $2$. This means that:




            1. $H,K$ are normal.


            2. $H,K$ are maximal.
              Indeed, let $M$ be a proper subgroup of $G$ containing $H$. Then $M/H$ is a proper subgroup of $G/H$. Since
              $G/H$ is the group with two elements, it follows that $M/H$ is trivial, i.e. $M=H$. The same holds for $K$.


            3. By maximality of $H$ and $K$, you have that either $HK=G$ or $HK=H$ or $HK=K$. But the latter two can hold only for $H=HK=K$ (and this contradicts the hypothesis).







            share|cite|improve this answer









            $endgroup$



            $H,K$ have index $2$. This means that:




            1. $H,K$ are normal.


            2. $H,K$ are maximal.
              Indeed, let $M$ be a proper subgroup of $G$ containing $H$. Then $M/H$ is a proper subgroup of $G/H$. Since
              $G/H$ is the group with two elements, it follows that $M/H$ is trivial, i.e. $M=H$. The same holds for $K$.


            3. By maximality of $H$ and $K$, you have that either $HK=G$ or $HK=H$ or $HK=K$. But the latter two can hold only for $H=HK=K$ (and this contradicts the hypothesis).








            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 30 '18 at 11:52









            CrostulCrostul

            27.7k22352




            27.7k22352












            • $begingroup$
              I like the style & format of this answer.
              $endgroup$
              – Shaun
              Dec 30 '18 at 12:01


















            • $begingroup$
              I like the style & format of this answer.
              $endgroup$
              – Shaun
              Dec 30 '18 at 12:01
















            $begingroup$
            I like the style & format of this answer.
            $endgroup$
            – Shaun
            Dec 30 '18 at 12:01




            $begingroup$
            I like the style & format of this answer.
            $endgroup$
            – Shaun
            Dec 30 '18 at 12:01


















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