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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056734%2fhow-is-it-possible-that-textrmhk-textrmg%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3056734%2fhow-is-it-possible-that-textrmhk-textrmg%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
            Read more