$operatorname{Aut}(mathbb Q^*)$ =?












2














Group $mathbb Q^*$ is rationals without $0$ under multiplication.



From The group $mathbb Q^*$ as a direct product/sum we know
$mathbb Q^*cong mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z$.



So what is $text{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$?



Edit:



One of my classmates told me that every element in $operatorname{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$ is one-to-one correspond to ordered pairs $(theta,phi)$, $theta in operatorname{Aut}(bigoplus_{i=1}^{infty} mathbb Z)$, and $phi in operatorname{Hom}(bigoplus_{i=1}^{infty} mathbb Z, mathbb Z_2 )$.



All $phi $ is $bigotimes_{i=1}^{infty} mathbb Z_2$, but all $theta$ is rather complicated, is part of $ bigotimes_{i=1}^{infty} (bigoplus_{i=1}^{infty} mathbb Z)$, is all invertible elements in this ring: matrix in $mathbb Z^{infty times infty} $ with every column only finite matrix element nonzero.



Are these correct?






abstract-algebra group-theory







share|cite|improve this question





























    2














    Group $mathbb Q^*$ is rationals without $0$ under multiplication.



    From The group $mathbb Q^*$ as a direct product/sum we know
    $mathbb Q^*cong mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z$.



    So what is $text{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$?



    Edit:



    One of my classmates told me that every element in $operatorname{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$ is one-to-one correspond to ordered pairs $(theta,phi)$, $theta in operatorname{Aut}(bigoplus_{i=1}^{infty} mathbb Z)$, and $phi in operatorname{Hom}(bigoplus_{i=1}^{infty} mathbb Z, mathbb Z_2 )$.



    All $phi $ is $bigotimes_{i=1}^{infty} mathbb Z_2$, but all $theta$ is rather complicated, is part of $ bigotimes_{i=1}^{infty} (bigoplus_{i=1}^{infty} mathbb Z)$, is all invertible elements in this ring: matrix in $mathbb Z^{infty times infty} $ with every column only finite matrix element nonzero.



    Are these correct?






    abstract-algebra group-theory







    share|cite|improve this question



























      2












      2








      2


      1





      Group $mathbb Q^*$ is rationals without $0$ under multiplication.



      From The group $mathbb Q^*$ as a direct product/sum we know
      $mathbb Q^*cong mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z$.



      So what is $text{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$?



      Edit:



      One of my classmates told me that every element in $operatorname{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$ is one-to-one correspond to ordered pairs $(theta,phi)$, $theta in operatorname{Aut}(bigoplus_{i=1}^{infty} mathbb Z)$, and $phi in operatorname{Hom}(bigoplus_{i=1}^{infty} mathbb Z, mathbb Z_2 )$.



      All $phi $ is $bigotimes_{i=1}^{infty} mathbb Z_2$, but all $theta$ is rather complicated, is part of $ bigotimes_{i=1}^{infty} (bigoplus_{i=1}^{infty} mathbb Z)$, is all invertible elements in this ring: matrix in $mathbb Z^{infty times infty} $ with every column only finite matrix element nonzero.



      Are these correct?






      abstract-algebra group-theory







      share|cite|improve this question















      Group $mathbb Q^*$ is rationals without $0$ under multiplication.



      From The group $mathbb Q^*$ as a direct product/sum we know
      $mathbb Q^*cong mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z$.



      So what is $text{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$?



      Edit:



      One of my classmates told me that every element in $operatorname{Aut}(mathbb Z_2 times bigoplus_{i=1}^{infty} mathbb Z)$ is one-to-one correspond to ordered pairs $(theta,phi)$, $theta in operatorname{Aut}(bigoplus_{i=1}^{infty} mathbb Z)$, and $phi in operatorname{Hom}(bigoplus_{i=1}^{infty} mathbb Z, mathbb Z_2 )$.



      All $phi $ is $bigotimes_{i=1}^{infty} mathbb Z_2$, but all $theta$ is rather complicated, is part of $ bigotimes_{i=1}^{infty} (bigoplus_{i=1}^{infty} mathbb Z)$, is all invertible elements in this ring: matrix in $mathbb Z^{infty times infty} $ with every column only finite matrix element nonzero.



      Are these correct?






      abstract-algebra group-theory




      abstract-algebra group-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 at 22:30

























      asked Nov 25 at 14:43









      Andrews

      352317




      352317






















          1 Answer
          1






          active

          oldest

          votes


















          8














          If we consider instead $mathbb{Z}/2mathbb{Z}times F$, where $F$ is a free abelian group on a countable basis, the endomorphism ring can be represented as
          $$DeclareMathOperator{Hom}{Hom}DeclareMathOperator{End}{End}
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          Hom(mathbb{Z}/2mathbb{Z},F) & End(F)
          end{bmatrix}=
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          0 & End(F)
          end{bmatrix}
          $$
          An endomorphism is invertible if and only if both components in the diagonal are invertible, as it easy to show.



          Now, as rings, $End(mathbb{Z}/2mathbb{Z})=mathbb{Z}/2mathbb{Z}$, so an automorphism has $1$ in the top left corner.



          The group of automorphisms is therefore $Hom(F,mathbb{Z}/2mathbb{Z})timesoperatorname{Aut}(F)$ with the operation
          $$
          (f_1,g_1)(f_2,g_2)=(f_2+f_1g_2,g_1g_2)
          $$






          share|cite|improve this answer





















          • I think you are right.And in this case End($F$) is just GL$(infty,mathbb Z)$
            – Andrews
            Nov 25 at 15:35












          • @Andrews Yes, but that's not a much better description: column finite matrices indexed by the natural numbers which are invertible.
            – egreg
            Nov 25 at 15:41






          • 2




            I think $text{GL}(infty,mathbb{Z})$ has a different meaning. At least in my experience, it denotes the direct limit of $text{GL}(n,mathbb{Z})$ under the inclusions $text{GL}(n,mathbb{Z})hookrightarrow text{GL}(n+1,mathbb{Z})$ sending $$textbf{X}mapsto begin{bmatrix}textbf{X}&textbf{0}\textbf{0}&1end{bmatrix}$$ for each $mathbf{X}intext{GL}(n,mathbb{Z})$ (i.e, each element of $text{GL}(infty,mathbb{Z})$ is finitary). There are elements of $text{Aut}(mathbb{Z}^omega)$ that is not finitary.
            – Batominovski
            Nov 26 at 10:30













          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%2f3012919%2foperatornameaut-mathbb-q%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8














          If we consider instead $mathbb{Z}/2mathbb{Z}times F$, where $F$ is a free abelian group on a countable basis, the endomorphism ring can be represented as
          $$DeclareMathOperator{Hom}{Hom}DeclareMathOperator{End}{End}
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          Hom(mathbb{Z}/2mathbb{Z},F) & End(F)
          end{bmatrix}=
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          0 & End(F)
          end{bmatrix}
          $$
          An endomorphism is invertible if and only if both components in the diagonal are invertible, as it easy to show.



          Now, as rings, $End(mathbb{Z}/2mathbb{Z})=mathbb{Z}/2mathbb{Z}$, so an automorphism has $1$ in the top left corner.



          The group of automorphisms is therefore $Hom(F,mathbb{Z}/2mathbb{Z})timesoperatorname{Aut}(F)$ with the operation
          $$
          (f_1,g_1)(f_2,g_2)=(f_2+f_1g_2,g_1g_2)
          $$






          share|cite|improve this answer





















          • I think you are right.And in this case End($F$) is just GL$(infty,mathbb Z)$
            – Andrews
            Nov 25 at 15:35












          • @Andrews Yes, but that's not a much better description: column finite matrices indexed by the natural numbers which are invertible.
            – egreg
            Nov 25 at 15:41






          • 2




            I think $text{GL}(infty,mathbb{Z})$ has a different meaning. At least in my experience, it denotes the direct limit of $text{GL}(n,mathbb{Z})$ under the inclusions $text{GL}(n,mathbb{Z})hookrightarrow text{GL}(n+1,mathbb{Z})$ sending $$textbf{X}mapsto begin{bmatrix}textbf{X}&textbf{0}\textbf{0}&1end{bmatrix}$$ for each $mathbf{X}intext{GL}(n,mathbb{Z})$ (i.e, each element of $text{GL}(infty,mathbb{Z})$ is finitary). There are elements of $text{Aut}(mathbb{Z}^omega)$ that is not finitary.
            – Batominovski
            Nov 26 at 10:30


















          8














          If we consider instead $mathbb{Z}/2mathbb{Z}times F$, where $F$ is a free abelian group on a countable basis, the endomorphism ring can be represented as
          $$DeclareMathOperator{Hom}{Hom}DeclareMathOperator{End}{End}
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          Hom(mathbb{Z}/2mathbb{Z},F) & End(F)
          end{bmatrix}=
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          0 & End(F)
          end{bmatrix}
          $$
          An endomorphism is invertible if and only if both components in the diagonal are invertible, as it easy to show.



          Now, as rings, $End(mathbb{Z}/2mathbb{Z})=mathbb{Z}/2mathbb{Z}$, so an automorphism has $1$ in the top left corner.



          The group of automorphisms is therefore $Hom(F,mathbb{Z}/2mathbb{Z})timesoperatorname{Aut}(F)$ with the operation
          $$
          (f_1,g_1)(f_2,g_2)=(f_2+f_1g_2,g_1g_2)
          $$






          share|cite|improve this answer





















          • I think you are right.And in this case End($F$) is just GL$(infty,mathbb Z)$
            – Andrews
            Nov 25 at 15:35












          • @Andrews Yes, but that's not a much better description: column finite matrices indexed by the natural numbers which are invertible.
            – egreg
            Nov 25 at 15:41






          • 2




            I think $text{GL}(infty,mathbb{Z})$ has a different meaning. At least in my experience, it denotes the direct limit of $text{GL}(n,mathbb{Z})$ under the inclusions $text{GL}(n,mathbb{Z})hookrightarrow text{GL}(n+1,mathbb{Z})$ sending $$textbf{X}mapsto begin{bmatrix}textbf{X}&textbf{0}\textbf{0}&1end{bmatrix}$$ for each $mathbf{X}intext{GL}(n,mathbb{Z})$ (i.e, each element of $text{GL}(infty,mathbb{Z})$ is finitary). There are elements of $text{Aut}(mathbb{Z}^omega)$ that is not finitary.
            – Batominovski
            Nov 26 at 10:30
















          8












          8








          8






          If we consider instead $mathbb{Z}/2mathbb{Z}times F$, where $F$ is a free abelian group on a countable basis, the endomorphism ring can be represented as
          $$DeclareMathOperator{Hom}{Hom}DeclareMathOperator{End}{End}
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          Hom(mathbb{Z}/2mathbb{Z},F) & End(F)
          end{bmatrix}=
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          0 & End(F)
          end{bmatrix}
          $$
          An endomorphism is invertible if and only if both components in the diagonal are invertible, as it easy to show.



          Now, as rings, $End(mathbb{Z}/2mathbb{Z})=mathbb{Z}/2mathbb{Z}$, so an automorphism has $1$ in the top left corner.



          The group of automorphisms is therefore $Hom(F,mathbb{Z}/2mathbb{Z})timesoperatorname{Aut}(F)$ with the operation
          $$
          (f_1,g_1)(f_2,g_2)=(f_2+f_1g_2,g_1g_2)
          $$






          share|cite|improve this answer












          If we consider instead $mathbb{Z}/2mathbb{Z}times F$, where $F$ is a free abelian group on a countable basis, the endomorphism ring can be represented as
          $$DeclareMathOperator{Hom}{Hom}DeclareMathOperator{End}{End}
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          Hom(mathbb{Z}/2mathbb{Z},F) & End(F)
          end{bmatrix}=
          begin{bmatrix}
          End(mathbb{Z}/2mathbb{Z}) & Hom(F,mathbb{Z}/2mathbb{Z}) \
          0 & End(F)
          end{bmatrix}
          $$
          An endomorphism is invertible if and only if both components in the diagonal are invertible, as it easy to show.



          Now, as rings, $End(mathbb{Z}/2mathbb{Z})=mathbb{Z}/2mathbb{Z}$, so an automorphism has $1$ in the top left corner.



          The group of automorphisms is therefore $Hom(F,mathbb{Z}/2mathbb{Z})timesoperatorname{Aut}(F)$ with the operation
          $$
          (f_1,g_1)(f_2,g_2)=(f_2+f_1g_2,g_1g_2)
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 15:07









          egreg

          177k1484200




          177k1484200












          • I think you are right.And in this case End($F$) is just GL$(infty,mathbb Z)$
            – Andrews
            Nov 25 at 15:35












          • @Andrews Yes, but that's not a much better description: column finite matrices indexed by the natural numbers which are invertible.
            – egreg
            Nov 25 at 15:41






          • 2




            I think $text{GL}(infty,mathbb{Z})$ has a different meaning. At least in my experience, it denotes the direct limit of $text{GL}(n,mathbb{Z})$ under the inclusions $text{GL}(n,mathbb{Z})hookrightarrow text{GL}(n+1,mathbb{Z})$ sending $$textbf{X}mapsto begin{bmatrix}textbf{X}&textbf{0}\textbf{0}&1end{bmatrix}$$ for each $mathbf{X}intext{GL}(n,mathbb{Z})$ (i.e, each element of $text{GL}(infty,mathbb{Z})$ is finitary). There are elements of $text{Aut}(mathbb{Z}^omega)$ that is not finitary.
            – Batominovski
            Nov 26 at 10:30




















          • I think you are right.And in this case End($F$) is just GL$(infty,mathbb Z)$
            – Andrews
            Nov 25 at 15:35












          • @Andrews Yes, but that's not a much better description: column finite matrices indexed by the natural numbers which are invertible.
            – egreg
            Nov 25 at 15:41






          • 2




            I think $text{GL}(infty,mathbb{Z})$ has a different meaning. At least in my experience, it denotes the direct limit of $text{GL}(n,mathbb{Z})$ under the inclusions $text{GL}(n,mathbb{Z})hookrightarrow text{GL}(n+1,mathbb{Z})$ sending $$textbf{X}mapsto begin{bmatrix}textbf{X}&textbf{0}\textbf{0}&1end{bmatrix}$$ for each $mathbf{X}intext{GL}(n,mathbb{Z})$ (i.e, each element of $text{GL}(infty,mathbb{Z})$ is finitary). There are elements of $text{Aut}(mathbb{Z}^omega)$ that is not finitary.
            – Batominovski
            Nov 26 at 10:30


















          I think you are right.And in this case End($F$) is just GL$(infty,mathbb Z)$
          – Andrews
          Nov 25 at 15:35






          I think you are right.And in this case End($F$) is just GL$(infty,mathbb Z)$
          – Andrews
          Nov 25 at 15:35














          @Andrews Yes, but that's not a much better description: column finite matrices indexed by the natural numbers which are invertible.
          – egreg
          Nov 25 at 15:41




          @Andrews Yes, but that's not a much better description: column finite matrices indexed by the natural numbers which are invertible.
          – egreg
          Nov 25 at 15:41




          2




          2




          I think $text{GL}(infty,mathbb{Z})$ has a different meaning. At least in my experience, it denotes the direct limit of $text{GL}(n,mathbb{Z})$ under the inclusions $text{GL}(n,mathbb{Z})hookrightarrow text{GL}(n+1,mathbb{Z})$ sending $$textbf{X}mapsto begin{bmatrix}textbf{X}&textbf{0}\textbf{0}&1end{bmatrix}$$ for each $mathbf{X}intext{GL}(n,mathbb{Z})$ (i.e, each element of $text{GL}(infty,mathbb{Z})$ is finitary). There are elements of $text{Aut}(mathbb{Z}^omega)$ that is not finitary.
          – Batominovski
          Nov 26 at 10:30






          I think $text{GL}(infty,mathbb{Z})$ has a different meaning. At least in my experience, it denotes the direct limit of $text{GL}(n,mathbb{Z})$ under the inclusions $text{GL}(n,mathbb{Z})hookrightarrow text{GL}(n+1,mathbb{Z})$ sending $$textbf{X}mapsto begin{bmatrix}textbf{X}&textbf{0}\textbf{0}&1end{bmatrix}$$ for each $mathbf{X}intext{GL}(n,mathbb{Z})$ (i.e, each element of $text{GL}(infty,mathbb{Z})$ is finitary). There are elements of $text{Aut}(mathbb{Z}^omega)$ that is not finitary.
          – Batominovski
          Nov 26 at 10:30




















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3012919%2foperatornameaut-mathbb-q%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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more