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












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

      339317




      339317






















          1 Answer
          1






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          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













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          1 Answer
          1






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          active

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          active

          oldest

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          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




















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