Multiplicative Group of a Field












6












$begingroup$


The multiplicative group $F^{times}=Fsetminus {0}$ of a field is abelian, and it may contain torsion elements, may contain torsion free elements, or both may occur, as can be seen from the examples of any finite field, $mathbb{Q},mathbb{R}$ and $mathbb{C}$. The Prüfer $p$-group is a (proper) subgroup of $mathbb{C}^{times}$. The question I would like to ask is



Question: Is there an infinite field $F$ such that $F^{times}$ is isomorphic to the Prüfer $p$-group?






group-theory field-theory







share|cite|improve this question











$endgroup$

















    6












    $begingroup$


    The multiplicative group $F^{times}=Fsetminus {0}$ of a field is abelian, and it may contain torsion elements, may contain torsion free elements, or both may occur, as can be seen from the examples of any finite field, $mathbb{Q},mathbb{R}$ and $mathbb{C}$. The Prüfer $p$-group is a (proper) subgroup of $mathbb{C}^{times}$. The question I would like to ask is



    Question: Is there an infinite field $F$ such that $F^{times}$ is isomorphic to the Prüfer $p$-group?






    group-theory field-theory







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      2



      $begingroup$


      The multiplicative group $F^{times}=Fsetminus {0}$ of a field is abelian, and it may contain torsion elements, may contain torsion free elements, or both may occur, as can be seen from the examples of any finite field, $mathbb{Q},mathbb{R}$ and $mathbb{C}$. The Prüfer $p$-group is a (proper) subgroup of $mathbb{C}^{times}$. The question I would like to ask is



      Question: Is there an infinite field $F$ such that $F^{times}$ is isomorphic to the Prüfer $p$-group?






      group-theory field-theory







      share|cite|improve this question











      $endgroup$




      The multiplicative group $F^{times}=Fsetminus {0}$ of a field is abelian, and it may contain torsion elements, may contain torsion free elements, or both may occur, as can be seen from the examples of any finite field, $mathbb{Q},mathbb{R}$ and $mathbb{C}$. The Prüfer $p$-group is a (proper) subgroup of $mathbb{C}^{times}$. The question I would like to ask is



      Question: Is there an infinite field $F$ such that $F^{times}$ is isomorphic to the Prüfer $p$-group?






      group-theory field-theory




      group-theory field-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jul 30 '13 at 11:58









      azimut

      16.3k1051100




      16.3k1051100










      asked May 30 '13 at 3:54









      RDKRDK

      38917




      38917






















          2 Answers
          2






          active

          oldest

          votes


















          5












          $begingroup$

          Nice question! The answer is no.



          If $F$ has characteristic zero it contains a copy of $mathbb{Q}$, so $F^{times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{times}$ is isomorphic to a Prüfer $ell$-group for some prime $ell$, then $F$ contains all $ell$-power roots of unity, and $mathbb{F}_p^{times}$ must consist of $ell$-power roots of unity, so in particular $p neq ell$.



          Let $k$ be a positive integer. Since $F$ contains all $ell^k$-th roots of unity, $F$ contains $mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $ell^k | p^n - 1$. Then $F^{times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Note that already the assumption that $F^{times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities.
            $endgroup$
            – Qiaochu Yuan
            May 30 '13 at 4:25






          • 1




            $begingroup$
            Could you explain how $mbox{F}^{times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious.
            $endgroup$
            – Zach L.
            May 30 '13 at 4:49








          • 1




            $begingroup$
            @Zach: if $L$ is a subfield of $F$, then $L^{times}$ is a subgroup of $F^{times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $mathbb{F}_p$ as a subfield, so in particular $F^{times}$ contains $mathbb{F}_p^{times}$ as a subgroup.
            $endgroup$
            – Qiaochu Yuan
            May 30 '13 at 4:52












          • $begingroup$
            That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason.
            $endgroup$
            – Zach L.
            May 30 '13 at 15:56










          • $begingroup$
            @Zach: this follows from standard facts about finite fields. If $alpha$ is algebraic over $mathbb{F}_p$, then the subfield of the algebraic closure generated by $alpha$ is $mathbb{F}_{p^n}$ where $n$ is the order of $alpha$ under the action of the Frobenius map. For a primitive $ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $ell^k | p^n - 1$.
            $endgroup$
            – Qiaochu Yuan
            Jun 1 '13 at 7:33



















          1












          $begingroup$

          The following more general statement is also true.



          There is no commutative ring $R$ whose unit group $R^*$ is the Prüfer $p$-group.



          For a proof, see https://arxiv.org/pdf/1505.03508.pdf



          In this paper, Keir Lockride and I gave a complete classification of all indecomposable abelian groups which occur as the group of units of a commutative ring. Note that a Prüfer $p$-group is indecomposable and it does not appear in our list.






          share|cite|improve this answer









          $endgroup$













            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%2f406339%2fmultiplicative-group-of-a-field%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            Nice question! The answer is no.



            If $F$ has characteristic zero it contains a copy of $mathbb{Q}$, so $F^{times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{times}$ is isomorphic to a Prüfer $ell$-group for some prime $ell$, then $F$ contains all $ell$-power roots of unity, and $mathbb{F}_p^{times}$ must consist of $ell$-power roots of unity, so in particular $p neq ell$.



            Let $k$ be a positive integer. Since $F$ contains all $ell^k$-th roots of unity, $F$ contains $mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $ell^k | p^n - 1$. Then $F^{times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Note that already the assumption that $F^{times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:25






            • 1




              $begingroup$
              Could you explain how $mbox{F}^{times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious.
              $endgroup$
              – Zach L.
              May 30 '13 at 4:49








            • 1




              $begingroup$
              @Zach: if $L$ is a subfield of $F$, then $L^{times}$ is a subgroup of $F^{times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $mathbb{F}_p$ as a subfield, so in particular $F^{times}$ contains $mathbb{F}_p^{times}$ as a subgroup.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:52












            • $begingroup$
              That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason.
              $endgroup$
              – Zach L.
              May 30 '13 at 15:56










            • $begingroup$
              @Zach: this follows from standard facts about finite fields. If $alpha$ is algebraic over $mathbb{F}_p$, then the subfield of the algebraic closure generated by $alpha$ is $mathbb{F}_{p^n}$ where $n$ is the order of $alpha$ under the action of the Frobenius map. For a primitive $ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $ell^k | p^n - 1$.
              $endgroup$
              – Qiaochu Yuan
              Jun 1 '13 at 7:33
















            5












            $begingroup$

            Nice question! The answer is no.



            If $F$ has characteristic zero it contains a copy of $mathbb{Q}$, so $F^{times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{times}$ is isomorphic to a Prüfer $ell$-group for some prime $ell$, then $F$ contains all $ell$-power roots of unity, and $mathbb{F}_p^{times}$ must consist of $ell$-power roots of unity, so in particular $p neq ell$.



            Let $k$ be a positive integer. Since $F$ contains all $ell^k$-th roots of unity, $F$ contains $mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $ell^k | p^n - 1$. Then $F^{times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Note that already the assumption that $F^{times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:25






            • 1




              $begingroup$
              Could you explain how $mbox{F}^{times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious.
              $endgroup$
              – Zach L.
              May 30 '13 at 4:49








            • 1




              $begingroup$
              @Zach: if $L$ is a subfield of $F$, then $L^{times}$ is a subgroup of $F^{times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $mathbb{F}_p$ as a subfield, so in particular $F^{times}$ contains $mathbb{F}_p^{times}$ as a subgroup.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:52












            • $begingroup$
              That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason.
              $endgroup$
              – Zach L.
              May 30 '13 at 15:56










            • $begingroup$
              @Zach: this follows from standard facts about finite fields. If $alpha$ is algebraic over $mathbb{F}_p$, then the subfield of the algebraic closure generated by $alpha$ is $mathbb{F}_{p^n}$ where $n$ is the order of $alpha$ under the action of the Frobenius map. For a primitive $ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $ell^k | p^n - 1$.
              $endgroup$
              – Qiaochu Yuan
              Jun 1 '13 at 7:33














            5












            5








            5





            $begingroup$

            Nice question! The answer is no.



            If $F$ has characteristic zero it contains a copy of $mathbb{Q}$, so $F^{times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{times}$ is isomorphic to a Prüfer $ell$-group for some prime $ell$, then $F$ contains all $ell$-power roots of unity, and $mathbb{F}_p^{times}$ must consist of $ell$-power roots of unity, so in particular $p neq ell$.



            Let $k$ be a positive integer. Since $F$ contains all $ell^k$-th roots of unity, $F$ contains $mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $ell^k | p^n - 1$. Then $F^{times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem.






            share|cite|improve this answer











            $endgroup$



            Nice question! The answer is no.



            If $F$ has characteristic zero it contains a copy of $mathbb{Q}$, so $F^{times}$ has torsion-free elements; hence $F$ has positive characteristic $p$. If $F^{times}$ is isomorphic to a Prüfer $ell$-group for some prime $ell$, then $F$ contains all $ell$-power roots of unity, and $mathbb{F}_p^{times}$ must consist of $ell$-power roots of unity, so in particular $p neq ell$.



            Let $k$ be a positive integer. Since $F$ contains all $ell^k$-th roots of unity, $F$ contains $mathbb{F}_{p^n}$ where $n$ is the smallest positive integer such that $ell^k | p^n - 1$. Then $F^{times}$ has a subgroup of order $p^n - 1$, which is necessarily a power of $ell$. By choosing $k$ large enough, this is impossible by Zsigmondy's theorem.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 30 '13 at 4:36

























            answered May 30 '13 at 4:15









            Qiaochu YuanQiaochu Yuan

            278k32585921




            278k32585921












            • $begingroup$
              Note that already the assumption that $F^{times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:25






            • 1




              $begingroup$
              Could you explain how $mbox{F}^{times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious.
              $endgroup$
              – Zach L.
              May 30 '13 at 4:49








            • 1




              $begingroup$
              @Zach: if $L$ is a subfield of $F$, then $L^{times}$ is a subgroup of $F^{times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $mathbb{F}_p$ as a subfield, so in particular $F^{times}$ contains $mathbb{F}_p^{times}$ as a subgroup.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:52












            • $begingroup$
              That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason.
              $endgroup$
              – Zach L.
              May 30 '13 at 15:56










            • $begingroup$
              @Zach: this follows from standard facts about finite fields. If $alpha$ is algebraic over $mathbb{F}_p$, then the subfield of the algebraic closure generated by $alpha$ is $mathbb{F}_{p^n}$ where $n$ is the order of $alpha$ under the action of the Frobenius map. For a primitive $ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $ell^k | p^n - 1$.
              $endgroup$
              – Qiaochu Yuan
              Jun 1 '13 at 7:33


















            • $begingroup$
              Note that already the assumption that $F^{times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:25






            • 1




              $begingroup$
              Could you explain how $mbox{F}^{times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious.
              $endgroup$
              – Zach L.
              May 30 '13 at 4:49








            • 1




              $begingroup$
              @Zach: if $L$ is a subfield of $F$, then $L^{times}$ is a subgroup of $F^{times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $mathbb{F}_p$ as a subfield, so in particular $F^{times}$ contains $mathbb{F}_p^{times}$ as a subgroup.
              $endgroup$
              – Qiaochu Yuan
              May 30 '13 at 4:52












            • $begingroup$
              That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason.
              $endgroup$
              – Zach L.
              May 30 '13 at 15:56










            • $begingroup$
              @Zach: this follows from standard facts about finite fields. If $alpha$ is algebraic over $mathbb{F}_p$, then the subfield of the algebraic closure generated by $alpha$ is $mathbb{F}_{p^n}$ where $n$ is the order of $alpha$ under the action of the Frobenius map. For a primitive $ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $ell^k | p^n - 1$.
              $endgroup$
              – Qiaochu Yuan
              Jun 1 '13 at 7:33
















            $begingroup$
            Note that already the assumption that $F^{times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities.
            $endgroup$
            – Qiaochu Yuan
            May 30 '13 at 4:25




            $begingroup$
            Note that already the assumption that $F^{times}$ is torsion implies that $F$ must be an algebraic extension of a finite field, and from here it is possible to use knowledge of the absolute Galois groups of finite fields to classify all of the possibilities.
            $endgroup$
            – Qiaochu Yuan
            May 30 '13 at 4:25




            1




            1




            $begingroup$
            Could you explain how $mbox{F}^{times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious.
            $endgroup$
            – Zach L.
            May 30 '13 at 4:49






            $begingroup$
            Could you explain how $mbox{F}^{times}_p$ enters in the first paragraph? I don't know much field theory, so it's probably something obvious.
            $endgroup$
            – Zach L.
            May 30 '13 at 4:49






            1




            1




            $begingroup$
            @Zach: if $L$ is a subfield of $F$, then $L^{times}$ is a subgroup of $F^{times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $mathbb{F}_p$ as a subfield, so in particular $F^{times}$ contains $mathbb{F}_p^{times}$ as a subgroup.
            $endgroup$
            – Qiaochu Yuan
            May 30 '13 at 4:52






            $begingroup$
            @Zach: if $L$ is a subfield of $F$, then $L^{times}$ is a subgroup of $F^{times}$. To say that $F$ has characteristic $p$ is equivalent to saying that $F$ contains $mathbb{F}_p$ as a subfield, so in particular $F^{times}$ contains $mathbb{F}_p^{times}$ as a subgroup.
            $endgroup$
            – Qiaochu Yuan
            May 30 '13 at 4:52














            $begingroup$
            That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason.
            $endgroup$
            – Zach L.
            May 30 '13 at 15:56




            $begingroup$
            That wasn't my issue, so my first comment was poorly worded. What I wasn't putting together was that if $p=l$, then $x^l-1$ has only one solution. Now I'm wondering why all $mbox{F}_{p^n}$ must lie in $F$ if $l^k | p^n - 1$. I assume it's for a similar reason.
            $endgroup$
            – Zach L.
            May 30 '13 at 15:56












            $begingroup$
            @Zach: this follows from standard facts about finite fields. If $alpha$ is algebraic over $mathbb{F}_p$, then the subfield of the algebraic closure generated by $alpha$ is $mathbb{F}_{p^n}$ where $n$ is the order of $alpha$ under the action of the Frobenius map. For a primitive $ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $ell^k | p^n - 1$.
            $endgroup$
            – Qiaochu Yuan
            Jun 1 '13 at 7:33




            $begingroup$
            @Zach: this follows from standard facts about finite fields. If $alpha$ is algebraic over $mathbb{F}_p$, then the subfield of the algebraic closure generated by $alpha$ is $mathbb{F}_{p^n}$ where $n$ is the order of $alpha$ under the action of the Frobenius map. For a primitive $ell^k$-th root of unity, it follows that $n$ is the smallest positive integer such that $ell^k | p^n - 1$.
            $endgroup$
            – Qiaochu Yuan
            Jun 1 '13 at 7:33











            1












            $begingroup$

            The following more general statement is also true.



            There is no commutative ring $R$ whose unit group $R^*$ is the Prüfer $p$-group.



            For a proof, see https://arxiv.org/pdf/1505.03508.pdf



            In this paper, Keir Lockride and I gave a complete classification of all indecomposable abelian groups which occur as the group of units of a commutative ring. Note that a Prüfer $p$-group is indecomposable and it does not appear in our list.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              The following more general statement is also true.



              There is no commutative ring $R$ whose unit group $R^*$ is the Prüfer $p$-group.



              For a proof, see https://arxiv.org/pdf/1505.03508.pdf



              In this paper, Keir Lockride and I gave a complete classification of all indecomposable abelian groups which occur as the group of units of a commutative ring. Note that a Prüfer $p$-group is indecomposable and it does not appear in our list.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                The following more general statement is also true.



                There is no commutative ring $R$ whose unit group $R^*$ is the Prüfer $p$-group.



                For a proof, see https://arxiv.org/pdf/1505.03508.pdf



                In this paper, Keir Lockride and I gave a complete classification of all indecomposable abelian groups which occur as the group of units of a commutative ring. Note that a Prüfer $p$-group is indecomposable and it does not appear in our list.






                share|cite|improve this answer









                $endgroup$



                The following more general statement is also true.



                There is no commutative ring $R$ whose unit group $R^*$ is the Prüfer $p$-group.



                For a proof, see https://arxiv.org/pdf/1505.03508.pdf



                In this paper, Keir Lockride and I gave a complete classification of all indecomposable abelian groups which occur as the group of units of a commutative ring. Note that a Prüfer $p$-group is indecomposable and it does not appear in our list.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 2 at 19:35









                CheboluChebolu

                111




                111






























                    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%2f406339%2fmultiplicative-group-of-a-field%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

                    Questions related to Moebius Transform of Characteristic Function of the Primes

                    Image
                    2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
                    Read more

                    List of scandals in India

                    Image
                    List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
                    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