Is $Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$? [closed]












0












$begingroup$


Is the following statement true/false?




Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$




My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$



Is my logic correct?



Any hints/solution will be appreciated






abstract-algebra







share|cite|improve this question











$endgroup$



closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Why would $gcd(3, 3)neq 1$ prove the result?
    $endgroup$
    – Arthur
    Dec 3 '18 at 8:39


















0












$begingroup$


Is the following statement true/false?




Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$




My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$



Is my logic correct?



Any hints/solution will be appreciated






abstract-algebra







share|cite|improve this question











$endgroup$



closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 1




    $begingroup$
    Why would $gcd(3, 3)neq 1$ prove the result?
    $endgroup$
    – Arthur
    Dec 3 '18 at 8:39
















0












0








0





$begingroup$


Is the following statement true/false?




Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$




My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$



Is my logic correct?



Any hints/solution will be appreciated






abstract-algebra







share|cite|improve this question











$endgroup$




Is the following statement true/false?




Let $(mathbb{F},+,cdot)$ be the finite field with $9$ elements. Let $K=(mathbb{F},+)$ denote the underlying additive group; then
$$Kcong (mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$$




My attempt: I thinks this statement is false because $gcd(3,3)= 3 neq 1$



Is my logic correct?



Any hints/solution will be appreciated






abstract-algebra




abstract-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 8:38









egreg

181k1485202




181k1485202










asked Dec 3 '18 at 8:31









jasminejasmine

1,701417




1,701417




closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician Dec 8 '18 at 15:19


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Jyrki Lahtonen, Saad, Brahadeesh, ancientmathematician

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    $begingroup$
    Why would $gcd(3, 3)neq 1$ prove the result?
    $endgroup$
    – Arthur
    Dec 3 '18 at 8:39
















  • 1




    $begingroup$
    Why would $gcd(3, 3)neq 1$ prove the result?
    $endgroup$
    – Arthur
    Dec 3 '18 at 8:39










1




1




$begingroup$
Why would $gcd(3, 3)neq 1$ prove the result?
$endgroup$
– Arthur
Dec 3 '18 at 8:39






$begingroup$
Why would $gcd(3, 3)neq 1$ prove the result?
$endgroup$
– Arthur
Dec 3 '18 at 8:39












2 Answers
2






active

oldest

votes


















2












$begingroup$

The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.



It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:42












  • $begingroup$
    @jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
    $endgroup$
    – freakish
    Dec 3 '18 at 8:47












  • $begingroup$
    @freakish that mean $mathbb{F_9}$ is True ??
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:48






  • 1




    $begingroup$
    @jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
    $endgroup$
    – egreg
    Dec 3 '18 at 8:53










  • $begingroup$
    @egreg thanks u .
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:56



















1












$begingroup$

It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).






share|cite|improve this answer









$endgroup$




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.



    It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:42












    • $begingroup$
      @jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
      $endgroup$
      – freakish
      Dec 3 '18 at 8:47












    • $begingroup$
      @freakish that mean $mathbb{F_9}$ is True ??
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:48






    • 1




      $begingroup$
      @jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
      $endgroup$
      – egreg
      Dec 3 '18 at 8:53










    • $begingroup$
      @egreg thanks u .
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:56
















    2












    $begingroup$

    The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.



    It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:42












    • $begingroup$
      @jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
      $endgroup$
      – freakish
      Dec 3 '18 at 8:47












    • $begingroup$
      @freakish that mean $mathbb{F_9}$ is True ??
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:48






    • 1




      $begingroup$
      @jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
      $endgroup$
      – egreg
      Dec 3 '18 at 8:53










    • $begingroup$
      @egreg thanks u .
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:56














    2












    2








    2





    $begingroup$

    The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.



    It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.






    share|cite|improve this answer









    $endgroup$



    The statement is true; $mathbb{F}$ is a vector space over the three-element field and has dimension $2$. The same holds for $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$.



    It is false that $mathbb{F}cong(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})$ as rings.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 3 '18 at 8:40









    egregegreg

    181k1485202




    181k1485202












    • $begingroup$
      @@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:42












    • $begingroup$
      @jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
      $endgroup$
      – freakish
      Dec 3 '18 at 8:47












    • $begingroup$
      @freakish that mean $mathbb{F_9}$ is True ??
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:48






    • 1




      $begingroup$
      @jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
      $endgroup$
      – egreg
      Dec 3 '18 at 8:53










    • $begingroup$
      @egreg thanks u .
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:56


















    • $begingroup$
      @@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:42












    • $begingroup$
      @jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
      $endgroup$
      – freakish
      Dec 3 '18 at 8:47












    • $begingroup$
      @freakish that mean $mathbb{F_9}$ is True ??
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:48






    • 1




      $begingroup$
      @jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
      $endgroup$
      – egreg
      Dec 3 '18 at 8:53










    • $begingroup$
      @egreg thanks u .
      $endgroup$
      – jasmine
      Dec 3 '18 at 8:56
















    $begingroup$
    @@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:42






    $begingroup$
    @@@egreq that mean $mathbb{Z} /3mathbb{Z} timesmathbb{Z} /3mathbb{Z}= mathbb{Z_9} $ or $mathbb{F_9} $ ? which one u r saying
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:42














    $begingroup$
    @jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
    $endgroup$
    – freakish
    Dec 3 '18 at 8:47






    $begingroup$
    @jasmine It is not true that $mathbb{Z}/3mathbb{Z}timesmathbb{Z}/3mathbb{Z}simeqmathbb{Z}_9$.
    $endgroup$
    – freakish
    Dec 3 '18 at 8:47














    $begingroup$
    @freakish that mean $mathbb{F_9}$ is True ??
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:48




    $begingroup$
    @freakish that mean $mathbb{F_9}$ is True ??
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:48




    1




    1




    $begingroup$
    @jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
    $endgroup$
    – egreg
    Dec 3 '18 at 8:53




    $begingroup$
    @jasmine The statement in the question (isomorphism as abelian group) is true. The statement that $(mathbb{Z}/3mathbb{Z})times(mathbb{Z}/3mathbb{Z})congmathbb{Z}/9mathbb{Z}$ is false.
    $endgroup$
    – egreg
    Dec 3 '18 at 8:53












    $begingroup$
    @egreg thanks u .
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:56




    $begingroup$
    @egreg thanks u .
    $endgroup$
    – jasmine
    Dec 3 '18 at 8:56











    1












    $begingroup$

    It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).






        share|cite|improve this answer









        $endgroup$



        It's an abelian group with nine elements, where each element has order $3$. There is only one such group (which I suppose is what you actually need to prove).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 8:40









        ArthurArthur

        113k7110193




        113k7110193















            -

            Popular posts from this blog

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more