Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ; forall g in G$, $g= h_{1}h_{2}ldots h_n$ where...











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Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
> forall g in G$
, $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
unique. Here, $H_{i}lhd G ; forall i$




I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.










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    up vote
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    down vote

    favorite













    Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
    > forall g in G$
    , $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
    unique. Here, $H_{i}lhd G ; forall i$




    I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
      > forall g in G$
      , $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
      unique. Here, $H_{i}lhd G ; forall i$




      I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.










      share|cite|improve this question














      Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
      > forall g in G$
      , $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
      unique. Here, $H_{i}lhd G ; forall i$




      I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.







      group-theory direct-product






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      asked Nov 21 at 4:51









      So Lo

      62218




      62218






















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          Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.



          Take base case for n=2 . which you had already proved



          Assume this result for n such normal group direct product



          then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now






          share|cite|improve this answer





















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            up vote
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            down vote













            Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.



            Take base case for n=2 . which you had already proved



            Assume this result for n such normal group direct product



            then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now






            share|cite|improve this answer

























              up vote
              1
              down vote













              Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.



              Take base case for n=2 . which you had already proved



              Assume this result for n such normal group direct product



              then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.



                Take base case for n=2 . which you had already proved



                Assume this result for n such normal group direct product



                then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now






                share|cite|improve this answer












                Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.



                Take base case for n=2 . which you had already proved



                Assume this result for n such normal group direct product



                then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 21 at 5:15









                Shubham

                1,3711518




                1,3711518






























                     

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