Proving the given $mathbb R^3/H$ $cong$ $mathbb R^2$ where $H$ = {$(y,0,0)|y in mathbb R$}












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$begingroup$


So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?






abstract-algebra group-isomorphism







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    1












    $begingroup$


    So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?






    abstract-algebra group-isomorphism







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    $endgroup$















      1












      1








      1





      $begingroup$


      So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?






      abstract-algebra group-isomorphism







      share|cite|improve this question











      $endgroup$




      So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?






      abstract-algebra group-isomorphism




      abstract-algebra group-isomorphism






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      share|cite|improve this question













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      edited yesterday









      YuiTo Cheng

      2,41341037




      2,41341037










      asked yesterday









      UfomammutUfomammut

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          2 Answers
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          $begingroup$

          We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
          If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
          Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
          The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
          Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
            $endgroup$
            – Ufomammut
            yesterday












          • $begingroup$
            Yes, that will also work.
            $endgroup$
            – Mayank Mishra
            yesterday



















          4












          $begingroup$

          The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.






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            2 Answers
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            $begingroup$

            We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
            If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
            Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
            The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
            Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
              $endgroup$
              – Ufomammut
              yesterday












            • $begingroup$
              Yes, that will also work.
              $endgroup$
              – Mayank Mishra
              yesterday
















            2












            $begingroup$

            We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
            If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
            Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
            The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
            Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
              $endgroup$
              – Ufomammut
              yesterday












            • $begingroup$
              Yes, that will also work.
              $endgroup$
              – Mayank Mishra
              yesterday














            2












            2








            2





            $begingroup$

            We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
            If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
            Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
            The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
            Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$






            share|cite|improve this answer











            $endgroup$



            We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = {(y,0,0) | y in mathbb{R} }$ is its kernel.
            If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
            Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
            The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
            Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited yesterday

























            answered yesterday









            Mayank MishraMayank Mishra

            1368




            1368








            • 1




              $begingroup$
              I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
              $endgroup$
              – Ufomammut
              yesterday












            • $begingroup$
              Yes, that will also work.
              $endgroup$
              – Mayank Mishra
              yesterday














            • 1




              $begingroup$
              I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
              $endgroup$
              – Ufomammut
              yesterday












            • $begingroup$
              Yes, that will also work.
              $endgroup$
              – Mayank Mishra
              yesterday








            1




            1




            $begingroup$
            I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
            $endgroup$
            – Ufomammut
            yesterday






            $begingroup$
            I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
            $endgroup$
            – Ufomammut
            yesterday














            $begingroup$
            Yes, that will also work.
            $endgroup$
            – Mayank Mishra
            yesterday




            $begingroup$
            Yes, that will also work.
            $endgroup$
            – Mayank Mishra
            yesterday











            4












            $begingroup$

            The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.






                share|cite|improve this answer









                $endgroup$



                The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                lEmlEm

                3,4971921




                3,4971921






























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