Proving the given two groups are isomorphic












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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 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 what is going on. Can anyone provide some help on this?






abstract-algebra group-isomorphism







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$endgroup$

















    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 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 what is going on. Can anyone provide some help on this?






    abstract-algebra group-isomorphism







    share|cite|improve this question









    $endgroup$















      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 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 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 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 what is going on. Can anyone provide some help on this?






      abstract-algebra group-isomorphism




      abstract-algebra group-isomorphism






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      asked 1 hour ago









      UfomammutUfomammut

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          $begingroup$

          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












            $begingroup$

            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











            $endgroup$













            • $begingroup$
              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
              27 mins ago












            • $begingroup$
              Yes, that will also work.
              $endgroup$
              – Mayank Mishra
              24 mins ago












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            2 Answers
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            $begingroup$

            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









            $endgroup$


















              3












              $begingroup$

              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









              $endgroup$
















                3












                3








                3





                $begingroup$

                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 35 mins ago









                lEmlEm

                3,4621921




                3,4621921























                    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











                    $endgroup$













                    • $begingroup$
                      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
                      27 mins ago












                    • $begingroup$
                      Yes, that will also work.
                      $endgroup$
                      – Mayank Mishra
                      24 mins ago
















                    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











                    $endgroup$













                    • $begingroup$
                      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
                      27 mins ago












                    • $begingroup$
                      Yes, that will also work.
                      $endgroup$
                      – Mayank Mishra
                      24 mins ago














                    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 9 mins ago

























                    answered 29 mins ago









                    Mayank MishraMayank Mishra

                    1068




                    1068












                    • $begingroup$
                      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
                      27 mins ago












                    • $begingroup$
                      Yes, that will also work.
                      $endgroup$
                      – Mayank Mishra
                      24 mins ago


















                    • $begingroup$
                      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
                      27 mins ago












                    • $begingroup$
                      Yes, that will also work.
                      $endgroup$
                      – Mayank Mishra
                      24 mins ago
















                    $begingroup$
                    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
                    27 mins ago






                    $begingroup$
                    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
                    27 mins ago














                    $begingroup$
                    Yes, that will also work.
                    $endgroup$
                    – Mayank Mishra
                    24 mins ago




                    $begingroup$
                    Yes, that will also work.
                    $endgroup$
                    – Mayank Mishra
                    24 mins ago


















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