Show that $G/Hcongmathbb{R}^*$.












3












$begingroup$



Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.

I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.




My attempt:



$$G/H={gHmid gin G}$$



$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$



And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.



$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$



$$Rightarrow G/Ncong mathbb{R}^*$$



Is it correct? is there another way to solve this?






abstract-algebra matrices group-theory group-isomorphism infinite-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
    $endgroup$
    – Roland
    Jul 7 '15 at 13:40










  • $begingroup$
    edited the matix multiplication
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:44










  • $begingroup$
    Is $R^*$ the group of real numbers under multiplication?
    $endgroup$
    – man_in_green_shirt
    Jul 7 '15 at 13:47








  • 1




    $begingroup$
    yes, multiplication without zero
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:48
















3












$begingroup$



Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.

I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.




My attempt:



$$G/H={gHmid gin G}$$



$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$



And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.



$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$



$$Rightarrow G/Ncong mathbb{R}^*$$



Is it correct? is there another way to solve this?






abstract-algebra matrices group-theory group-isomorphism infinite-groups







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
    $endgroup$
    – Roland
    Jul 7 '15 at 13:40










  • $begingroup$
    edited the matix multiplication
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:44










  • $begingroup$
    Is $R^*$ the group of real numbers under multiplication?
    $endgroup$
    – man_in_green_shirt
    Jul 7 '15 at 13:47








  • 1




    $begingroup$
    yes, multiplication without zero
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:48














3












3








3


0



$begingroup$



Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.

I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.




My attempt:



$$G/H={gHmid gin G}$$



$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$



And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.



$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$



$$Rightarrow G/Ncong mathbb{R}^*$$



Is it correct? is there another way to solve this?






abstract-algebra matrices group-theory group-isomorphism infinite-groups







share|cite|improve this question











$endgroup$





Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.

I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.




My attempt:



$$G/H={gHmid gin G}$$



$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$



And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.



$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$



$$Rightarrow G/Ncong mathbb{R}^*$$



Is it correct? is there another way to solve this?






abstract-algebra matrices group-theory group-isomorphism infinite-groups




abstract-algebra matrices group-theory group-isomorphism infinite-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 20:47









Batominovski

33.1k33293




33.1k33293










asked Jul 7 '15 at 13:36









3SAT3SAT

6,35121334




6,35121334








  • 1




    $begingroup$
    No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
    $endgroup$
    – Roland
    Jul 7 '15 at 13:40










  • $begingroup$
    edited the matix multiplication
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:44










  • $begingroup$
    Is $R^*$ the group of real numbers under multiplication?
    $endgroup$
    – man_in_green_shirt
    Jul 7 '15 at 13:47








  • 1




    $begingroup$
    yes, multiplication without zero
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:48














  • 1




    $begingroup$
    No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
    $endgroup$
    – Roland
    Jul 7 '15 at 13:40










  • $begingroup$
    edited the matix multiplication
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:44










  • $begingroup$
    Is $R^*$ the group of real numbers under multiplication?
    $endgroup$
    – man_in_green_shirt
    Jul 7 '15 at 13:47








  • 1




    $begingroup$
    yes, multiplication without zero
    $endgroup$
    – 3SAT
    Jul 7 '15 at 13:48








1




1




$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40




$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40












$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44




$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44












$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47






$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47






1




1




$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48




$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48










3 Answers
3






active

oldest

votes


















7












$begingroup$

I don't quite understand your solution, but here is another way to do it:



You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
$$
phi: G to mathbb{R}^times
$$
given by
$$
phipmatrix{a & b \ 0 & a} = a.
$$
You have to show that this map




  1. is a homomorphism,

  2. is surjective,

  3. has $H$ as kernel.


Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.




If you don't know the First Isomorphism Theorem, you can define the map
$$
psi: G/H to mathbb{R}^times
$$
by
$$
psi pmatrix{a & b \ 0 & a}H = a
$$
and show that this map is




  1. well defined,

  2. a homomorphism,
    3, surjective,

  3. injective.


(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Here is the "backwards" proof:



      Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.



      Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$



      (Note how we use the fact that $a' neq 0$).



      Note as well that $AH = (aI)H$.



      Thus $a mapsto (aI)H$ is the desired isomorphism.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        I don't quite understand your solution, but here is another way to do it:



        You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
        $$
        phi: G to mathbb{R}^times
        $$
        given by
        $$
        phipmatrix{a & b \ 0 & a} = a.
        $$
        You have to show that this map




        1. is a homomorphism,

        2. is surjective,

        3. has $H$ as kernel.


        Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.




        If you don't know the First Isomorphism Theorem, you can define the map
        $$
        psi: G/H to mathbb{R}^times
        $$
        by
        $$
        psi pmatrix{a & b \ 0 & a}H = a
        $$
        and show that this map is




        1. well defined,

        2. a homomorphism,
          3, surjective,

        3. injective.


        (Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          I don't quite understand your solution, but here is another way to do it:



          You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
          $$
          phi: G to mathbb{R}^times
          $$
          given by
          $$
          phipmatrix{a & b \ 0 & a} = a.
          $$
          You have to show that this map




          1. is a homomorphism,

          2. is surjective,

          3. has $H$ as kernel.


          Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.




          If you don't know the First Isomorphism Theorem, you can define the map
          $$
          psi: G/H to mathbb{R}^times
          $$
          by
          $$
          psi pmatrix{a & b \ 0 & a}H = a
          $$
          and show that this map is




          1. well defined,

          2. a homomorphism,
            3, surjective,

          3. injective.


          (Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            I don't quite understand your solution, but here is another way to do it:



            You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
            $$
            phi: G to mathbb{R}^times
            $$
            given by
            $$
            phipmatrix{a & b \ 0 & a} = a.
            $$
            You have to show that this map




            1. is a homomorphism,

            2. is surjective,

            3. has $H$ as kernel.


            Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.




            If you don't know the First Isomorphism Theorem, you can define the map
            $$
            psi: G/H to mathbb{R}^times
            $$
            by
            $$
            psi pmatrix{a & b \ 0 & a}H = a
            $$
            and show that this map is




            1. well defined,

            2. a homomorphism,
              3, surjective,

            3. injective.


            (Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)






            share|cite|improve this answer









            $endgroup$



            I don't quite understand your solution, but here is another way to do it:



            You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
            $$
            phi: G to mathbb{R}^times
            $$
            given by
            $$
            phipmatrix{a & b \ 0 & a} = a.
            $$
            You have to show that this map




            1. is a homomorphism,

            2. is surjective,

            3. has $H$ as kernel.


            Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.




            If you don't know the First Isomorphism Theorem, you can define the map
            $$
            psi: G/H to mathbb{R}^times
            $$
            by
            $$
            psi pmatrix{a & b \ 0 & a}H = a
            $$
            and show that this map is




            1. well defined,

            2. a homomorphism,
              3, surjective,

            3. injective.


            (Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jul 7 '15 at 13:41









            ThomasThomas

            35.6k1058117




            35.6k1058117























                3












                $begingroup$

                How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?






                share|cite|improve this answer











                $endgroup$


















                  3












                  $begingroup$

                  How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?






                  share|cite|improve this answer











                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?






                    share|cite|improve this answer











                    $endgroup$



                    How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 14 '18 at 20:51

























                    answered Jul 7 '15 at 13:40









                    BatominovskiBatominovski

                    33.1k33293




                    33.1k33293























                        0












                        $begingroup$

                        Here is the "backwards" proof:



                        Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.



                        Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$



                        (Note how we use the fact that $a' neq 0$).



                        Note as well that $AH = (aI)H$.



                        Thus $a mapsto (aI)H$ is the desired isomorphism.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Here is the "backwards" proof:



                          Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.



                          Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$



                          (Note how we use the fact that $a' neq 0$).



                          Note as well that $AH = (aI)H$.



                          Thus $a mapsto (aI)H$ is the desired isomorphism.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Here is the "backwards" proof:



                            Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.



                            Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$



                            (Note how we use the fact that $a' neq 0$).



                            Note as well that $AH = (aI)H$.



                            Thus $a mapsto (aI)H$ is the desired isomorphism.






                            share|cite|improve this answer









                            $endgroup$



                            Here is the "backwards" proof:



                            Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.



                            Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$



                            (Note how we use the fact that $a' neq 0$).



                            Note as well that $AH = (aI)H$.



                            Thus $a mapsto (aI)H$ is the desired isomorphism.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jul 7 '15 at 15:34









                            David WheelerDavid Wheeler

                            12.7k11730




                            12.7k11730






























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

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

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                                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
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