Construct a group isomorphism $phi : U(20) to mathbb Z_2 oplus mathbb Z_4$.












0














Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.



EDIT: removed mapping because it was not an isomorphism










share|cite|improve this question




















  • 1




    Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
    – Denis
    Apr 8 '14 at 14:52












  • It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
    – user49685
    Apr 8 '14 at 14:54












  • Oops, you're right. How could I start to find an isomorphism then?
    – appel
    Apr 8 '14 at 15:01






  • 1




    @David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
    – user49685
    Apr 8 '14 at 15:10








  • 1




    With $U(20)$ I mean the unitary group of degree 20
    – appel
    Apr 8 '14 at 15:13
















0














Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.



EDIT: removed mapping because it was not an isomorphism










share|cite|improve this question




















  • 1




    Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
    – Denis
    Apr 8 '14 at 14:52












  • It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
    – user49685
    Apr 8 '14 at 14:54












  • Oops, you're right. How could I start to find an isomorphism then?
    – appel
    Apr 8 '14 at 15:01






  • 1




    @David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
    – user49685
    Apr 8 '14 at 15:10








  • 1




    With $U(20)$ I mean the unitary group of degree 20
    – appel
    Apr 8 '14 at 15:13














0












0








0







Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.



EDIT: removed mapping because it was not an isomorphism










share|cite|improve this question















Construct a group isomorphism $phi : G_1 to G_2$, with $G_1 = U(20)$ and $G_2 = mathbb Z_2 oplus mathbb Z_4$.



EDIT: removed mapping because it was not an isomorphism







linear-algebra group-theory group-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 0:16









Shaun

8,805113680




8,805113680










asked Apr 8 '14 at 14:47









appel

1265




1265








  • 1




    Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
    – Denis
    Apr 8 '14 at 14:52












  • It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
    – user49685
    Apr 8 '14 at 14:54












  • Oops, you're right. How could I start to find an isomorphism then?
    – appel
    Apr 8 '14 at 15:01






  • 1




    @David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
    – user49685
    Apr 8 '14 at 15:10








  • 1




    With $U(20)$ I mean the unitary group of degree 20
    – appel
    Apr 8 '14 at 15:13














  • 1




    Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
    – Denis
    Apr 8 '14 at 14:52












  • It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
    – user49685
    Apr 8 '14 at 14:54












  • Oops, you're right. How could I start to find an isomorphism then?
    – appel
    Apr 8 '14 at 15:01






  • 1




    @David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
    – user49685
    Apr 8 '14 at 15:10








  • 1




    With $U(20)$ I mean the unitary group of degree 20
    – appel
    Apr 8 '14 at 15:13








1




1




Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52






Why do you want a formula ? Is this really an isomorphism ? What about 3*7 ?
– Denis
Apr 8 '14 at 14:52














It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54






It's not even a homomorphism: $3.7 = 1$ in $U(20)$, but their image $(0; 1) + (0; 2) = (0; 3) neq (0; 0)$ in $mathbb{Z}_2 oplusmathbb{Z}_4$. What I'm trying to show is that $f(3.7) neq f(3) + f(7)$. Hence it's not a homomorphism.
– user49685
Apr 8 '14 at 14:54














Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01




Oops, you're right. How could I start to find an isomorphism then?
– appel
Apr 8 '14 at 15:01




1




1




@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10






@David: It's the set of all units in the group $(mathbb{Z}_{20}; times)$
– user49685
Apr 8 '14 at 15:10






1




1




With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13




With $U(20)$ I mean the unitary group of degree 20
– appel
Apr 8 '14 at 15:13










2 Answers
2






active

oldest

votes


















1














Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.



details:
define
$$
F(1,0)=19, F(0,1)=17
$$

such as $F$ is an homomorphism. This is possible because the orders of
$19$ and $17$ are 2 and 4.



You then get
$$
F(0,2) = 17times 17 = 9\
F(0,3) = 17times 9 = 13\
F(1,1) = 19times 17 = 3\
F(1,2) = 19times 9 = 11\
F(1,3) = 19times 13 = 7
$$
hence $F$ is an isomorphism.





This was the naïve approach.



As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
you know that the subgroup
$langle arangle$ is of order 4 and does not contain $b$.



So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.



Then the morphism has to be onto (without exhaustive check).






share|cite|improve this answer























  • I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
    – appel
    Apr 8 '14 at 15:10












  • I wrote the complete details.
    – mookid
    Apr 8 '14 at 15:13










  • Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
    – appel
    Apr 8 '14 at 15:16










  • These are the first I found (see that 19=-1).
    – mookid
    Apr 8 '14 at 15:18










  • I added another solution.
    – mookid
    Apr 8 '14 at 15:21



















0














$U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.



Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.



To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.



Thus to get an isomorphism just match up elements of the same order.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.



    details:
    define
    $$
    F(1,0)=19, F(0,1)=17
    $$

    such as $F$ is an homomorphism. This is possible because the orders of
    $19$ and $17$ are 2 and 4.



    You then get
    $$
    F(0,2) = 17times 17 = 9\
    F(0,3) = 17times 9 = 13\
    F(1,1) = 19times 17 = 3\
    F(1,2) = 19times 9 = 11\
    F(1,3) = 19times 13 = 7
    $$
    hence $F$ is an isomorphism.





    This was the naïve approach.



    As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
    you know that the subgroup
    $langle arangle$ is of order 4 and does not contain $b$.



    So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.



    Then the morphism has to be onto (without exhaustive check).






    share|cite|improve this answer























    • I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
      – appel
      Apr 8 '14 at 15:10












    • I wrote the complete details.
      – mookid
      Apr 8 '14 at 15:13










    • Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
      – appel
      Apr 8 '14 at 15:16










    • These are the first I found (see that 19=-1).
      – mookid
      Apr 8 '14 at 15:18










    • I added another solution.
      – mookid
      Apr 8 '14 at 15:21
















    1














    Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.



    details:
    define
    $$
    F(1,0)=19, F(0,1)=17
    $$

    such as $F$ is an homomorphism. This is possible because the orders of
    $19$ and $17$ are 2 and 4.



    You then get
    $$
    F(0,2) = 17times 17 = 9\
    F(0,3) = 17times 9 = 13\
    F(1,1) = 19times 17 = 3\
    F(1,2) = 19times 9 = 11\
    F(1,3) = 19times 13 = 7
    $$
    hence $F$ is an isomorphism.





    This was the naïve approach.



    As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
    you know that the subgroup
    $langle arangle$ is of order 4 and does not contain $b$.



    So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.



    Then the morphism has to be onto (without exhaustive check).






    share|cite|improve this answer























    • I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
      – appel
      Apr 8 '14 at 15:10












    • I wrote the complete details.
      – mookid
      Apr 8 '14 at 15:13










    • Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
      – appel
      Apr 8 '14 at 15:16










    • These are the first I found (see that 19=-1).
      – mookid
      Apr 8 '14 at 15:18










    • I added another solution.
      – mookid
      Apr 8 '14 at 15:21














    1












    1








    1






    Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.



    details:
    define
    $$
    F(1,0)=19, F(0,1)=17
    $$

    such as $F$ is an homomorphism. This is possible because the orders of
    $19$ and $17$ are 2 and 4.



    You then get
    $$
    F(0,2) = 17times 17 = 9\
    F(0,3) = 17times 9 = 13\
    F(1,1) = 19times 17 = 3\
    F(1,2) = 19times 9 = 11\
    F(1,3) = 19times 13 = 7
    $$
    hence $F$ is an isomorphism.





    This was the naïve approach.



    As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
    you know that the subgroup
    $langle arangle$ is of order 4 and does not contain $b$.



    So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.



    Then the morphism has to be onto (without exhaustive check).






    share|cite|improve this answer














    Hint: $Bbb Z _2 oplusBbb Z_4$ is generated by two elements, of orders $2$ and $4$.



    details:
    define
    $$
    F(1,0)=19, F(0,1)=17
    $$

    such as $F$ is an homomorphism. This is possible because the orders of
    $19$ and $17$ are 2 and 4.



    You then get
    $$
    F(0,2) = 17times 17 = 9\
    F(0,3) = 17times 9 = 13\
    F(1,1) = 19times 17 = 3\
    F(1,2) = 19times 9 = 11\
    F(1,3) = 19times 13 = 7
    $$
    hence $F$ is an isomorphism.





    This was the naïve approach.



    As soon as you found $a$ of order 4 and $b$ of order 2 and $bneq a^2$,
    you know that the subgroup
    $langle arangle$ is of order 4 and does not contain $b$.



    So the subgroup $langle a,brangle$ has no other choice than being $U(20)$.



    Then the morphism has to be onto (without exhaustive check).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 29 '18 at 0:17









    Shaun

    8,805113680




    8,805113680










    answered Apr 8 '14 at 15:03









    mookid

    25.5k52447




    25.5k52447












    • I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
      – appel
      Apr 8 '14 at 15:10












    • I wrote the complete details.
      – mookid
      Apr 8 '14 at 15:13










    • Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
      – appel
      Apr 8 '14 at 15:16










    • These are the first I found (see that 19=-1).
      – mookid
      Apr 8 '14 at 15:18










    • I added another solution.
      – mookid
      Apr 8 '14 at 15:21


















    • I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
      – appel
      Apr 8 '14 at 15:10












    • I wrote the complete details.
      – mookid
      Apr 8 '14 at 15:13










    • Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
      – appel
      Apr 8 '14 at 15:16










    • These are the first I found (see that 19=-1).
      – mookid
      Apr 8 '14 at 15:18










    • I added another solution.
      – mookid
      Apr 8 '14 at 15:21
















    I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
    – appel
    Apr 8 '14 at 15:10






    I understand there should be a mapping $phi(u) = (a,b)$ with $u in U, a in mathbb Z_2, b in mathbb Z_4$, but I don't see what the function could be.
    – appel
    Apr 8 '14 at 15:10














    I wrote the complete details.
    – mookid
    Apr 8 '14 at 15:13




    I wrote the complete details.
    – mookid
    Apr 8 '14 at 15:13












    Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
    – appel
    Apr 8 '14 at 15:16




    Thank you, but doesn't 3 also have order 4? and 9 order 2? Why do you take these numbers.
    – appel
    Apr 8 '14 at 15:16












    These are the first I found (see that 19=-1).
    – mookid
    Apr 8 '14 at 15:18




    These are the first I found (see that 19=-1).
    – mookid
    Apr 8 '14 at 15:18












    I added another solution.
    – mookid
    Apr 8 '14 at 15:21




    I added another solution.
    – mookid
    Apr 8 '14 at 15:21











    0














    $U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.



    Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.



    To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.



    Thus to get an isomorphism just match up elements of the same order.






    share|cite|improve this answer


























      0














      $U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.



      Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.



      To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.



      Thus to get an isomorphism just match up elements of the same order.






      share|cite|improve this answer
























        0












        0








        0






        $U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.



        Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.



        To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.



        Thus to get an isomorphism just match up elements of the same order.






        share|cite|improve this answer












        $U(20)={1,3,7,9,11,13,17,19}$ with the group operation multiplication modulo $20$.



        Both $U(20)$ and $mathbb Z_2oplusmathbb Z_4$ have presentation $langle a,bmid a^2=e,b^4=e, ab=barangle$.



        To see that this is true for $U(20)$ , check the orders of the elements: $o(1)=1,o(3)=o(7)=o(13)=o(17)=4$ and $o(9)=o(11)=o(19)=2$.



        Thus to get an isomorphism just match up elements of the same order.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 1:09









        Chris Custer

        10.9k3824




        10.9k3824






























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