Finding all group homomorphisms from $mathbb{Z}_7$ to $mathbb{Z}_{12}$












1












$begingroup$


Suppose that I'm interested in finding all group homomorphisms from $mathbb{Z}_7$ to $mathbb{Z}_{12}$. The textbook has provided a brief explanation:




Let $phi$ be such a homomorphism. Since the kernel of ϕ must be a subgroup of $mathbb{Z}_7$, there are only two possible
kernels, ${0}$ and all of $mathbb{Z}_7$. The image of a subgroup of $mathbb{Z}_7$ must be a subgroup of $mathbb{Z}_{12}$.
Hence, there is no injective homomorphism; otherwise, $mathbb{Z}_{12}$ would have a subgroup of order
$7$, which is impossible. Consequently, the only possible homomorphism from $mathbb{Z}_{7}$ to $mathbb{Z}_{12}$ is
the one mapping all elements to zero.




The explanation makes sense except for the conclusion. Why does non-injectivity imply the above conclusion?






abstract-algebra group-homomorphism







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$endgroup$

















    1












    $begingroup$


    Suppose that I'm interested in finding all group homomorphisms from $mathbb{Z}_7$ to $mathbb{Z}_{12}$. The textbook has provided a brief explanation:




    Let $phi$ be such a homomorphism. Since the kernel of ϕ must be a subgroup of $mathbb{Z}_7$, there are only two possible
    kernels, ${0}$ and all of $mathbb{Z}_7$. The image of a subgroup of $mathbb{Z}_7$ must be a subgroup of $mathbb{Z}_{12}$.
    Hence, there is no injective homomorphism; otherwise, $mathbb{Z}_{12}$ would have a subgroup of order
    $7$, which is impossible. Consequently, the only possible homomorphism from $mathbb{Z}_{7}$ to $mathbb{Z}_{12}$ is
    the one mapping all elements to zero.




    The explanation makes sense except for the conclusion. Why does non-injectivity imply the above conclusion?






    abstract-algebra group-homomorphism







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Suppose that I'm interested in finding all group homomorphisms from $mathbb{Z}_7$ to $mathbb{Z}_{12}$. The textbook has provided a brief explanation:




      Let $phi$ be such a homomorphism. Since the kernel of ϕ must be a subgroup of $mathbb{Z}_7$, there are only two possible
      kernels, ${0}$ and all of $mathbb{Z}_7$. The image of a subgroup of $mathbb{Z}_7$ must be a subgroup of $mathbb{Z}_{12}$.
      Hence, there is no injective homomorphism; otherwise, $mathbb{Z}_{12}$ would have a subgroup of order
      $7$, which is impossible. Consequently, the only possible homomorphism from $mathbb{Z}_{7}$ to $mathbb{Z}_{12}$ is
      the one mapping all elements to zero.




      The explanation makes sense except for the conclusion. Why does non-injectivity imply the above conclusion?






      abstract-algebra group-homomorphism







      share|cite|improve this question











      $endgroup$




      Suppose that I'm interested in finding all group homomorphisms from $mathbb{Z}_7$ to $mathbb{Z}_{12}$. The textbook has provided a brief explanation:




      Let $phi$ be such a homomorphism. Since the kernel of ϕ must be a subgroup of $mathbb{Z}_7$, there are only two possible
      kernels, ${0}$ and all of $mathbb{Z}_7$. The image of a subgroup of $mathbb{Z}_7$ must be a subgroup of $mathbb{Z}_{12}$.
      Hence, there is no injective homomorphism; otherwise, $mathbb{Z}_{12}$ would have a subgroup of order
      $7$, which is impossible. Consequently, the only possible homomorphism from $mathbb{Z}_{7}$ to $mathbb{Z}_{12}$ is
      the one mapping all elements to zero.




      The explanation makes sense except for the conclusion. Why does non-injectivity imply the above conclusion?






      abstract-algebra group-homomorphism




      abstract-algebra group-homomorphism






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 5 '18 at 13:32









      Chinnapparaj R

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      asked Dec 5 '18 at 13:24









      TedTed

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          $begingroup$

          There are only two possibilites for the homomorphism: trivial or injective (this corresponds to the only two possibilities for the kernel: kernel $Bbb Z_7$ gives trivial homomorphism and kernel ${0}$ gives injective homomorphism). They have concluded that injective isn't possible, so trivial is the only option left.



          A common alternative approach works on an element-by-element level: We know $0_7$ must be mapped to $0_{12}$. What about any of the other elements? Well, for any $nin Bbb Z_7$, we have $7cdot n = 0_7$, so whatever we map $n$ to in $Bbb Z_{12}$, that property must be upheld. Which $kin Bbb Z_{12}$ are such that $7cdot k = 0_{12}$? Those are the only possible images of elements in $Bbb Z_7$.






          share|cite|improve this answer











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            $begingroup$

            There are only two possibilites for the homomorphism: trivial or injective (this corresponds to the only two possibilities for the kernel: kernel $Bbb Z_7$ gives trivial homomorphism and kernel ${0}$ gives injective homomorphism). They have concluded that injective isn't possible, so trivial is the only option left.



            A common alternative approach works on an element-by-element level: We know $0_7$ must be mapped to $0_{12}$. What about any of the other elements? Well, for any $nin Bbb Z_7$, we have $7cdot n = 0_7$, so whatever we map $n$ to in $Bbb Z_{12}$, that property must be upheld. Which $kin Bbb Z_{12}$ are such that $7cdot k = 0_{12}$? Those are the only possible images of elements in $Bbb Z_7$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              There are only two possibilites for the homomorphism: trivial or injective (this corresponds to the only two possibilities for the kernel: kernel $Bbb Z_7$ gives trivial homomorphism and kernel ${0}$ gives injective homomorphism). They have concluded that injective isn't possible, so trivial is the only option left.



              A common alternative approach works on an element-by-element level: We know $0_7$ must be mapped to $0_{12}$. What about any of the other elements? Well, for any $nin Bbb Z_7$, we have $7cdot n = 0_7$, so whatever we map $n$ to in $Bbb Z_{12}$, that property must be upheld. Which $kin Bbb Z_{12}$ are such that $7cdot k = 0_{12}$? Those are the only possible images of elements in $Bbb Z_7$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                There are only two possibilites for the homomorphism: trivial or injective (this corresponds to the only two possibilities for the kernel: kernel $Bbb Z_7$ gives trivial homomorphism and kernel ${0}$ gives injective homomorphism). They have concluded that injective isn't possible, so trivial is the only option left.



                A common alternative approach works on an element-by-element level: We know $0_7$ must be mapped to $0_{12}$. What about any of the other elements? Well, for any $nin Bbb Z_7$, we have $7cdot n = 0_7$, so whatever we map $n$ to in $Bbb Z_{12}$, that property must be upheld. Which $kin Bbb Z_{12}$ are such that $7cdot k = 0_{12}$? Those are the only possible images of elements in $Bbb Z_7$.






                share|cite|improve this answer











                $endgroup$



                There are only two possibilites for the homomorphism: trivial or injective (this corresponds to the only two possibilities for the kernel: kernel $Bbb Z_7$ gives trivial homomorphism and kernel ${0}$ gives injective homomorphism). They have concluded that injective isn't possible, so trivial is the only option left.



                A common alternative approach works on an element-by-element level: We know $0_7$ must be mapped to $0_{12}$. What about any of the other elements? Well, for any $nin Bbb Z_7$, we have $7cdot n = 0_7$, so whatever we map $n$ to in $Bbb Z_{12}$, that property must be upheld. Which $kin Bbb Z_{12}$ are such that $7cdot k = 0_{12}$? Those are the only possible images of elements in $Bbb Z_7$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 5 '18 at 13:33

























                answered Dec 5 '18 at 13:27









                ArthurArthur

                113k7109193




                113k7109193






























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