The only function that forms a homomorphism from $mathbb Z_n$ to $mathbb Z_n$












0












$begingroup$


Define $M_a : Z_n → Z_n$ by $M_a([x]) = [ax]$



Prove that



(a) $M_a$ : $(Z_n, +) → (Z_n, +)$ is a homomorphism.



(b) Let φ : $(Z_n, +) → (Z_n, +)$ be a homomorphism. Prove that φ = $M_a$ for some a ∈ Z.



(c) For two homomorphism φ, ψ : $(Z_n, +) → (Z_n, +)$, prove that if φ = $M_a and ψ = M_b$, then for
their composition φ ◦ ψ = $M_{ab}$.



My attempt:
(a) is just using the definition of homomorphism and knowing that $[ax+ay]_n=[ax]_n +[ay]_n$



(c) involved simply using definitions of these functions as well.



I do not know how to prove (b):
Using the fact that the function has been given to be a homomorphism, what I do is:



$ψ(0 + a) = ψ(0) + ψ (a)$



This means $ψ (a) = [0]_n$
But I do not see how this will help me conclude the function must be of the form $M_a$
Could someone please help or provide an alternative?






group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $varphi([k]) = varphi([1]+cdots+[1]) = varphi([1])+cdots+varphi([1])$, where the sums have $k$ summands.
    $endgroup$
    – Christoph
    Dec 12 '18 at 17:30
















0












$begingroup$


Define $M_a : Z_n → Z_n$ by $M_a([x]) = [ax]$



Prove that



(a) $M_a$ : $(Z_n, +) → (Z_n, +)$ is a homomorphism.



(b) Let φ : $(Z_n, +) → (Z_n, +)$ be a homomorphism. Prove that φ = $M_a$ for some a ∈ Z.



(c) For two homomorphism φ, ψ : $(Z_n, +) → (Z_n, +)$, prove that if φ = $M_a and ψ = M_b$, then for
their composition φ ◦ ψ = $M_{ab}$.



My attempt:
(a) is just using the definition of homomorphism and knowing that $[ax+ay]_n=[ax]_n +[ay]_n$



(c) involved simply using definitions of these functions as well.



I do not know how to prove (b):
Using the fact that the function has been given to be a homomorphism, what I do is:



$ψ(0 + a) = ψ(0) + ψ (a)$



This means $ψ (a) = [0]_n$
But I do not see how this will help me conclude the function must be of the form $M_a$
Could someone please help or provide an alternative?






group-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Note that $varphi([k]) = varphi([1]+cdots+[1]) = varphi([1])+cdots+varphi([1])$, where the sums have $k$ summands.
    $endgroup$
    – Christoph
    Dec 12 '18 at 17:30














0












0








0





$begingroup$


Define $M_a : Z_n → Z_n$ by $M_a([x]) = [ax]$



Prove that



(a) $M_a$ : $(Z_n, +) → (Z_n, +)$ is a homomorphism.



(b) Let φ : $(Z_n, +) → (Z_n, +)$ be a homomorphism. Prove that φ = $M_a$ for some a ∈ Z.



(c) For two homomorphism φ, ψ : $(Z_n, +) → (Z_n, +)$, prove that if φ = $M_a and ψ = M_b$, then for
their composition φ ◦ ψ = $M_{ab}$.



My attempt:
(a) is just using the definition of homomorphism and knowing that $[ax+ay]_n=[ax]_n +[ay]_n$



(c) involved simply using definitions of these functions as well.



I do not know how to prove (b):
Using the fact that the function has been given to be a homomorphism, what I do is:



$ψ(0 + a) = ψ(0) + ψ (a)$



This means $ψ (a) = [0]_n$
But I do not see how this will help me conclude the function must be of the form $M_a$
Could someone please help or provide an alternative?






group-theory







share|cite|improve this question











$endgroup$




Define $M_a : Z_n → Z_n$ by $M_a([x]) = [ax]$



Prove that



(a) $M_a$ : $(Z_n, +) → (Z_n, +)$ is a homomorphism.



(b) Let φ : $(Z_n, +) → (Z_n, +)$ be a homomorphism. Prove that φ = $M_a$ for some a ∈ Z.



(c) For two homomorphism φ, ψ : $(Z_n, +) → (Z_n, +)$, prove that if φ = $M_a and ψ = M_b$, then for
their composition φ ◦ ψ = $M_{ab}$.



My attempt:
(a) is just using the definition of homomorphism and knowing that $[ax+ay]_n=[ax]_n +[ay]_n$



(c) involved simply using definitions of these functions as well.



I do not know how to prove (b):
Using the fact that the function has been given to be a homomorphism, what I do is:



$ψ(0 + a) = ψ(0) + ψ (a)$



This means $ψ (a) = [0]_n$
But I do not see how this will help me conclude the function must be of the form $M_a$
Could someone please help or provide an alternative?






group-theory




group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 17:48









user43210

406




406










asked Dec 12 '18 at 17:21









childishsadbinochildishsadbino

1148




1148








  • 1




    $begingroup$
    Note that $varphi([k]) = varphi([1]+cdots+[1]) = varphi([1])+cdots+varphi([1])$, where the sums have $k$ summands.
    $endgroup$
    – Christoph
    Dec 12 '18 at 17:30














  • 1




    $begingroup$
    Note that $varphi([k]) = varphi([1]+cdots+[1]) = varphi([1])+cdots+varphi([1])$, where the sums have $k$ summands.
    $endgroup$
    – Christoph
    Dec 12 '18 at 17:30








1




1




$begingroup$
Note that $varphi([k]) = varphi([1]+cdots+[1]) = varphi([1])+cdots+varphi([1])$, where the sums have $k$ summands.
$endgroup$
– Christoph
Dec 12 '18 at 17:30




$begingroup$
Note that $varphi([k]) = varphi([1]+cdots+[1]) = varphi([1])+cdots+varphi([1])$, where the sums have $k$ summands.
$endgroup$
– Christoph
Dec 12 '18 at 17:30










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$begingroup$

Let $ainmathbb Z$ be such that $varphi(1)=[a]$. Then $varphi(1)=M_a(1)$. Since $mathbb{Z}_n=langle1rangle$, you deduce from this that $varphi=M_a$.






share|cite|improve this answer









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    $begingroup$

    Let $ainmathbb Z$ be such that $varphi(1)=[a]$. Then $varphi(1)=M_a(1)$. Since $mathbb{Z}_n=langle1rangle$, you deduce from this that $varphi=M_a$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Let $ainmathbb Z$ be such that $varphi(1)=[a]$. Then $varphi(1)=M_a(1)$. Since $mathbb{Z}_n=langle1rangle$, you deduce from this that $varphi=M_a$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Let $ainmathbb Z$ be such that $varphi(1)=[a]$. Then $varphi(1)=M_a(1)$. Since $mathbb{Z}_n=langle1rangle$, you deduce from this that $varphi=M_a$.






        share|cite|improve this answer









        $endgroup$



        Let $ainmathbb Z$ be such that $varphi(1)=[a]$. Then $varphi(1)=M_a(1)$. Since $mathbb{Z}_n=langle1rangle$, you deduce from this that $varphi=M_a$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 17:25









        José Carlos SantosJosé Carlos Santos

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