The only function that forms a homomorphism from $mathbb Z_n$ to $mathbb Z_n$
$begingroup$
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
edited Dec 12 '18 at 17:48
user43210
406
asked Dec 12 '18 at 17:21
childishsadbinochildishsadbino
1148
$endgroup$
add a comment |
$begingroup$
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
edited Dec 12 '18 at 17:48
user43210
406
asked Dec 12 '18 at 17:21
childishsadbinochildishsadbino
1148
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 12 '18 at 17:48
user43210
406
asked Dec 12 '18 at 17:21
childishsadbinochildishsadbino
1148
$endgroup$
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
edited Dec 12 '18 at 17:48
user43210
406
asked Dec 12 '18 at 17:21
childishsadbinochildishsadbino
1148
edited Dec 12 '18 at 17:48
user43210
406
asked Dec 12 '18 at 17:21
childishsadbinochildishsadbino
1148
edited Dec 12 '18 at 17:48
user43210
406
edited Dec 12 '18 at 17:48
user43210
406
edited Dec 12 '18 at 17:48
user43210
406
406
asked Dec 12 '18 at 17:21
childishsadbinochildishsadbino
1148
asked Dec 12 '18 at 17:21
childishsadbinochildishsadbino
1148
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
add a comment |
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
add a comment |
1 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 $varphi=M_a$.
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
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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 $varphi=M_a$.
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
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 $varphi=M_a$.
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
add a comment |
$begingroup$
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 $varphi=M_a$.
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
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 $varphi=M_a$.
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
answered Dec 12 '18 at 17:25
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
add a comment |
add a comment |
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$begingroup$
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