Finding all group homomorphisms from $mathbb{Z}_7$ to $mathbb{Z}_{12}$
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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
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
asked Dec 5 at 13:24
Ted
738
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up vote
1
down vote
favorite
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
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
asked Dec 5 at 13:24
Ted
738
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
asked Dec 5 at 13:24
Ted
738
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
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
asked Dec 5 at 13:24
Ted
738
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
asked Dec 5 at 13:24
Ted
738
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
edited Dec 5 at 13:32
Chinnapparaj R
4,9951825
4,9951825
asked Dec 5 at 13:24
Ted
738
asked Dec 5 at 13:24
Ted
738
asked Dec 5 at 13:24
Ted
738
738
add a comment |
add a comment |
1 Answer
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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 $Bbb Z_7$.
answered Dec 5 at 13:27
Arthur
110k7104186
add a comment |
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1 Answer
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active
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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 $Bbb Z_7$.
answered Dec 5 at 13:27
Arthur
110k7104186
add a comment |
up vote
2
down vote
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 $Bbb Z_7$.
answered Dec 5 at 13:27
Arthur
110k7104186
add a comment |
up vote
2
down vote
up vote
2
down vote
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 $Bbb Z_7$.
answered Dec 5 at 13:27
Arthur
110k7104186
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 $Bbb Z_7$.
answered Dec 5 at 13:27
Arthur
110k7104186
edited Dec 5 at 13:33
answered Dec 5 at 13:27
Arthur
110k7104186
answered Dec 5 at 13:27
Arthur
110k7104186
answered Dec 5 at 13:27
Arthur
110k7104186
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