The number of homomorphism f from $(Z,+)$ to $(Q^{*},.)$ such that $f(2)=frac{1}{3}$ .
$begingroup$
How to count the number of homomorphism f from $(Z,+)$ to $(Q^{*},.)$ such that $f(2)=frac{1}{3}$ ?
Can u tell me the way or algorithm to do find that ?
abstract-algebra group-theory
asked Jan 25 at 9:57
sejysejy
1629
$endgroup$
add a comment |
$begingroup$
How to count the number of homomorphism f from $(Z,+)$ to $(Q^{*},.)$ such that $f(2)=frac{1}{3}$ ?
Can u tell me the way or algorithm to do find that ?
abstract-algebra group-theory
asked Jan 25 at 9:57
sejysejy
1629
$endgroup$
add a comment |
$begingroup$
How to count the number of homomorphism f from $(Z,+)$ to $(Q^{*},.)$ such that $f(2)=frac{1}{3}$ ?
Can u tell me the way or algorithm to do find that ?
abstract-algebra group-theory
asked Jan 25 at 9:57
sejysejy
1629
$endgroup$
How to count the number of homomorphism f from $(Z,+)$ to $(Q^{*},.)$ such that $f(2)=frac{1}{3}$ ?
Can u tell me the way or algorithm to do find that ?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 25 at 9:57
sejysejy
1629
asked Jan 25 at 9:57
sejysejy
1629
asked Jan 25 at 9:57
sejysejy
1629
asked Jan 25 at 9:57
sejysejy
1629
asked Jan 25 at 9:57
sejysejy
1629
1629
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
The group $(mathbb{Z},+)$ is generated by the single element $1$. If we know where it goes, we know everything about the homomorphism. So, then, what can we send $1$ to in $(mathbb{Q}^*,cdot)$ so that $2$ goes to $frac13$? Well, $f(1+1)=f(1)cdot f(1)$, so we would then need $f(1)^2=frac13$. There are no rational square roots of $frac13$, so there are no such homomorphisms.
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The group $(mathbb{Z},+)$ is generated by the single element $1$. If we know where it goes, we know everything about the homomorphism. So, then, what can we send $1$ to in $(mathbb{Q}^*,cdot)$ so that $2$ goes to $frac13$? Well, $f(1+1)=f(1)cdot f(1)$, so we would then need $f(1)^2=frac13$. There are no rational square roots of $frac13$, so there are no such homomorphisms.
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
$endgroup$
add a comment |
$begingroup$
The group $(mathbb{Z},+)$ is generated by the single element $1$. If we know where it goes, we know everything about the homomorphism. So, then, what can we send $1$ to in $(mathbb{Q}^*,cdot)$ so that $2$ goes to $frac13$? Well, $f(1+1)=f(1)cdot f(1)$, so we would then need $f(1)^2=frac13$. There are no rational square roots of $frac13$, so there are no such homomorphisms.
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
$endgroup$
add a comment |
$begingroup$
The group $(mathbb{Z},+)$ is generated by the single element $1$. If we know where it goes, we know everything about the homomorphism. So, then, what can we send $1$ to in $(mathbb{Q}^*,cdot)$ so that $2$ goes to $frac13$? Well, $f(1+1)=f(1)cdot f(1)$, so we would then need $f(1)^2=frac13$. There are no rational square roots of $frac13$, so there are no such homomorphisms.
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
$endgroup$
The group $(mathbb{Z},+)$ is generated by the single element $1$. If we know where it goes, we know everything about the homomorphism. So, then, what can we send $1$ to in $(mathbb{Q}^*,cdot)$ so that $2$ goes to $frac13$? Well, $f(1+1)=f(1)cdot f(1)$, so we would then need $f(1)^2=frac13$. There are no rational square roots of $frac13$, so there are no such homomorphisms.
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
answered Jan 25 at 10:09
jmerryjmerry
13.3k1628
13.3k1628
add a comment |
add a comment |
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