$mathbb{Z}leq G leq mathbb{Q}$ and $varphi:Grightarrow H$ a map which behaves nicely on fractions.
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Let $(H,+)$ be any abelian group, and let $mathbb{Z}leq Gleq mathbb{Q}$ a subgroup of the rational numbers with respect to addition. Let $varphi:Grightarrow H$ be a map with the following properties
(1) The restriction of $varphi$ to $mathbb{Z}$ is a homomorphism. (i.e. $varphi(n) = nvarphi(1)$ and $varphi(0)=0_H$)
(2) For every rational number $qin G$, we can find $a,binmathbb{Z}$ such that $q=frac{a}{b}$ and $varphi(q)^b = varphi(a)$.
Is it necessarily true that $varphi$ is an (additive) homomorphism. That is for all $q,pin G$ we have $varphi(p+q) = varphi(p) + varphi(q)$?
group-theory rational-numbers group-homomorphism
asked Nov 4 at 14:38
Yanko
5,750723
add a comment |
up vote
1
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favorite
Let $(H,+)$ be any abelian group, and let $mathbb{Z}leq Gleq mathbb{Q}$ a subgroup of the rational numbers with respect to addition. Let $varphi:Grightarrow H$ be a map with the following properties
(1) The restriction of $varphi$ to $mathbb{Z}$ is a homomorphism. (i.e. $varphi(n) = nvarphi(1)$ and $varphi(0)=0_H$)
(2) For every rational number $qin G$, we can find $a,binmathbb{Z}$ such that $q=frac{a}{b}$ and $varphi(q)^b = varphi(a)$.
Is it necessarily true that $varphi$ is an (additive) homomorphism. That is for all $q,pin G$ we have $varphi(p+q) = varphi(p) + varphi(q)$?
group-theory rational-numbers group-homomorphism
asked Nov 4 at 14:38
Yanko
5,750723
I might need help with the title as well. Any ideas?
– Yanko
Nov 4 at 14:42
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $(H,+)$ be any abelian group, and let $mathbb{Z}leq Gleq mathbb{Q}$ a subgroup of the rational numbers with respect to addition. Let $varphi:Grightarrow H$ be a map with the following properties
(1) The restriction of $varphi$ to $mathbb{Z}$ is a homomorphism. (i.e. $varphi(n) = nvarphi(1)$ and $varphi(0)=0_H$)
(2) For every rational number $qin G$, we can find $a,binmathbb{Z}$ such that $q=frac{a}{b}$ and $varphi(q)^b = varphi(a)$.
Is it necessarily true that $varphi$ is an (additive) homomorphism. That is for all $q,pin G$ we have $varphi(p+q) = varphi(p) + varphi(q)$?
group-theory rational-numbers group-homomorphism
asked Nov 4 at 14:38
Yanko
5,750723
Let $(H,+)$ be any abelian group, and let $mathbb{Z}leq Gleq mathbb{Q}$ a subgroup of the rational numbers with respect to addition. Let $varphi:Grightarrow H$ be a map with the following properties
(1) The restriction of $varphi$ to $mathbb{Z}$ is a homomorphism. (i.e. $varphi(n) = nvarphi(1)$ and $varphi(0)=0_H$)
(2) For every rational number $qin G$, we can find $a,binmathbb{Z}$ such that $q=frac{a}{b}$ and $varphi(q)^b = varphi(a)$.
Is it necessarily true that $varphi$ is an (additive) homomorphism. That is for all $q,pin G$ we have $varphi(p+q) = varphi(p) + varphi(q)$?
group-theory rational-numbers group-homomorphism
group-theory rational-numbers group-homomorphism
asked Nov 4 at 14:38
Yanko
5,750723
asked Nov 4 at 14:38
Yanko
5,750723
asked Nov 4 at 14:38
Yanko
5,750723
asked Nov 4 at 14:38
Yanko
5,750723
asked Nov 4 at 14:38
Yanko
5,750723
5,750723
I might need help with the title as well. Any ideas?
– Yanko
Nov 4 at 14:42
add a comment |
I might need help with the title as well. Any ideas?
– Yanko
Nov 4 at 14:42
I might need help with the title as well. Any ideas?
– Yanko
Nov 4 at 14:42
I might need help with the title as well. Any ideas?
– Yanko
Nov 4 at 14:42
add a comment |
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You should have written $bvarphi(q)$ instead of $varphi(q)^b$ in (2) as you write groups additively.
The answer is negative (even if we replace "we can find" with "for any").
Take $G={n/3 : ninmathbb{Z}}$, $H=mathbb{Z}times(mathbb{Z}/3mathbb{Z})$ (the second "multiplier" is ${0,1,2}$ with addition modulo $3$) and $varphi(n/3)=(n,d(n))$, where $d(n)=0$ if $3|n$ and $d(n)=1$ otherwise. Then $varphi$ meets (1) and (2), but $$(2,1)=varphi(2/3)neqvarphi(1/3)+varphi(1/3)=(2,2).$$
answered Nov 4 at 16:05
metamorphy
3,0071520
1
$Bbb{Z}[1/3]$ is also not good, since that conventionally means ${n/3^k:ninBbb{Z},kinBbb{N}}$. Instead I would suggest $frac{1}{3}Bbb{Z}$. Nonetheless, good answer +1.
– jgon
Nov 4 at 16:27
@jgon: Everywhere I look $A[a]$ means "the extension generated by $A$ and $a$". As our context is additive groups (not multiplicative groups or fields), this actually looks good I think. (Oh, maybe I should just remove it... ;)
– metamorphy
Nov 4 at 16:34
@metamorphy While I agree that that's what is meant in words by that notation, I've never seen it used in the context of abelian groups, and it definitely surprised me when I read what set it was meant to denote. (That being said I am very much a rings person, so that might just be me.)
– jgon
Nov 4 at 16:37
add a comment |
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1 Answer
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You should have written $bvarphi(q)$ instead of $varphi(q)^b$ in (2) as you write groups additively.
The answer is negative (even if we replace "we can find" with "for any").
Take $G={n/3 : ninmathbb{Z}}$, $H=mathbb{Z}times(mathbb{Z}/3mathbb{Z})$ (the second "multiplier" is ${0,1,2}$ with addition modulo $3$) and $varphi(n/3)=(n,d(n))$, where $d(n)=0$ if $3|n$ and $d(n)=1$ otherwise. Then $varphi$ meets (1) and (2), but $$(2,1)=varphi(2/3)neqvarphi(1/3)+varphi(1/3)=(2,2).$$
answered Nov 4 at 16:05
metamorphy
3,0071520
1
$Bbb{Z}[1/3]$ is also not good, since that conventionally means ${n/3^k:ninBbb{Z},kinBbb{N}}$. Instead I would suggest $frac{1}{3}Bbb{Z}$. Nonetheless, good answer +1.
– jgon
Nov 4 at 16:27
@jgon: Everywhere I look $A[a]$ means "the extension generated by $A$ and $a$". As our context is additive groups (not multiplicative groups or fields), this actually looks good I think. (Oh, maybe I should just remove it... ;)
– metamorphy
Nov 4 at 16:34
@metamorphy While I agree that that's what is meant in words by that notation, I've never seen it used in the context of abelian groups, and it definitely surprised me when I read what set it was meant to denote. (That being said I am very much a rings person, so that might just be me.)
– jgon
Nov 4 at 16:37
add a comment |
up vote
3
down vote
accepted
You should have written $bvarphi(q)$ instead of $varphi(q)^b$ in (2) as you write groups additively.
The answer is negative (even if we replace "we can find" with "for any").
Take $G={n/3 : ninmathbb{Z}}$, $H=mathbb{Z}times(mathbb{Z}/3mathbb{Z})$ (the second "multiplier" is ${0,1,2}$ with addition modulo $3$) and $varphi(n/3)=(n,d(n))$, where $d(n)=0$ if $3|n$ and $d(n)=1$ otherwise. Then $varphi$ meets (1) and (2), but $$(2,1)=varphi(2/3)neqvarphi(1/3)+varphi(1/3)=(2,2).$$
answered Nov 4 at 16:05
metamorphy
3,0071520
1
$Bbb{Z}[1/3]$ is also not good, since that conventionally means ${n/3^k:ninBbb{Z},kinBbb{N}}$. Instead I would suggest $frac{1}{3}Bbb{Z}$. Nonetheless, good answer +1.
– jgon
Nov 4 at 16:27
@jgon: Everywhere I look $A[a]$ means "the extension generated by $A$ and $a$". As our context is additive groups (not multiplicative groups or fields), this actually looks good I think. (Oh, maybe I should just remove it... ;)
– metamorphy
Nov 4 at 16:34
@metamorphy While I agree that that's what is meant in words by that notation, I've never seen it used in the context of abelian groups, and it definitely surprised me when I read what set it was meant to denote. (That being said I am very much a rings person, so that might just be me.)
– jgon
Nov 4 at 16:37
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You should have written $bvarphi(q)$ instead of $varphi(q)^b$ in (2) as you write groups additively.
The answer is negative (even if we replace "we can find" with "for any").
Take $G={n/3 : ninmathbb{Z}}$, $H=mathbb{Z}times(mathbb{Z}/3mathbb{Z})$ (the second "multiplier" is ${0,1,2}$ with addition modulo $3$) and $varphi(n/3)=(n,d(n))$, where $d(n)=0$ if $3|n$ and $d(n)=1$ otherwise. Then $varphi$ meets (1) and (2), but $$(2,1)=varphi(2/3)neqvarphi(1/3)+varphi(1/3)=(2,2).$$
answered Nov 4 at 16:05
metamorphy
3,0071520
You should have written $bvarphi(q)$ instead of $varphi(q)^b$ in (2) as you write groups additively.
The answer is negative (even if we replace "we can find" with "for any").
Take $G={n/3 : ninmathbb{Z}}$, $H=mathbb{Z}times(mathbb{Z}/3mathbb{Z})$ (the second "multiplier" is ${0,1,2}$ with addition modulo $3$) and $varphi(n/3)=(n,d(n))$, where $d(n)=0$ if $3|n$ and $d(n)=1$ otherwise. Then $varphi$ meets (1) and (2), but $$(2,1)=varphi(2/3)neqvarphi(1/3)+varphi(1/3)=(2,2).$$
answered Nov 4 at 16:05
metamorphy
3,0071520
edited Nov 29 at 3:09
answered Nov 4 at 16:05
metamorphy
3,0071520
answered Nov 4 at 16:05
metamorphy
3,0071520
answered Nov 4 at 16:05
metamorphy
3,0071520
3,0071520
1
$Bbb{Z}[1/3]$ is also not good, since that conventionally means ${n/3^k:ninBbb{Z},kinBbb{N}}$. Instead I would suggest $frac{1}{3}Bbb{Z}$. Nonetheless, good answer +1.
– jgon
Nov 4 at 16:27
@jgon: Everywhere I look $A[a]$ means "the extension generated by $A$ and $a$". As our context is additive groups (not multiplicative groups or fields), this actually looks good I think. (Oh, maybe I should just remove it... ;)
– metamorphy
Nov 4 at 16:34
@metamorphy While I agree that that's what is meant in words by that notation, I've never seen it used in the context of abelian groups, and it definitely surprised me when I read what set it was meant to denote. (That being said I am very much a rings person, so that might just be me.)
– jgon
Nov 4 at 16:37
add a comment |
1
$Bbb{Z}[1/3]$ is also not good, since that conventionally means ${n/3^k:ninBbb{Z},kinBbb{N}}$. Instead I would suggest $frac{1}{3}Bbb{Z}$. Nonetheless, good answer +1.
– jgon
Nov 4 at 16:27
@jgon: Everywhere I look $A[a]$ means "the extension generated by $A$ and $a$". As our context is additive groups (not multiplicative groups or fields), this actually looks good I think. (Oh, maybe I should just remove it... ;)
– metamorphy
Nov 4 at 16:34
@metamorphy While I agree that that's what is meant in words by that notation, I've never seen it used in the context of abelian groups, and it definitely surprised me when I read what set it was meant to denote. (That being said I am very much a rings person, so that might just be me.)
– jgon
Nov 4 at 16:37
1
1
$Bbb{Z}[1/3]$ is also not good, since that conventionally means ${n/3^k:ninBbb{Z},kinBbb{N}}$. Instead I would suggest $frac{1}{3}Bbb{Z}$. Nonetheless, good answer +1.
– jgon
Nov 4 at 16:27
$Bbb{Z}[1/3]$ is also not good, since that conventionally means ${n/3^k:ninBbb{Z},kinBbb{N}}$. Instead I would suggest $frac{1}{3}Bbb{Z}$. Nonetheless, good answer +1.
– jgon
Nov 4 at 16:27
@jgon: Everywhere I look $A[a]$ means "the extension generated by $A$ and $a$". As our context is additive groups (not multiplicative groups or fields), this actually looks good I think. (Oh, maybe I should just remove it... ;)
– metamorphy
Nov 4 at 16:34
@jgon: Everywhere I look $A[a]$ means "the extension generated by $A$ and $a$". As our context is additive groups (not multiplicative groups or fields), this actually looks good I think. (Oh, maybe I should just remove it... ;)
– metamorphy
Nov 4 at 16:34
@metamorphy While I agree that that's what is meant in words by that notation, I've never seen it used in the context of abelian groups, and it definitely surprised me when I read what set it was meant to denote. (That being said I am very much a rings person, so that might just be me.)
– jgon
Nov 4 at 16:37
@metamorphy While I agree that that's what is meant in words by that notation, I've never seen it used in the context of abelian groups, and it definitely surprised me when I read what set it was meant to denote. (That being said I am very much a rings person, so that might just be me.)
– jgon
Nov 4 at 16:37
add a comment |
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I might need help with the title as well. Any ideas?
– Yanko
Nov 4 at 14:42