Show that the property holds for the group homomorphism

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We have $(mathbbQ^2,+)$ with the operation $(x,y)+(x',y')=(x+x',y+y')$. For $ain mathbbQ$ and $(x,y)in mathbbQ^2$ we have that $acdot (x,y)=(acdot x, acdot y)$.



Let $f:mathbbQ^2rightarrow mathbbQ^2$ be a group homomorphism.



I want to show that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbZ$ and $(x,y)in mathbbQ^2$.



For that do we have to write $a(x,y)$ as an addition of each component a times to use the operation that's given?



$$$$



EDIT:



I have shown that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbN_0$. To show that it holds for $mathbbZ$ we have to show that $f((−a)cdot (x,y))=−acdot f(x,y)$, or not? But how?




abstract-algebra group-theory




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  • is $ain mathbbQ$ or $mathbbZ$? One time you say the former and another the latter?
    – CPM
    Aug 16 at 9:35















up vote
-1
down vote

favorite












We have $(mathbbQ^2,+)$ with the operation $(x,y)+(x',y')=(x+x',y+y')$. For $ain mathbbQ$ and $(x,y)in mathbbQ^2$ we have that $acdot (x,y)=(acdot x, acdot y)$.



Let $f:mathbbQ^2rightarrow mathbbQ^2$ be a group homomorphism.



I want to show that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbZ$ and $(x,y)in mathbbQ^2$.



For that do we have to write $a(x,y)$ as an addition of each component a times to use the operation that's given?



$$$$



EDIT:



I have shown that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbN_0$. To show that it holds for $mathbbZ$ we have to show that $f((−a)cdot (x,y))=−acdot f(x,y)$, or not? But how?




abstract-algebra group-theory




share|cite|improve this question






















  • is $ain mathbbQ$ or $mathbbZ$? One time you say the former and another the latter?
    – CPM
    Aug 16 at 9:35













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











We have $(mathbbQ^2,+)$ with the operation $(x,y)+(x',y')=(x+x',y+y')$. For $ain mathbbQ$ and $(x,y)in mathbbQ^2$ we have that $acdot (x,y)=(acdot x, acdot y)$.



Let $f:mathbbQ^2rightarrow mathbbQ^2$ be a group homomorphism.



I want to show that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbZ$ and $(x,y)in mathbbQ^2$.



For that do we have to write $a(x,y)$ as an addition of each component a times to use the operation that's given?



$$$$



EDIT:



I have shown that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbN_0$. To show that it holds for $mathbbZ$ we have to show that $f((−a)cdot (x,y))=−acdot f(x,y)$, or not? But how?




abstract-algebra group-theory




share|cite|improve this question














We have $(mathbbQ^2,+)$ with the operation $(x,y)+(x',y')=(x+x',y+y')$. For $ain mathbbQ$ and $(x,y)in mathbbQ^2$ we have that $acdot (x,y)=(acdot x, acdot y)$.



Let $f:mathbbQ^2rightarrow mathbbQ^2$ be a group homomorphism.



I want to show that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbZ$ and $(x,y)in mathbbQ^2$.



For that do we have to write $a(x,y)$ as an addition of each component a times to use the operation that's given?



$$$$



EDIT:



I have shown that $f(acdot (x,y))=acdot f(x,y)$ for all $ain mathbbN_0$. To show that it holds for $mathbbZ$ we have to show that $f((−a)cdot (x,y))=−acdot f(x,y)$, or not? But how?





abstract-algebra group-theory





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 25 at 23:41

























asked Aug 16 at 9:27









Mary Star

2,73282056




2,73282056











  • is $ain mathbbQ$ or $mathbbZ$? One time you say the former and another the latter?
    – CPM
    Aug 16 at 9:35

















  • is $ain mathbbQ$ or $mathbbZ$? One time you say the former and another the latter?
    – CPM
    Aug 16 at 9:35
















is $ain mathbbQ$ or $mathbbZ$? One time you say the former and another the latter?
– CPM
Aug 16 at 9:35





is $ain mathbbQ$ or $mathbbZ$? One time you say the former and another the latter?
– CPM
Aug 16 at 9:35











1 Answer
1






active

oldest

votes

















up vote
2
down vote













I assume it is supposed to be the case that $ain mathbbZ$. My advice would be to think of a small example and see if you can generalize $f(3cdot (x,y))=f((x,y)+(x,y)+(x,y))=f(x,y)+f(x,y)+f(x,y)=3cdot f(x,y)$. What if it were $f((-3)cdot (x,y))$? Induction to prove it carefully.






share|cite|improve this answer




















  • So do we show by induction that it holds for every natural a and then that the negative values are symmetrical? What would be if a is rational?
    – Mary Star
    Aug 16 at 11:06










  • It works also for $a in mathbb Q$.
    – Derek Holt
    Aug 16 at 16:18










  • How can we show that for $ain mathbbQ$ ? @DerekHolt
    – Mary Star
    Aug 16 at 16:47






  • 1




    Prove it first for $a in mathbb Z$. Let $f(frac1n(a,b)) = (c,d)$. Then by the result for $a in mathbb Z$, $f(a,b) = (nc,nd)$, so $f(frac1n(a,b)) = frac1nf(a,b)$.
    – Derek Holt
    Aug 16 at 18:12






  • 1




    $0 = f(0) = f(a(x,y) + (-a)(x,y)) = f(a(x,y)) + f(-a(x,y))$.
    – Derek Holt
    2 days ago










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1 Answer
1






active

oldest

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active

oldest

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up vote
2
down vote













I assume it is supposed to be the case that $ain mathbbZ$. My advice would be to think of a small example and see if you can generalize $f(3cdot (x,y))=f((x,y)+(x,y)+(x,y))=f(x,y)+f(x,y)+f(x,y)=3cdot f(x,y)$. What if it were $f((-3)cdot (x,y))$? Induction to prove it carefully.






share|cite|improve this answer




















  • So do we show by induction that it holds for every natural a and then that the negative values are symmetrical? What would be if a is rational?
    – Mary Star
    Aug 16 at 11:06










  • It works also for $a in mathbb Q$.
    – Derek Holt
    Aug 16 at 16:18










  • How can we show that for $ain mathbbQ$ ? @DerekHolt
    – Mary Star
    Aug 16 at 16:47






  • 1




    Prove it first for $a in mathbb Z$. Let $f(frac1n(a,b)) = (c,d)$. Then by the result for $a in mathbb Z$, $f(a,b) = (nc,nd)$, so $f(frac1n(a,b)) = frac1nf(a,b)$.
    – Derek Holt
    Aug 16 at 18:12






  • 1




    $0 = f(0) = f(a(x,y) + (-a)(x,y)) = f(a(x,y)) + f(-a(x,y))$.
    – Derek Holt
    2 days ago














up vote
2
down vote













I assume it is supposed to be the case that $ain mathbbZ$. My advice would be to think of a small example and see if you can generalize $f(3cdot (x,y))=f((x,y)+(x,y)+(x,y))=f(x,y)+f(x,y)+f(x,y)=3cdot f(x,y)$. What if it were $f((-3)cdot (x,y))$? Induction to prove it carefully.






share|cite|improve this answer




















  • So do we show by induction that it holds for every natural a and then that the negative values are symmetrical? What would be if a is rational?
    – Mary Star
    Aug 16 at 11:06










  • It works also for $a in mathbb Q$.
    – Derek Holt
    Aug 16 at 16:18










  • How can we show that for $ain mathbbQ$ ? @DerekHolt
    – Mary Star
    Aug 16 at 16:47






  • 1




    Prove it first for $a in mathbb Z$. Let $f(frac1n(a,b)) = (c,d)$. Then by the result for $a in mathbb Z$, $f(a,b) = (nc,nd)$, so $f(frac1n(a,b)) = frac1nf(a,b)$.
    – Derek Holt
    Aug 16 at 18:12






  • 1




    $0 = f(0) = f(a(x,y) + (-a)(x,y)) = f(a(x,y)) + f(-a(x,y))$.
    – Derek Holt
    2 days ago












up vote
2
down vote










up vote
2
down vote









I assume it is supposed to be the case that $ain mathbbZ$. My advice would be to think of a small example and see if you can generalize $f(3cdot (x,y))=f((x,y)+(x,y)+(x,y))=f(x,y)+f(x,y)+f(x,y)=3cdot f(x,y)$. What if it were $f((-3)cdot (x,y))$? Induction to prove it carefully.






share|cite|improve this answer












I assume it is supposed to be the case that $ain mathbbZ$. My advice would be to think of a small example and see if you can generalize $f(3cdot (x,y))=f((x,y)+(x,y)+(x,y))=f(x,y)+f(x,y)+f(x,y)=3cdot f(x,y)$. What if it were $f((-3)cdot (x,y))$? Induction to prove it carefully.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 16 at 9:39









CPM

3,045921




3,045921











  • So do we show by induction that it holds for every natural a and then that the negative values are symmetrical? What would be if a is rational?
    – Mary Star
    Aug 16 at 11:06










  • It works also for $a in mathbb Q$.
    – Derek Holt
    Aug 16 at 16:18










  • How can we show that for $ain mathbbQ$ ? @DerekHolt
    – Mary Star
    Aug 16 at 16:47






  • 1




    Prove it first for $a in mathbb Z$. Let $f(frac1n(a,b)) = (c,d)$. Then by the result for $a in mathbb Z$, $f(a,b) = (nc,nd)$, so $f(frac1n(a,b)) = frac1nf(a,b)$.
    – Derek Holt
    Aug 16 at 18:12






  • 1




    $0 = f(0) = f(a(x,y) + (-a)(x,y)) = f(a(x,y)) + f(-a(x,y))$.
    – Derek Holt
    2 days ago
















  • So do we show by induction that it holds for every natural a and then that the negative values are symmetrical? What would be if a is rational?
    – Mary Star
    Aug 16 at 11:06










  • It works also for $a in mathbb Q$.
    – Derek Holt
    Aug 16 at 16:18










  • How can we show that for $ain mathbbQ$ ? @DerekHolt
    – Mary Star
    Aug 16 at 16:47






  • 1




    Prove it first for $a in mathbb Z$. Let $f(frac1n(a,b)) = (c,d)$. Then by the result for $a in mathbb Z$, $f(a,b) = (nc,nd)$, so $f(frac1n(a,b)) = frac1nf(a,b)$.
    – Derek Holt
    Aug 16 at 18:12






  • 1




    $0 = f(0) = f(a(x,y) + (-a)(x,y)) = f(a(x,y)) + f(-a(x,y))$.
    – Derek Holt
    2 days ago















So do we show by induction that it holds for every natural a and then that the negative values are symmetrical? What would be if a is rational?
– Mary Star
Aug 16 at 11:06




So do we show by induction that it holds for every natural a and then that the negative values are symmetrical? What would be if a is rational?
– Mary Star
Aug 16 at 11:06












It works also for $a in mathbb Q$.
– Derek Holt
Aug 16 at 16:18




It works also for $a in mathbb Q$.
– Derek Holt
Aug 16 at 16:18












How can we show that for $ain mathbbQ$ ? @DerekHolt
– Mary Star
Aug 16 at 16:47




How can we show that for $ain mathbbQ$ ? @DerekHolt
– Mary Star
Aug 16 at 16:47




1




1




Prove it first for $a in mathbb Z$. Let $f(frac1n(a,b)) = (c,d)$. Then by the result for $a in mathbb Z$, $f(a,b) = (nc,nd)$, so $f(frac1n(a,b)) = frac1nf(a,b)$.
– Derek Holt
Aug 16 at 18:12




Prove it first for $a in mathbb Z$. Let $f(frac1n(a,b)) = (c,d)$. Then by the result for $a in mathbb Z$, $f(a,b) = (nc,nd)$, so $f(frac1n(a,b)) = frac1nf(a,b)$.
– Derek Holt
Aug 16 at 18:12




1




1




$0 = f(0) = f(a(x,y) + (-a)(x,y)) = f(a(x,y)) + f(-a(x,y))$.
– Derek Holt
2 days ago




$0 = f(0) = f(a(x,y) + (-a)(x,y)) = f(a(x,y)) + f(-a(x,y))$.
– Derek Holt
2 days ago












 

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