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Every surjective homomorphism $alpha : mathbbZ^3 rightarrow mathbbZ^3$ is isomorphism [closed]
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Let $ alpha : mathbbZ^3 rightarrow mathbbZ^3$ be a surjective homomorphism. Prove that $alpha$ is isomorphism.
I'm not sure how to approach this... Why does every surjective homomorphism has to be isomorphism?
group-theory group-isomorphism group-homomorphism
asked Jul 21 at 8:14
ChikChak
617216
closed as off-topic by Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson Jul 23 at 0:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson
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up vote
2
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favorite
Let $ alpha : mathbbZ^3 rightarrow mathbbZ^3$ be a surjective homomorphism. Prove that $alpha$ is isomorphism.
I'm not sure how to approach this... Why does every surjective homomorphism has to be isomorphism?
group-theory group-isomorphism group-homomorphism
asked Jul 21 at 8:14
ChikChak
617216
closed as off-topic by Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson Jul 23 at 0:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson
3
A finitely generated free group is Hopfian.
– Alex Provost
Jul 21 at 8:41
2
@AlexProvost That does not seem particularly relevant! A finitely generated free abelian group is also Hopfian, but that is just a rewording of the problem.
– Derek Holt
Jul 21 at 14:10
@DerekHolt You are right! I merely wanted to introduce the terminology to folks unaware of such a thing. (And I was missing the word Abelian, too, although the statement above is still true, if irrelevant.)
– Alex Provost
Jul 21 at 16:34
@ChikChak: Since both domain and range of $alpha$ have equal and finite cardinality, it does imply that $alpha$ must also be injective.
– Moritz
Jul 21 at 22:28
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $ alpha : mathbbZ^3 rightarrow mathbbZ^3$ be a surjective homomorphism. Prove that $alpha$ is isomorphism.
I'm not sure how to approach this... Why does every surjective homomorphism has to be isomorphism?
group-theory group-isomorphism group-homomorphism
asked Jul 21 at 8:14
ChikChak
617216
Let $ alpha : mathbbZ^3 rightarrow mathbbZ^3$ be a surjective homomorphism. Prove that $alpha$ is isomorphism.
I'm not sure how to approach this... Why does every surjective homomorphism has to be isomorphism?
group-theory group-isomorphism group-homomorphism
asked Jul 21 at 8:14
ChikChak
617216
asked Jul 21 at 8:14
ChikChak
617216
asked Jul 21 at 8:14
ChikChak
617216
asked Jul 21 at 8:14
ChikChak
617216
617216
closed as off-topic by Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson Jul 23 at 0:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson
closed as off-topic by Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson Jul 23 at 0:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Derek Holt, amWhy, Mostafa Ayaz, Parcly Taxel, Xander Henderson
3
A finitely generated free group is Hopfian.
– Alex Provost
Jul 21 at 8:41
2
@AlexProvost That does not seem particularly relevant! A finitely generated free abelian group is also Hopfian, but that is just a rewording of the problem.
– Derek Holt
Jul 21 at 14:10
@DerekHolt You are right! I merely wanted to introduce the terminology to folks unaware of such a thing. (And I was missing the word Abelian, too, although the statement above is still true, if irrelevant.)
– Alex Provost
Jul 21 at 16:34
@ChikChak: Since both domain and range of $alpha$ have equal and finite cardinality, it does imply that $alpha$ must also be injective.
– Moritz
Jul 21 at 22:28
add a comment |Â
3
A finitely generated free group is Hopfian.
– Alex Provost
Jul 21 at 8:41
2
@AlexProvost That does not seem particularly relevant! A finitely generated free abelian group is also Hopfian, but that is just a rewording of the problem.
– Derek Holt
Jul 21 at 14:10
@DerekHolt You are right! I merely wanted to introduce the terminology to folks unaware of such a thing. (And I was missing the word Abelian, too, although the statement above is still true, if irrelevant.)
– Alex Provost
Jul 21 at 16:34
@ChikChak: Since both domain and range of $alpha$ have equal and finite cardinality, it does imply that $alpha$ must also be injective.
– Moritz
Jul 21 at 22:28
3
3
A finitely generated free group is Hopfian.
– Alex Provost
Jul 21 at 8:41
A finitely generated free group is Hopfian.
– Alex Provost
Jul 21 at 8:41
2
2
@AlexProvost That does not seem particularly relevant! A finitely generated free abelian group is also Hopfian, but that is just a rewording of the problem.
– Derek Holt
Jul 21 at 14:10
@AlexProvost That does not seem particularly relevant! A finitely generated free abelian group is also Hopfian, but that is just a rewording of the problem.
– Derek Holt
Jul 21 at 14:10
@DerekHolt You are right! I merely wanted to introduce the terminology to folks unaware of such a thing. (And I was missing the word Abelian, too, although the statement above is still true, if irrelevant.)
– Alex Provost
Jul 21 at 16:34
@DerekHolt You are right! I merely wanted to introduce the terminology to folks unaware of such a thing. (And I was missing the word Abelian, too, although the statement above is still true, if irrelevant.)
– Alex Provost
Jul 21 at 16:34
@ChikChak: Since both domain and range of $alpha$ have equal and finite cardinality, it does imply that $alpha$ must also be injective.
– Moritz
Jul 21 at 22:28
@ChikChak: Since both domain and range of $alpha$ have equal and finite cardinality, it does imply that $alpha$ must also be injective.
– Moritz
Jul 21 at 22:28
add a comment |Â
2 Answers
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accepted
Take $v_1=alpha(1,0,0)$, $v_2=alpha(0,1,0)$, and $v_3=alpha(0,0,1)$. Since $alpha$ is surjective, $mathbbZ^3=langle v_1,v_2,v_3rangle$ and therefore $v_1$, $v_2$, and $v_3$ are linearly independent over $mathbb Q$. But then$$(forallalpha_1,alpha_2,alpha_3inmathbbQ):alpha_1v_1+alpha_2v_2+alpha_3v_3=0impliesalpha_1=alpha_2=alpha_3=0.$$In other words, $keralpha=(0,0,0)$.
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
add a comment |Â
up vote
3
down vote
Here's a more general proof than the one using $mathbbQ$.
Consider $(mathrmker(alpha^n))_n$: it's an increasing sequence of subgroups of $mathbbZ^3$, therefore it is stationary : for some $n>0$, $mathrmker(alpha^2n) = mathrmker(alpha^n)$.
But $alpha^n$ is also surjective, because $alpha$ is. So if $alpha (x)=0$, $alpha^n(x)=0$. But also $x=alpha^n(y) $ for some $y$, so $alpha^2n(y)=0$, so $alpha^n(y) =0$, so $x=0$.
The only thing I used is that an increasing sequence of subgroups of $mathbbZ^3$ is stationary. So there are two interesting generalizations : this will work for any group that has this property; but also if $R$ is a noetherian ring, this will work for $R$-linear maps $R^nto R^n$.
answered Jul 21 at 9:16
Max
9,8801635
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Take $v_1=alpha(1,0,0)$, $v_2=alpha(0,1,0)$, and $v_3=alpha(0,0,1)$. Since $alpha$ is surjective, $mathbbZ^3=langle v_1,v_2,v_3rangle$ and therefore $v_1$, $v_2$, and $v_3$ are linearly independent over $mathbb Q$. But then$$(forallalpha_1,alpha_2,alpha_3inmathbbQ):alpha_1v_1+alpha_2v_2+alpha_3v_3=0impliesalpha_1=alpha_2=alpha_3=0.$$In other words, $keralpha=(0,0,0)$.
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
add a comment |Â
up vote
9
down vote
accepted
Take $v_1=alpha(1,0,0)$, $v_2=alpha(0,1,0)$, and $v_3=alpha(0,0,1)$. Since $alpha$ is surjective, $mathbbZ^3=langle v_1,v_2,v_3rangle$ and therefore $v_1$, $v_2$, and $v_3$ are linearly independent over $mathbb Q$. But then$$(forallalpha_1,alpha_2,alpha_3inmathbbQ):alpha_1v_1+alpha_2v_2+alpha_3v_3=0impliesalpha_1=alpha_2=alpha_3=0.$$In other words, $keralpha=(0,0,0)$.
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Take $v_1=alpha(1,0,0)$, $v_2=alpha(0,1,0)$, and $v_3=alpha(0,0,1)$. Since $alpha$ is surjective, $mathbbZ^3=langle v_1,v_2,v_3rangle$ and therefore $v_1$, $v_2$, and $v_3$ are linearly independent over $mathbb Q$. But then$$(forallalpha_1,alpha_2,alpha_3inmathbbQ):alpha_1v_1+alpha_2v_2+alpha_3v_3=0impliesalpha_1=alpha_2=alpha_3=0.$$In other words, $keralpha=(0,0,0)$.
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
Take $v_1=alpha(1,0,0)$, $v_2=alpha(0,1,0)$, and $v_3=alpha(0,0,1)$. Since $alpha$ is surjective, $mathbbZ^3=langle v_1,v_2,v_3rangle$ and therefore $v_1$, $v_2$, and $v_3$ are linearly independent over $mathbb Q$. But then$$(forallalpha_1,alpha_2,alpha_3inmathbbQ):alpha_1v_1+alpha_2v_2+alpha_3v_3=0impliesalpha_1=alpha_2=alpha_3=0.$$In other words, $keralpha=(0,0,0)$.
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
edited Jul 27 at 13:03
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
answered Jul 21 at 8:30
José Carlos Santos
114k1698177
114k1698177
add a comment |Â
add a comment |Â
up vote
3
down vote
Here's a more general proof than the one using $mathbbQ$.
Consider $(mathrmker(alpha^n))_n$: it's an increasing sequence of subgroups of $mathbbZ^3$, therefore it is stationary : for some $n>0$, $mathrmker(alpha^2n) = mathrmker(alpha^n)$.
But $alpha^n$ is also surjective, because $alpha$ is. So if $alpha (x)=0$, $alpha^n(x)=0$. But also $x=alpha^n(y) $ for some $y$, so $alpha^2n(y)=0$, so $alpha^n(y) =0$, so $x=0$.
The only thing I used is that an increasing sequence of subgroups of $mathbbZ^3$ is stationary. So there are two interesting generalizations : this will work for any group that has this property; but also if $R$ is a noetherian ring, this will work for $R$-linear maps $R^nto R^n$.
answered Jul 21 at 9:16
Max
9,8801635
add a comment |Â
up vote
3
down vote
Here's a more general proof than the one using $mathbbQ$.
Consider $(mathrmker(alpha^n))_n$: it's an increasing sequence of subgroups of $mathbbZ^3$, therefore it is stationary : for some $n>0$, $mathrmker(alpha^2n) = mathrmker(alpha^n)$.
But $alpha^n$ is also surjective, because $alpha$ is. So if $alpha (x)=0$, $alpha^n(x)=0$. But also $x=alpha^n(y) $ for some $y$, so $alpha^2n(y)=0$, so $alpha^n(y) =0$, so $x=0$.
The only thing I used is that an increasing sequence of subgroups of $mathbbZ^3$ is stationary. So there are two interesting generalizations : this will work for any group that has this property; but also if $R$ is a noetherian ring, this will work for $R$-linear maps $R^nto R^n$.
answered Jul 21 at 9:16
Max
9,8801635
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here's a more general proof than the one using $mathbbQ$.
Consider $(mathrmker(alpha^n))_n$: it's an increasing sequence of subgroups of $mathbbZ^3$, therefore it is stationary : for some $n>0$, $mathrmker(alpha^2n) = mathrmker(alpha^n)$.
But $alpha^n$ is also surjective, because $alpha$ is. So if $alpha (x)=0$, $alpha^n(x)=0$. But also $x=alpha^n(y) $ for some $y$, so $alpha^2n(y)=0$, so $alpha^n(y) =0$, so $x=0$.
The only thing I used is that an increasing sequence of subgroups of $mathbbZ^3$ is stationary. So there are two interesting generalizations : this will work for any group that has this property; but also if $R$ is a noetherian ring, this will work for $R$-linear maps $R^nto R^n$.
answered Jul 21 at 9:16
Max
9,8801635
Here's a more general proof than the one using $mathbbQ$.
Consider $(mathrmker(alpha^n))_n$: it's an increasing sequence of subgroups of $mathbbZ^3$, therefore it is stationary : for some $n>0$, $mathrmker(alpha^2n) = mathrmker(alpha^n)$.
But $alpha^n$ is also surjective, because $alpha$ is. So if $alpha (x)=0$, $alpha^n(x)=0$. But also $x=alpha^n(y) $ for some $y$, so $alpha^2n(y)=0$, so $alpha^n(y) =0$, so $x=0$.
The only thing I used is that an increasing sequence of subgroups of $mathbbZ^3$ is stationary. So there are two interesting generalizations : this will work for any group that has this property; but also if $R$ is a noetherian ring, this will work for $R$-linear maps $R^nto R^n$.
answered Jul 21 at 9:16
Max
9,8801635
answered Jul 21 at 9:16
Max
9,8801635
answered Jul 21 at 9:16
Max
9,8801635
answered Jul 21 at 9:16
Max
9,8801635
9,8801635
add a comment |Â
add a comment |Â
3
A finitely generated free group is Hopfian.
– Alex Provost
Jul 21 at 8:41
2
@AlexProvost That does not seem particularly relevant! A finitely generated free abelian group is also Hopfian, but that is just a rewording of the problem.
– Derek Holt
Jul 21 at 14:10
@DerekHolt You are right! I merely wanted to introduce the terminology to folks unaware of such a thing. (And I was missing the word Abelian, too, although the statement above is still true, if irrelevant.)
– Alex Provost
Jul 21 at 16:34
@ChikChak: Since both domain and range of $alpha$ have equal and finite cardinality, it does imply that $alpha$ must also be injective.
– Moritz
Jul 21 at 22:28