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Homomorphisms from $mathbb Q$ to a finitely generated abelian group
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Let $G$ be a finitely generated abelian group. Prove that there is no non-trivial homomophism from $mathbb Q$ to $G$.
I believe this has to do with divisibility. $mathbb Q$ is divisible. The image of it under a group homomorphism must be divisible. If $G$ were finite, this would imply that the image is the trivial subgroup since any divisible finite group is trivial. But $G$ is not necessarily finite. What should the argument be like instead?
abstract-algebra group-theory group-homomorphism
asked Jul 26 at 17:58
user437309
558212
 |Â
show 2 more comments
up vote
6
down vote
favorite
Let $G$ be a finitely generated abelian group. Prove that there is no non-trivial homomophism from $mathbb Q$ to $G$.
I believe this has to do with divisibility. $mathbb Q$ is divisible. The image of it under a group homomorphism must be divisible. If $G$ were finite, this would imply that the image is the trivial subgroup since any divisible finite group is trivial. But $G$ is not necessarily finite. What should the argument be like instead?
abstract-algebra group-theory group-homomorphism
asked Jul 26 at 17:58
user437309
558212
2
Maybe you should try solving this first when $G$ is cyclic.
– Gal Porat
Jul 26 at 18:01
Also, are you aware of the classification theoren for finitely generated abelian groups?
– Gal Porat
Jul 26 at 18:08
1
@GalPorat Yes, I am aware of the classification theorem. For the cyclic case, if $G$ is infinite cyclic, then it's isomorphic to $mathbb Z$, and all its subgroups are of the form $mmathbb Z$, and such a groups are not divisible for $mne 0$. If $G$ is finite cyclic, then the only divisible subgroup is the trivial group.
– user437309
Jul 26 at 18:14
Great, so I think you should try to proceed from this point.
– Gal Porat
Jul 26 at 18:16
1
A homomorphism of $mathbbQ$ into a finite product of groups is determined by the homomorphisms into each component, and hence...
– Gal Porat
Jul 26 at 18:38
 |Â
show 2 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $G$ be a finitely generated abelian group. Prove that there is no non-trivial homomophism from $mathbb Q$ to $G$.
I believe this has to do with divisibility. $mathbb Q$ is divisible. The image of it under a group homomorphism must be divisible. If $G$ were finite, this would imply that the image is the trivial subgroup since any divisible finite group is trivial. But $G$ is not necessarily finite. What should the argument be like instead?
abstract-algebra group-theory group-homomorphism
asked Jul 26 at 17:58
user437309
558212
Let $G$ be a finitely generated abelian group. Prove that there is no non-trivial homomophism from $mathbb Q$ to $G$.
I believe this has to do with divisibility. $mathbb Q$ is divisible. The image of it under a group homomorphism must be divisible. If $G$ were finite, this would imply that the image is the trivial subgroup since any divisible finite group is trivial. But $G$ is not necessarily finite. What should the argument be like instead?
abstract-algebra group-theory group-homomorphism
asked Jul 26 at 17:58
user437309
558212
asked Jul 26 at 17:58
user437309
558212
asked Jul 26 at 17:58
user437309
558212
asked Jul 26 at 17:58
user437309
558212
558212
2
Maybe you should try solving this first when $G$ is cyclic.
– Gal Porat
Jul 26 at 18:01
Also, are you aware of the classification theoren for finitely generated abelian groups?
– Gal Porat
Jul 26 at 18:08
1
@GalPorat Yes, I am aware of the classification theorem. For the cyclic case, if $G$ is infinite cyclic, then it's isomorphic to $mathbb Z$, and all its subgroups are of the form $mmathbb Z$, and such a groups are not divisible for $mne 0$. If $G$ is finite cyclic, then the only divisible subgroup is the trivial group.
– user437309
Jul 26 at 18:14
Great, so I think you should try to proceed from this point.
– Gal Porat
Jul 26 at 18:16
1
A homomorphism of $mathbbQ$ into a finite product of groups is determined by the homomorphisms into each component, and hence...
– Gal Porat
Jul 26 at 18:38
 |Â
show 2 more comments
2
Maybe you should try solving this first when $G$ is cyclic.
– Gal Porat
Jul 26 at 18:01
Also, are you aware of the classification theoren for finitely generated abelian groups?
– Gal Porat
Jul 26 at 18:08
1
@GalPorat Yes, I am aware of the classification theorem. For the cyclic case, if $G$ is infinite cyclic, then it's isomorphic to $mathbb Z$, and all its subgroups are of the form $mmathbb Z$, and such a groups are not divisible for $mne 0$. If $G$ is finite cyclic, then the only divisible subgroup is the trivial group.
– user437309
Jul 26 at 18:14
Great, so I think you should try to proceed from this point.
– Gal Porat
Jul 26 at 18:16
1
A homomorphism of $mathbbQ$ into a finite product of groups is determined by the homomorphisms into each component, and hence...
– Gal Porat
Jul 26 at 18:38
2
2
Maybe you should try solving this first when $G$ is cyclic.
– Gal Porat
Jul 26 at 18:01
Maybe you should try solving this first when $G$ is cyclic.
– Gal Porat
Jul 26 at 18:01
Also, are you aware of the classification theoren for finitely generated abelian groups?
– Gal Porat
Jul 26 at 18:08
Also, are you aware of the classification theoren for finitely generated abelian groups?
– Gal Porat
Jul 26 at 18:08
1
1
@GalPorat Yes, I am aware of the classification theorem. For the cyclic case, if $G$ is infinite cyclic, then it's isomorphic to $mathbb Z$, and all its subgroups are of the form $mmathbb Z$, and such a groups are not divisible for $mne 0$. If $G$ is finite cyclic, then the only divisible subgroup is the trivial group.
– user437309
Jul 26 at 18:14
@GalPorat Yes, I am aware of the classification theorem. For the cyclic case, if $G$ is infinite cyclic, then it's isomorphic to $mathbb Z$, and all its subgroups are of the form $mmathbb Z$, and such a groups are not divisible for $mne 0$. If $G$ is finite cyclic, then the only divisible subgroup is the trivial group.
– user437309
Jul 26 at 18:14
Great, so I think you should try to proceed from this point.
– Gal Porat
Jul 26 at 18:16
Great, so I think you should try to proceed from this point.
– Gal Porat
Jul 26 at 18:16
1
1
A homomorphism of $mathbbQ$ into a finite product of groups is determined by the homomorphisms into each component, and hence...
– Gal Porat
Jul 26 at 18:38
A homomorphism of $mathbbQ$ into a finite product of groups is determined by the homomorphisms into each component, and hence...
– Gal Porat
Jul 26 at 18:38
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
A nonzero divisible group is not finitely generated.
Consider a homomorphism $fcolonmathbbQto G$, where $G$ is a finitely generated abelian group. If you compose it with the projection $picolon Gto G/t(G)$, where $t(G)$ is the torsion part of $G$, you get a homomorphism $picirc fcolonmathbbQto G/t(G)$. Since $G/t(G)$ is finitely generated and torsion-free, also the image of $picirc f$ is finitely generated and torsion free. But it's divisible, so it's isomorphic to $mathbbQ^n$, for some $n$. This group is not finitely generated when $n>0$. Thus $picirc f=0$ and therefore the image of $f$ is contained in $t(G)$, which is a finitely generated torsion group, hence finite.
No nonzero divisible group is finite.
answered Jul 26 at 20:30
egreg
164k1180187
add a comment |Â
up vote
2
down vote
A finite generated abelian group $G$ is isomorphic to $oplus_i=1^i=nmathbbZ/n_imathbbZ$. Let $xin G$, and $f:mathbbQ rightarrow G$ such that $f(1)=x, f(n.1over n) =nf(1over n) =x$. Write $x=(x_1,...,x_n), x_iinmathbbZ/n_imathbbZ $, $x_i$ is divisible by every non zero integer implies that $x_i=0$.
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
add a comment |Â
up vote
2
down vote
Since $G$ is f.g., it's sufficient to prove the result for $G = mathbbZ$ and $G = mathbbZ_p^n$. But neither of these groups contains a nontrivial divisible subgroup.
answered Jul 26 at 19:20
anomaly
16.3k42561
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
A nonzero divisible group is not finitely generated.
Consider a homomorphism $fcolonmathbbQto G$, where $G$ is a finitely generated abelian group. If you compose it with the projection $picolon Gto G/t(G)$, where $t(G)$ is the torsion part of $G$, you get a homomorphism $picirc fcolonmathbbQto G/t(G)$. Since $G/t(G)$ is finitely generated and torsion-free, also the image of $picirc f$ is finitely generated and torsion free. But it's divisible, so it's isomorphic to $mathbbQ^n$, for some $n$. This group is not finitely generated when $n>0$. Thus $picirc f=0$ and therefore the image of $f$ is contained in $t(G)$, which is a finitely generated torsion group, hence finite.
No nonzero divisible group is finite.
answered Jul 26 at 20:30
egreg
164k1180187
add a comment |Â
up vote
3
down vote
A nonzero divisible group is not finitely generated.
Consider a homomorphism $fcolonmathbbQto G$, where $G$ is a finitely generated abelian group. If you compose it with the projection $picolon Gto G/t(G)$, where $t(G)$ is the torsion part of $G$, you get a homomorphism $picirc fcolonmathbbQto G/t(G)$. Since $G/t(G)$ is finitely generated and torsion-free, also the image of $picirc f$ is finitely generated and torsion free. But it's divisible, so it's isomorphic to $mathbbQ^n$, for some $n$. This group is not finitely generated when $n>0$. Thus $picirc f=0$ and therefore the image of $f$ is contained in $t(G)$, which is a finitely generated torsion group, hence finite.
No nonzero divisible group is finite.
answered Jul 26 at 20:30
egreg
164k1180187
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A nonzero divisible group is not finitely generated.
Consider a homomorphism $fcolonmathbbQto G$, where $G$ is a finitely generated abelian group. If you compose it with the projection $picolon Gto G/t(G)$, where $t(G)$ is the torsion part of $G$, you get a homomorphism $picirc fcolonmathbbQto G/t(G)$. Since $G/t(G)$ is finitely generated and torsion-free, also the image of $picirc f$ is finitely generated and torsion free. But it's divisible, so it's isomorphic to $mathbbQ^n$, for some $n$. This group is not finitely generated when $n>0$. Thus $picirc f=0$ and therefore the image of $f$ is contained in $t(G)$, which is a finitely generated torsion group, hence finite.
No nonzero divisible group is finite.
answered Jul 26 at 20:30
egreg
164k1180187
A nonzero divisible group is not finitely generated.
Consider a homomorphism $fcolonmathbbQto G$, where $G$ is a finitely generated abelian group. If you compose it with the projection $picolon Gto G/t(G)$, where $t(G)$ is the torsion part of $G$, you get a homomorphism $picirc fcolonmathbbQto G/t(G)$. Since $G/t(G)$ is finitely generated and torsion-free, also the image of $picirc f$ is finitely generated and torsion free. But it's divisible, so it's isomorphic to $mathbbQ^n$, for some $n$. This group is not finitely generated when $n>0$. Thus $picirc f=0$ and therefore the image of $f$ is contained in $t(G)$, which is a finitely generated torsion group, hence finite.
No nonzero divisible group is finite.
answered Jul 26 at 20:30
egreg
164k1180187
answered Jul 26 at 20:30
egreg
164k1180187
answered Jul 26 at 20:30
egreg
164k1180187
answered Jul 26 at 20:30
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
up vote
2
down vote
A finite generated abelian group $G$ is isomorphic to $oplus_i=1^i=nmathbbZ/n_imathbbZ$. Let $xin G$, and $f:mathbbQ rightarrow G$ such that $f(1)=x, f(n.1over n) =nf(1over n) =x$. Write $x=(x_1,...,x_n), x_iinmathbbZ/n_imathbbZ $, $x_i$ is divisible by every non zero integer implies that $x_i=0$.
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
add a comment |Â
up vote
2
down vote
A finite generated abelian group $G$ is isomorphic to $oplus_i=1^i=nmathbbZ/n_imathbbZ$. Let $xin G$, and $f:mathbbQ rightarrow G$ such that $f(1)=x, f(n.1over n) =nf(1over n) =x$. Write $x=(x_1,...,x_n), x_iinmathbbZ/n_imathbbZ $, $x_i$ is divisible by every non zero integer implies that $x_i=0$.
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A finite generated abelian group $G$ is isomorphic to $oplus_i=1^i=nmathbbZ/n_imathbbZ$. Let $xin G$, and $f:mathbbQ rightarrow G$ such that $f(1)=x, f(n.1over n) =nf(1over n) =x$. Write $x=(x_1,...,x_n), x_iinmathbbZ/n_imathbbZ $, $x_i$ is divisible by every non zero integer implies that $x_i=0$.
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
A finite generated abelian group $G$ is isomorphic to $oplus_i=1^i=nmathbbZ/n_imathbbZ$. Let $xin G$, and $f:mathbbQ rightarrow G$ such that $f(1)=x, f(n.1over n) =nf(1over n) =x$. Write $x=(x_1,...,x_n), x_iinmathbbZ/n_imathbbZ $, $x_i$ is divisible by every non zero integer implies that $x_i=0$.
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
answered Jul 26 at 19:00
Tsemo Aristide
51k11143
51k11143
add a comment |Â
add a comment |Â
up vote
2
down vote
Since $G$ is f.g., it's sufficient to prove the result for $G = mathbbZ$ and $G = mathbbZ_p^n$. But neither of these groups contains a nontrivial divisible subgroup.
answered Jul 26 at 19:20
anomaly
16.3k42561
add a comment |Â
up vote
2
down vote
Since $G$ is f.g., it's sufficient to prove the result for $G = mathbbZ$ and $G = mathbbZ_p^n$. But neither of these groups contains a nontrivial divisible subgroup.
answered Jul 26 at 19:20
anomaly
16.3k42561
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since $G$ is f.g., it's sufficient to prove the result for $G = mathbbZ$ and $G = mathbbZ_p^n$. But neither of these groups contains a nontrivial divisible subgroup.
answered Jul 26 at 19:20
anomaly
16.3k42561
Since $G$ is f.g., it's sufficient to prove the result for $G = mathbbZ$ and $G = mathbbZ_p^n$. But neither of these groups contains a nontrivial divisible subgroup.
answered Jul 26 at 19:20
anomaly
16.3k42561
edited Jul 29 at 17:38
answered Jul 26 at 19:20
anomaly
16.3k42561
answered Jul 26 at 19:20
anomaly
16.3k42561
answered Jul 26 at 19:20
anomaly
16.3k42561
16.3k42561
add a comment |Â
add a comment |Â
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2
Maybe you should try solving this first when $G$ is cyclic.
– Gal Porat
Jul 26 at 18:01
Also, are you aware of the classification theoren for finitely generated abelian groups?
– Gal Porat
Jul 26 at 18:08
1
@GalPorat Yes, I am aware of the classification theorem. For the cyclic case, if $G$ is infinite cyclic, then it's isomorphic to $mathbb Z$, and all its subgroups are of the form $mmathbb Z$, and such a groups are not divisible for $mne 0$. If $G$ is finite cyclic, then the only divisible subgroup is the trivial group.
– user437309
Jul 26 at 18:14
Great, so I think you should try to proceed from this point.
– Gal Porat
Jul 26 at 18:16
1
A homomorphism of $mathbbQ$ into a finite product of groups is determined by the homomorphisms into each component, and hence...
– Gal Porat
Jul 26 at 18:38