finite abelian groups tensor product.
$begingroup$
Is the following question obvious ?
Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
$Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?
abstract-algebra group-theory finite-groups tensor-products abelian-groups
asked 4 hours ago
lablab
183
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$endgroup$
add a comment |
$begingroup$
Is the following question obvious ?
Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
$Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?
abstract-algebra group-theory finite-groups tensor-products abelian-groups
asked 4 hours ago
lablab
183
New contributor
lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Is the following question obvious ?
Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
$Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?
abstract-algebra group-theory finite-groups tensor-products abelian-groups
asked 4 hours ago
lablab
183
New contributor
lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Is the following question obvious ?
Let $G$ be an abelian group, such that for any finite abelian group $A$, we have
$Gotimes_{mathbf{Z}}A=0$, does it mean that $G$ is a $mathbf{Q}$-vector space ?
abstract-algebra group-theory finite-groups tensor-products abelian-groups
abstract-algebra group-theory finite-groups tensor-products abelian-groups
asked 4 hours ago
lablab
183
New contributor
lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
lablab
183
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lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 4 hours ago
lablab
183
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asked 4 hours ago
lablab
183
asked 4 hours ago
lablab
183
183
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$begingroup$
You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
$nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
$Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
Abelian groups.
But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
As an example, let $G=Bbb Q/Bbb Z$.
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
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$begingroup$
You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
$nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
$Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
Abelian groups.
But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
As an example, let $G=Bbb Q/Bbb Z$.
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
$nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
$Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
Abelian groups.
But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
As an example, let $G=Bbb Q/Bbb Z$.
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
$nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
$Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
Abelian groups.
But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
As an example, let $G=Bbb Q/Bbb Z$.
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
You have $Gotimes(Bbb Z/nBbb Z)=0$ for all
$nin Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
$Gotimes(Bbb Z/nBbb Z)=0$ and so $Gotimes A=0$ for all finitely generated
Abelian groups.
But not all divisible Abelian groups are $Bbb Q$-modules: they may have torsion.
As an example, let $G=Bbb Q/Bbb Z$.
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 4 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
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