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 11 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 11 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 11 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 11 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 11 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 11 hours ago
lablab
183
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asked 11 hours ago
lablab
183
asked 11 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 11 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 11 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 11 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 11 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 11 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 11 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 11 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 11 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
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