finite abelian groups tensor product.Tensor-commutative abelian groupson finite abelian groupsProduct and...
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finite abelian groups tensor product.
Tensor-commutative abelian groupson finite abelian groupsProduct and quotient in Abelian groupsIsomorphic finite abelian groupsExplicitly computing the isomorphism class of the tensor product of two finite abelian groupsFinite rank Abelian groupIsomorphism between tensor products of abelian groupsProduct in finite abelian groupsClassification of finite rank Abelian groupsTorsion-free abelian groups, tensor product and $p$-adic integers
$begingroup$
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 7 hours ago
lablab
183
New contributor
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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 7 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 7 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 7 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 7 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 7 hours ago
lablab
183
New contributor
lab is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 7 hours ago
lablab
183
asked 7 hours ago
lablab
183
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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$begingroup$
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 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
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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 7 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 7 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 7 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 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered 7 hours ago
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
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