Injective map of abelian group and product of cyclic quotientsEvery cyclic group $G$ is abelian; is every abelian group $G$ cyclic?Examples of non-cyclic group with a cyclic automorphism groupFinite abelian groups in which quotients of same order are isomorphicHow can there exist an isomorphism between this group and the cyclic group $(mathbb Z,+)$?A question about cyclic Abelian groupAbelian group is not cyclicProblem related to cyclic and abelian groupGroups whose proper quotients are cyclicHomomorphism Between Non-Abelian Group and Abelian GroupEvery Abelian group with order a product of two different primes is cyclic.
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Injective map of abelian group and product of cyclic quotients
Every cyclic group $G$ is abelian; is every abelian group $G$ cyclic?Examples of non-cyclic group with a cyclic automorphism groupFinite abelian groups in which quotients of same order are isomorphicHow can there exist an isomorphism between this group and the cyclic group $(mathbb Z,+)$?A question about cyclic Abelian groupAbelian group is not cyclicProblem related to cyclic and abelian groupGroups whose proper quotients are cyclicHomomorphism Between Non-Abelian Group and Abelian GroupEvery Abelian group with order a product of two different primes is cyclic.
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$begingroup$
Let $A$ be an abelian group. We have a map from $A to prod(A/I)$ where $A/I$ varies over the cyclic quotients of $A$. This map is given by sending $x$ to $prod (x text mod I)$. Is this map injective?
abstract-algebra group-theory
asked 9 hours ago
M. WangM. Wang
365 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $A$ be an abelian group. We have a map from $A to prod(A/I)$ where $A/I$ varies over the cyclic quotients of $A$. This map is given by sending $x$ to $prod (x text mod I)$. Is this map injective?
abstract-algebra group-theory
asked 9 hours ago
M. WangM. Wang
365 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $A$ be an abelian group. We have a map from $A to prod(A/I)$ where $A/I$ varies over the cyclic quotients of $A$. This map is given by sending $x$ to $prod (x text mod I)$. Is this map injective?
abstract-algebra group-theory
asked 9 hours ago
M. WangM. Wang
365 bronze badges
$endgroup$
Let $A$ be an abelian group. We have a map from $A to prod(A/I)$ where $A/I$ varies over the cyclic quotients of $A$. This map is given by sending $x$ to $prod (x text mod I)$. Is this map injective?
abstract-algebra group-theory
abstract-algebra group-theory
asked 9 hours ago
M. WangM. Wang
365 bronze badges
asked 9 hours ago
M. WangM. Wang
365 bronze badges
asked 9 hours ago
M. WangM. Wang
365 bronze badges
asked 9 hours ago
M. WangM. Wang
365 bronze badges
asked 9 hours ago
M. WangM. Wang
365 bronze badges
365 bronze badges
add a comment |
add a comment |
2 Answers
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No.
Let $A=mathbbQ$. Since $A$ is divisible, all quotients of $A$ are. In particular, $A$ has no nontrivial cyclic quotients. So $prod (A/I)$ is trivial.
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
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To complement @user10354138's answer:
It's true iff $A$ is residually finite.
If $A$ is a torsion-free abelian group, it is residually finite if and only if it does not contain any subgroup isomorphic to $mathbfQ$. Actually, a torsion-free abelian group always decomposes (non-canonically) as direct sum $A_mathrmdivoplus B$, where $A_mathrmdiv$ is its subgroup of divisible elements (this is a vector space over $mathbfQ$), and a residually finite abelian group $B$.
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
$endgroup$
$begingroup$
Thank you! I just found out that it's also assumed that A is a profinite group in my notes. Then it works.
$endgroup$
– M. Wang
5 hours ago
$begingroup$
In this case you can even consider the embedding into the product of cyclic Hausdorff quotients (i.e., those finite cyclic quotients for which the kernel is an open subgroup), and get a closed embedding, which is a homeomorphism onto its image.
$endgroup$
– YCor
5 hours ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No.
Let $A=mathbbQ$. Since $A$ is divisible, all quotients of $A$ are. In particular, $A$ has no nontrivial cyclic quotients. So $prod (A/I)$ is trivial.
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
$endgroup$
add a comment |
$begingroup$
No.
Let $A=mathbbQ$. Since $A$ is divisible, all quotients of $A$ are. In particular, $A$ has no nontrivial cyclic quotients. So $prod (A/I)$ is trivial.
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
$endgroup$
add a comment |
$begingroup$
No.
Let $A=mathbbQ$. Since $A$ is divisible, all quotients of $A$ are. In particular, $A$ has no nontrivial cyclic quotients. So $prod (A/I)$ is trivial.
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
$endgroup$
No.
Let $A=mathbbQ$. Since $A$ is divisible, all quotients of $A$ are. In particular, $A$ has no nontrivial cyclic quotients. So $prod (A/I)$ is trivial.
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
answered 8 hours ago
user10354138user10354138
17.8k2 gold badges12 silver badges32 bronze badges
17.8k2 gold badges12 silver badges32 bronze badges
add a comment |
add a comment |
$begingroup$
To complement @user10354138's answer:
It's true iff $A$ is residually finite.
If $A$ is a torsion-free abelian group, it is residually finite if and only if it does not contain any subgroup isomorphic to $mathbfQ$. Actually, a torsion-free abelian group always decomposes (non-canonically) as direct sum $A_mathrmdivoplus B$, where $A_mathrmdiv$ is its subgroup of divisible elements (this is a vector space over $mathbfQ$), and a residually finite abelian group $B$.
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
$endgroup$
$begingroup$
Thank you! I just found out that it's also assumed that A is a profinite group in my notes. Then it works.
$endgroup$
– M. Wang
5 hours ago
$begingroup$
In this case you can even consider the embedding into the product of cyclic Hausdorff quotients (i.e., those finite cyclic quotients for which the kernel is an open subgroup), and get a closed embedding, which is a homeomorphism onto its image.
$endgroup$
– YCor
5 hours ago
add a comment |
$begingroup$
To complement @user10354138's answer:
It's true iff $A$ is residually finite.
If $A$ is a torsion-free abelian group, it is residually finite if and only if it does not contain any subgroup isomorphic to $mathbfQ$. Actually, a torsion-free abelian group always decomposes (non-canonically) as direct sum $A_mathrmdivoplus B$, where $A_mathrmdiv$ is its subgroup of divisible elements (this is a vector space over $mathbfQ$), and a residually finite abelian group $B$.
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
$endgroup$
$begingroup$
Thank you! I just found out that it's also assumed that A is a profinite group in my notes. Then it works.
$endgroup$
– M. Wang
5 hours ago
$begingroup$
In this case you can even consider the embedding into the product of cyclic Hausdorff quotients (i.e., those finite cyclic quotients for which the kernel is an open subgroup), and get a closed embedding, which is a homeomorphism onto its image.
$endgroup$
– YCor
5 hours ago
add a comment |
$begingroup$
To complement @user10354138's answer:
It's true iff $A$ is residually finite.
If $A$ is a torsion-free abelian group, it is residually finite if and only if it does not contain any subgroup isomorphic to $mathbfQ$. Actually, a torsion-free abelian group always decomposes (non-canonically) as direct sum $A_mathrmdivoplus B$, where $A_mathrmdiv$ is its subgroup of divisible elements (this is a vector space over $mathbfQ$), and a residually finite abelian group $B$.
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
$endgroup$
To complement @user10354138's answer:
It's true iff $A$ is residually finite.
If $A$ is a torsion-free abelian group, it is residually finite if and only if it does not contain any subgroup isomorphic to $mathbfQ$. Actually, a torsion-free abelian group always decomposes (non-canonically) as direct sum $A_mathrmdivoplus B$, where $A_mathrmdiv$ is its subgroup of divisible elements (this is a vector space over $mathbfQ$), and a residually finite abelian group $B$.
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
answered 6 hours ago
YCorYCor
9,51613 silver badges31 bronze badges
9,51613 silver badges31 bronze badges
$begingroup$
Thank you! I just found out that it's also assumed that A is a profinite group in my notes. Then it works.
$endgroup$
– M. Wang
5 hours ago
$begingroup$
In this case you can even consider the embedding into the product of cyclic Hausdorff quotients (i.e., those finite cyclic quotients for which the kernel is an open subgroup), and get a closed embedding, which is a homeomorphism onto its image.
$endgroup$
– YCor
5 hours ago
add a comment |
$begingroup$
Thank you! I just found out that it's also assumed that A is a profinite group in my notes. Then it works.
$endgroup$
– M. Wang
5 hours ago
$begingroup$
In this case you can even consider the embedding into the product of cyclic Hausdorff quotients (i.e., those finite cyclic quotients for which the kernel is an open subgroup), and get a closed embedding, which is a homeomorphism onto its image.
$endgroup$
– YCor
5 hours ago
$begingroup$
Thank you! I just found out that it's also assumed that A is a profinite group in my notes. Then it works.
$endgroup$
– M. Wang
5 hours ago
$begingroup$
Thank you! I just found out that it's also assumed that A is a profinite group in my notes. Then it works.
$endgroup$
– M. Wang
5 hours ago
$begingroup$
In this case you can even consider the embedding into the product of cyclic Hausdorff quotients (i.e., those finite cyclic quotients for which the kernel is an open subgroup), and get a closed embedding, which is a homeomorphism onto its image.
$endgroup$
– YCor
5 hours ago
$begingroup$
In this case you can even consider the embedding into the product of cyclic Hausdorff quotients (i.e., those finite cyclic quotients for which the kernel is an open subgroup), and get a closed embedding, which is a homeomorphism onto its image.
$endgroup$
– YCor
5 hours ago
add a comment |
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