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A finitely generated abelian group is free abelian if its Betti number equals the number of elements in some generating set.
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Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $$BbbZ_(p_1)^r_1times BbbZ_(p_2)^r_2timesdotstimes BbbZ_(p_n)^r_ntimesunderbraceBbbZtimesBbbZdotstimesBbbZ_textr times, r : betti number$$ where $p_i$ are primes, not necessarily distinct, and $r_i$ are positive integers. The prime powers $(p_i)^r_i$ are unique.
Given the Betti number equals the number of elements in some generating set in a finitely generated abelian group $G$.
To show that $G$ is free abelian, I want to show that $G$ is torsion-free.
Let $x_1,dots,x_r$ be a generating set of $G$ and $T$ be a torsion group of $G$.
Then $G/T$ is of rank $r$ and is generated by $x_1T,dots,x_rT$. But here we are not sure that whether $x_1T,dots,x_rT$ forms a basis for $G/T$. Also I still can't get any relevant information to show that $G$ is torsion-free.
abstract-algebra group-theory
asked Aug 3 at 6:59
Alan Wang
4,018930
add a comment |Â
up vote
2
down vote
favorite
Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $$BbbZ_(p_1)^r_1times BbbZ_(p_2)^r_2timesdotstimes BbbZ_(p_n)^r_ntimesunderbraceBbbZtimesBbbZdotstimesBbbZ_textr times, r : betti number$$ where $p_i$ are primes, not necessarily distinct, and $r_i$ are positive integers. The prime powers $(p_i)^r_i$ are unique.
Given the Betti number equals the number of elements in some generating set in a finitely generated abelian group $G$.
To show that $G$ is free abelian, I want to show that $G$ is torsion-free.
Let $x_1,dots,x_r$ be a generating set of $G$ and $T$ be a torsion group of $G$.
Then $G/T$ is of rank $r$ and is generated by $x_1T,dots,x_rT$. But here we are not sure that whether $x_1T,dots,x_rT$ forms a basis for $G/T$. Also I still can't get any relevant information to show that $G$ is torsion-free.
abstract-algebra group-theory
asked Aug 3 at 6:59
Alan Wang
4,018930
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $$BbbZ_(p_1)^r_1times BbbZ_(p_2)^r_2timesdotstimes BbbZ_(p_n)^r_ntimesunderbraceBbbZtimesBbbZdotstimesBbbZ_textr times, r : betti number$$ where $p_i$ are primes, not necessarily distinct, and $r_i$ are positive integers. The prime powers $(p_i)^r_i$ are unique.
Given the Betti number equals the number of elements in some generating set in a finitely generated abelian group $G$.
To show that $G$ is free abelian, I want to show that $G$ is torsion-free.
Let $x_1,dots,x_r$ be a generating set of $G$ and $T$ be a torsion group of $G$.
Then $G/T$ is of rank $r$ and is generated by $x_1T,dots,x_rT$. But here we are not sure that whether $x_1T,dots,x_rT$ forms a basis for $G/T$. Also I still can't get any relevant information to show that $G$ is torsion-free.
abstract-algebra group-theory
asked Aug 3 at 6:59
Alan Wang
4,018930
Every finitely generated abelian group $G$ is isomorphic to a direct product of cyclic groups of the form $$BbbZ_(p_1)^r_1times BbbZ_(p_2)^r_2timesdotstimes BbbZ_(p_n)^r_ntimesunderbraceBbbZtimesBbbZdotstimesBbbZ_textr times, r : betti number$$ where $p_i$ are primes, not necessarily distinct, and $r_i$ are positive integers. The prime powers $(p_i)^r_i$ are unique.
Given the Betti number equals the number of elements in some generating set in a finitely generated abelian group $G$.
To show that $G$ is free abelian, I want to show that $G$ is torsion-free.
Let $x_1,dots,x_r$ be a generating set of $G$ and $T$ be a torsion group of $G$.
Then $G/T$ is of rank $r$ and is generated by $x_1T,dots,x_rT$. But here we are not sure that whether $x_1T,dots,x_rT$ forms a basis for $G/T$. Also I still can't get any relevant information to show that $G$ is torsion-free.
abstract-algebra group-theory
asked Aug 3 at 6:59
Alan Wang
4,018930
asked Aug 3 at 6:59
Alan Wang
4,018930
asked Aug 3 at 6:59
Alan Wang
4,018930
asked Aug 3 at 6:59
Alan Wang
4,018930
4,018930
add a comment |Â
add a comment |Â
1 Answer
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2
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You have $Gcong TtimesBbb Z^r$ where $T$ is torsion.
Then $G$ has $r$ generators
if there's an onto homomorphism $phi:Bbb Z^rto G$. We might
as well assume that $G=Ttimes Bbb Z^r$. There's the projection
map $pi_2:GtoBbb Z^r$. Then $pi_2circphi:Bbb Z^rtoBbb Z^r$
is onto. Thinking about matrices and determinants, this implies that
$pi_2circphi$ is one-to-one. If $T$ is nonzero, then there is
$ain Bbb Z^r$ such that $phi(a)ne0$ but $mphi(a)=0$ for $m>1$.
Then $phi(ma)=0$, but this is a contradiction: $ane0$ in $Bbb Z^r$
so $mane0$, but $phi$ is injective.
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
May I know what do you mean by "Thinking about matrices and determinants, this implies that $pi_2circphi$ is one-to-one"?
– Alan Wang
Aug 3 at 7:57
@AlanWang A homomorphism on $Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank....
– Lord Shark the Unknown
Aug 3 at 8:01
Just to make sure that I think correctly, do you mean that we view a homomorphism on $BbbZ^r$ as a linear transformation and apply Rank-Nullity Theorem here?
– Alan Wang
Aug 3 at 8:03
@AlanWang That's basically it: just employ some elementary linear algebra.
– Lord Shark the Unknown
Aug 3 at 8:04
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have $Gcong TtimesBbb Z^r$ where $T$ is torsion.
Then $G$ has $r$ generators
if there's an onto homomorphism $phi:Bbb Z^rto G$. We might
as well assume that $G=Ttimes Bbb Z^r$. There's the projection
map $pi_2:GtoBbb Z^r$. Then $pi_2circphi:Bbb Z^rtoBbb Z^r$
is onto. Thinking about matrices and determinants, this implies that
$pi_2circphi$ is one-to-one. If $T$ is nonzero, then there is
$ain Bbb Z^r$ such that $phi(a)ne0$ but $mphi(a)=0$ for $m>1$.
Then $phi(ma)=0$, but this is a contradiction: $ane0$ in $Bbb Z^r$
so $mane0$, but $phi$ is injective.
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
May I know what do you mean by "Thinking about matrices and determinants, this implies that $pi_2circphi$ is one-to-one"?
– Alan Wang
Aug 3 at 7:57
@AlanWang A homomorphism on $Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank....
– Lord Shark the Unknown
Aug 3 at 8:01
Just to make sure that I think correctly, do you mean that we view a homomorphism on $BbbZ^r$ as a linear transformation and apply Rank-Nullity Theorem here?
– Alan Wang
Aug 3 at 8:03
@AlanWang That's basically it: just employ some elementary linear algebra.
– Lord Shark the Unknown
Aug 3 at 8:04
add a comment |Â
up vote
2
down vote
accepted
You have $Gcong TtimesBbb Z^r$ where $T$ is torsion.
Then $G$ has $r$ generators
if there's an onto homomorphism $phi:Bbb Z^rto G$. We might
as well assume that $G=Ttimes Bbb Z^r$. There's the projection
map $pi_2:GtoBbb Z^r$. Then $pi_2circphi:Bbb Z^rtoBbb Z^r$
is onto. Thinking about matrices and determinants, this implies that
$pi_2circphi$ is one-to-one. If $T$ is nonzero, then there is
$ain Bbb Z^r$ such that $phi(a)ne0$ but $mphi(a)=0$ for $m>1$.
Then $phi(ma)=0$, but this is a contradiction: $ane0$ in $Bbb Z^r$
so $mane0$, but $phi$ is injective.
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
May I know what do you mean by "Thinking about matrices and determinants, this implies that $pi_2circphi$ is one-to-one"?
– Alan Wang
Aug 3 at 7:57
@AlanWang A homomorphism on $Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank....
– Lord Shark the Unknown
Aug 3 at 8:01
Just to make sure that I think correctly, do you mean that we view a homomorphism on $BbbZ^r$ as a linear transformation and apply Rank-Nullity Theorem here?
– Alan Wang
Aug 3 at 8:03
@AlanWang That's basically it: just employ some elementary linear algebra.
– Lord Shark the Unknown
Aug 3 at 8:04
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have $Gcong TtimesBbb Z^r$ where $T$ is torsion.
Then $G$ has $r$ generators
if there's an onto homomorphism $phi:Bbb Z^rto G$. We might
as well assume that $G=Ttimes Bbb Z^r$. There's the projection
map $pi_2:GtoBbb Z^r$. Then $pi_2circphi:Bbb Z^rtoBbb Z^r$
is onto. Thinking about matrices and determinants, this implies that
$pi_2circphi$ is one-to-one. If $T$ is nonzero, then there is
$ain Bbb Z^r$ such that $phi(a)ne0$ but $mphi(a)=0$ for $m>1$.
Then $phi(ma)=0$, but this is a contradiction: $ane0$ in $Bbb Z^r$
so $mane0$, but $phi$ is injective.
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
You have $Gcong TtimesBbb Z^r$ where $T$ is torsion.
Then $G$ has $r$ generators
if there's an onto homomorphism $phi:Bbb Z^rto G$. We might
as well assume that $G=Ttimes Bbb Z^r$. There's the projection
map $pi_2:GtoBbb Z^r$. Then $pi_2circphi:Bbb Z^rtoBbb Z^r$
is onto. Thinking about matrices and determinants, this implies that
$pi_2circphi$ is one-to-one. If $T$ is nonzero, then there is
$ain Bbb Z^r$ such that $phi(a)ne0$ but $mphi(a)=0$ for $m>1$.
Then $phi(ma)=0$, but this is a contradiction: $ane0$ in $Bbb Z^r$
so $mane0$, but $phi$ is injective.
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
answered Aug 3 at 7:22
Lord Shark the Unknown
84.1k950111
84.1k950111
May I know what do you mean by "Thinking about matrices and determinants, this implies that $pi_2circphi$ is one-to-one"?
– Alan Wang
Aug 3 at 7:57
@AlanWang A homomorphism on $Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank....
– Lord Shark the Unknown
Aug 3 at 8:01
Just to make sure that I think correctly, do you mean that we view a homomorphism on $BbbZ^r$ as a linear transformation and apply Rank-Nullity Theorem here?
– Alan Wang
Aug 3 at 8:03
@AlanWang That's basically it: just employ some elementary linear algebra.
– Lord Shark the Unknown
Aug 3 at 8:04
add a comment |Â
May I know what do you mean by "Thinking about matrices and determinants, this implies that $pi_2circphi$ is one-to-one"?
– Alan Wang
Aug 3 at 7:57
@AlanWang A homomorphism on $Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank....
– Lord Shark the Unknown
Aug 3 at 8:01
Just to make sure that I think correctly, do you mean that we view a homomorphism on $BbbZ^r$ as a linear transformation and apply Rank-Nullity Theorem here?
– Alan Wang
Aug 3 at 8:03
@AlanWang That's basically it: just employ some elementary linear algebra.
– Lord Shark the Unknown
Aug 3 at 8:04
May I know what do you mean by "Thinking about matrices and determinants, this implies that $pi_2circphi$ is one-to-one"?
– Alan Wang
Aug 3 at 7:57
May I know what do you mean by "Thinking about matrices and determinants, this implies that $pi_2circphi$ is one-to-one"?
– Alan Wang
Aug 3 at 7:57
@AlanWang A homomorphism on $Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank....
– Lord Shark the Unknown
Aug 3 at 8:01
@AlanWang A homomorphism on $Bbb Z^r$ is represented by a matrix with integer entries: this matrix must have full rank....
– Lord Shark the Unknown
Aug 3 at 8:01
Just to make sure that I think correctly, do you mean that we view a homomorphism on $BbbZ^r$ as a linear transformation and apply Rank-Nullity Theorem here?
– Alan Wang
Aug 3 at 8:03
Just to make sure that I think correctly, do you mean that we view a homomorphism on $BbbZ^r$ as a linear transformation and apply Rank-Nullity Theorem here?
– Alan Wang
Aug 3 at 8:03
@AlanWang That's basically it: just employ some elementary linear algebra.
– Lord Shark the Unknown
Aug 3 at 8:04
@AlanWang That's basically it: just employ some elementary linear algebra.
– Lord Shark the Unknown
Aug 3 at 8:04
add a comment |Â
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