If $bigoplus_i=1^infty mathbb Z cong Y oplus mathbb Z$, is $Y cong bigoplus_i=1^infty mathbb Z$?$H_0(X) cong tildeH_0(X) oplus mathbbZ$$H_0(X) cong tildeH_0(X) oplus mathbbZ$ isomorphism in algebraic topologyUsing $displaystyle mathbbC[G]cong bigoplus_irreducible rhorho^dim rho$ for $S_3$Proving that $H_0(X)=tildeH_0(X)oplusmathbbZ$$tilde H_0 oplus mathbb Z =mathbb Zoplus mathbb Z$$(bigoplus_i in ImathbbZ)/H cong mathbbZ Rightarrow bigoplus_i in I mathbbZ cong H oplus mathbbZ$?Does $Aoplus mathbbZcong Boplus mathbbZ$ imply $Acong B$?Why is $H_0(X)cong tildeH_0(X)oplus mathbbZ$?$bigoplus_i=1^inftymathbbZnotcongprod_i=1^inftymathbbZ$Why is $H_0(X,x_0) ne H_0(X)$
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If $bigoplus_i=1^infty mathbb Z cong Y oplus mathbb Z$, is $Y cong bigoplus_i=1^infty mathbb Z$?
$H_0(X) cong tildeH_0(X) oplus mathbbZ$$H_0(X) cong tildeH_0(X) oplus mathbbZ$ isomorphism in algebraic topologyUsing $displaystyle mathbbC[G]cong bigoplus_irreducible rhorho^dim rho$ for $S_3$Proving that $H_0(X)=tildeH_0(X)oplusmathbbZ$$tilde H_0 oplus mathbb Z =mathbb Zoplus mathbb Z$$(bigoplus_i in ImathbbZ)/H cong mathbbZ Rightarrow bigoplus_i in I mathbbZ cong H oplus mathbbZ$?Does $Aoplus mathbbZcong Boplus mathbbZ$ imply $Acong B$?Why is $H_0(X)cong tildeH_0(X)oplus mathbbZ$?$bigoplus_i=1^inftymathbbZnotcongprod_i=1^inftymathbbZ$Why is $H_0(X,x_0) ne H_0(X)$
$begingroup$
Since $mathbb Q$ is not path-connected, I know that $H_0(mathbb Q) cong bigoplus_i=1^infty mathbb Z$.
Also, $H_0(mathbb Q) cong tilde H_0(mathbb Q) oplus mathbb Z$.
So, is it true that $H_0(mathbb Q) cong tilde H_0(mathbb Q)$?
In other words, if $X cong bigoplus_i=1^infty mathbb Z$ and $X=Y oplus mathbb Z$, is $Y cong bigoplus_i=1^infty mathbb Z$?
abstract-algebra group-theory algebraic-topology direct-sum
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
$endgroup$
add a comment |
$begingroup$
Since $mathbb Q$ is not path-connected, I know that $H_0(mathbb Q) cong bigoplus_i=1^infty mathbb Z$.
Also, $H_0(mathbb Q) cong tilde H_0(mathbb Q) oplus mathbb Z$.
So, is it true that $H_0(mathbb Q) cong tilde H_0(mathbb Q)$?
In other words, if $X cong bigoplus_i=1^infty mathbb Z$ and $X=Y oplus mathbb Z$, is $Y cong bigoplus_i=1^infty mathbb Z$?
abstract-algebra group-theory algebraic-topology direct-sum
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
$endgroup$
$begingroup$
Actually, every subgroup of a free abelian group is free abelian. This implies that if $A$ is an infinitely generated free abelian group and $B$ is a subgroup of $A$ such that $A/B$ is finitely generated, then $B$ is isomorphic to $A$.
$endgroup$
– YCor
Mar 17 at 0:14
add a comment |
$begingroup$
Since $mathbb Q$ is not path-connected, I know that $H_0(mathbb Q) cong bigoplus_i=1^infty mathbb Z$.
Also, $H_0(mathbb Q) cong tilde H_0(mathbb Q) oplus mathbb Z$.
So, is it true that $H_0(mathbb Q) cong tilde H_0(mathbb Q)$?
In other words, if $X cong bigoplus_i=1^infty mathbb Z$ and $X=Y oplus mathbb Z$, is $Y cong bigoplus_i=1^infty mathbb Z$?
abstract-algebra group-theory algebraic-topology direct-sum
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
$endgroup$
Since $mathbb Q$ is not path-connected, I know that $H_0(mathbb Q) cong bigoplus_i=1^infty mathbb Z$.
Also, $H_0(mathbb Q) cong tilde H_0(mathbb Q) oplus mathbb Z$.
So, is it true that $H_0(mathbb Q) cong tilde H_0(mathbb Q)$?
In other words, if $X cong bigoplus_i=1^infty mathbb Z$ and $X=Y oplus mathbb Z$, is $Y cong bigoplus_i=1^infty mathbb Z$?
abstract-algebra group-theory algebraic-topology direct-sum
abstract-algebra group-theory algebraic-topology direct-sum
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
asked Mar 16 at 23:13
Al JebrAl Jebr
4,37443377
4,37443377
$begingroup$
Actually, every subgroup of a free abelian group is free abelian. This implies that if $A$ is an infinitely generated free abelian group and $B$ is a subgroup of $A$ such that $A/B$ is finitely generated, then $B$ is isomorphic to $A$.
$endgroup$
– YCor
Mar 17 at 0:14
add a comment |
$begingroup$
Actually, every subgroup of a free abelian group is free abelian. This implies that if $A$ is an infinitely generated free abelian group and $B$ is a subgroup of $A$ such that $A/B$ is finitely generated, then $B$ is isomorphic to $A$.
$endgroup$
– YCor
Mar 17 at 0:14
$begingroup$
Actually, every subgroup of a free abelian group is free abelian. This implies that if $A$ is an infinitely generated free abelian group and $B$ is a subgroup of $A$ such that $A/B$ is finitely generated, then $B$ is isomorphic to $A$.
$endgroup$
– YCor
Mar 17 at 0:14
$begingroup$
Actually, every subgroup of a free abelian group is free abelian. This implies that if $A$ is an infinitely generated free abelian group and $B$ is a subgroup of $A$ such that $A/B$ is finitely generated, then $B$ is isomorphic to $A$.
$endgroup$
– YCor
Mar 17 at 0:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The answer to your question about reduced homology is yes. The answer to your titular question is also yes: Since $Y$ is a direct summand of a free abelian group, it is a free abelian group. Additionally, we have that $bigopluslimits_i in mathbbN mathbbZ$ is free on the set consisting of the basis of $Y$, lets call $B$, union a disjoint element. So the cardinality of $B cup 1$ is that of $mathbbN$. Since the naturals are the lowest infinite cardinal, and clearly $B$ must be infinite, the cardinality of $B$ is that of the naturals which means $Y cong bigopluslimits_i in mathbbN mathbbZ$.
The argument can be easily adapted to show that that you can always solve for the isomorphism class of $Y$ from the isomorphism $F cong Y bigoplus mathbbZ^n$, for finite $n$ if $F$ is a free abelian group where you know its rank.
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
$endgroup$
add a comment |
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1 Answer
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$begingroup$
The answer to your question about reduced homology is yes. The answer to your titular question is also yes: Since $Y$ is a direct summand of a free abelian group, it is a free abelian group. Additionally, we have that $bigopluslimits_i in mathbbN mathbbZ$ is free on the set consisting of the basis of $Y$, lets call $B$, union a disjoint element. So the cardinality of $B cup 1$ is that of $mathbbN$. Since the naturals are the lowest infinite cardinal, and clearly $B$ must be infinite, the cardinality of $B$ is that of the naturals which means $Y cong bigopluslimits_i in mathbbN mathbbZ$.
The argument can be easily adapted to show that that you can always solve for the isomorphism class of $Y$ from the isomorphism $F cong Y bigoplus mathbbZ^n$, for finite $n$ if $F$ is a free abelian group where you know its rank.
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
$endgroup$
add a comment |
$begingroup$
The answer to your question about reduced homology is yes. The answer to your titular question is also yes: Since $Y$ is a direct summand of a free abelian group, it is a free abelian group. Additionally, we have that $bigopluslimits_i in mathbbN mathbbZ$ is free on the set consisting of the basis of $Y$, lets call $B$, union a disjoint element. So the cardinality of $B cup 1$ is that of $mathbbN$. Since the naturals are the lowest infinite cardinal, and clearly $B$ must be infinite, the cardinality of $B$ is that of the naturals which means $Y cong bigopluslimits_i in mathbbN mathbbZ$.
The argument can be easily adapted to show that that you can always solve for the isomorphism class of $Y$ from the isomorphism $F cong Y bigoplus mathbbZ^n$, for finite $n$ if $F$ is a free abelian group where you know its rank.
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
$endgroup$
add a comment |
$begingroup$
The answer to your question about reduced homology is yes. The answer to your titular question is also yes: Since $Y$ is a direct summand of a free abelian group, it is a free abelian group. Additionally, we have that $bigopluslimits_i in mathbbN mathbbZ$ is free on the set consisting of the basis of $Y$, lets call $B$, union a disjoint element. So the cardinality of $B cup 1$ is that of $mathbbN$. Since the naturals are the lowest infinite cardinal, and clearly $B$ must be infinite, the cardinality of $B$ is that of the naturals which means $Y cong bigopluslimits_i in mathbbN mathbbZ$.
The argument can be easily adapted to show that that you can always solve for the isomorphism class of $Y$ from the isomorphism $F cong Y bigoplus mathbbZ^n$, for finite $n$ if $F$ is a free abelian group where you know its rank.
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
$endgroup$
The answer to your question about reduced homology is yes. The answer to your titular question is also yes: Since $Y$ is a direct summand of a free abelian group, it is a free abelian group. Additionally, we have that $bigopluslimits_i in mathbbN mathbbZ$ is free on the set consisting of the basis of $Y$, lets call $B$, union a disjoint element. So the cardinality of $B cup 1$ is that of $mathbbN$. Since the naturals are the lowest infinite cardinal, and clearly $B$ must be infinite, the cardinality of $B$ is that of the naturals which means $Y cong bigopluslimits_i in mathbbN mathbbZ$.
The argument can be easily adapted to show that that you can always solve for the isomorphism class of $Y$ from the isomorphism $F cong Y bigoplus mathbbZ^n$, for finite $n$ if $F$ is a free abelian group where you know its rank.
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
edited Mar 16 at 23:41
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
answered Mar 16 at 23:26
Connor MalinConnor Malin
559111
559111
add a comment |
add a comment |
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$begingroup$
Actually, every subgroup of a free abelian group is free abelian. This implies that if $A$ is an infinitely generated free abelian group and $B$ is a subgroup of $A$ such that $A/B$ is finitely generated, then $B$ is isomorphic to $A$.
$endgroup$
– YCor
Mar 17 at 0:14