Relation between abelian and free abelian group
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I am trying to understand the basic theory of free abelian groups and let me ask you the following question. This is the excerpt of the book which I'm reading.
Lemma (Universal property of free abelian group): Let $X$- theset of generators of abelian group $F$, $|X|=n$. Then the following
statements are equivalent:
(i) $F=F^{ab}_n$ - free abelian groups and $X$- the set of its
generators.
(ii) Any mapping $varphi:Xto A$ where $A$ - abelian group
induces the homomorphism $tilde{varphi}:Fto A$.
Corollary: Any abelian group with $n$ generators is homomorphic image of some free abelian group $F^{ab}_n$.
Proof: Follows immediately from univeral property.
However, for me is not clear how it follows from lemma. But I have done in the following way: suppose that $(A,+)$ is abelian group with generators ${a_1,a_2,dots, a_n}$. Let $F^{ab}_n$ be some free abelian group with generators $X={x_1,dots,x_n}$. Consider the mapping $varphi:Xto A$ by $varphi: x_imapsto a_i$ then we can consider the homomorphism $tilde varphi: F^{ab}_nto A$ defined via $$tilde varphi:sum k_ix_i mapsto sum k_ia_i.$$
Then easy to see that $tilde varphi(F^{ab}_n)=A$.
Is my proof looks correct?
But I was not able to derive it immediately from Universal property.
Would be very grateful for help!
group-theory
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up vote
2
down vote
favorite
I am trying to understand the basic theory of free abelian groups and let me ask you the following question. This is the excerpt of the book which I'm reading.
Lemma (Universal property of free abelian group): Let $X$- theset of generators of abelian group $F$, $|X|=n$. Then the following
statements are equivalent:
(i) $F=F^{ab}_n$ - free abelian groups and $X$- the set of its
generators.
(ii) Any mapping $varphi:Xto A$ where $A$ - abelian group
induces the homomorphism $tilde{varphi}:Fto A$.
Corollary: Any abelian group with $n$ generators is homomorphic image of some free abelian group $F^{ab}_n$.
Proof: Follows immediately from univeral property.
However, for me is not clear how it follows from lemma. But I have done in the following way: suppose that $(A,+)$ is abelian group with generators ${a_1,a_2,dots, a_n}$. Let $F^{ab}_n$ be some free abelian group with generators $X={x_1,dots,x_n}$. Consider the mapping $varphi:Xto A$ by $varphi: x_imapsto a_i$ then we can consider the homomorphism $tilde varphi: F^{ab}_nto A$ defined via $$tilde varphi:sum k_ix_i mapsto sum k_ia_i.$$
Then easy to see that $tilde varphi(F^{ab}_n)=A$.
Is my proof looks correct?
But I was not able to derive it immediately from Universal property.
Would be very grateful for help!
group-theory
1
The condition (ii) seems to be completely out of way. The usual (for me) universal property (ii) states that any map $;phi: Xto A;$ , with $;A;$ any abelian group, can be uniquely extended to a homomorphism of abelian groups $;overlinephi: F_n^{ab}to A;$ ...Nothing about bijection.
– DonAntonio
Nov 17 at 22:06
@DonAntonio, Is my proof looks correct?
– ZFR
Nov 17 at 22:13
Yes, it is. Extremely easy...and useful. Delete that "bijective" there. Completely nonsense.
– DonAntonio
Nov 17 at 22:15
@DonAntonio, thanks a lot! Firstly I thought that without bijection the induced homomorphism will not be surjective.
– ZFR
Nov 17 at 22:17
If you leave the word "bijective, it'd mean $;|A|=n;$ , which of course not necessarily is true. It'd be true if you'd ask that the map be from $;X;$ to a generating set with $;n;$ elements of $;A;$ ...! Observe you also teh very important property of uniqueness ...which, btw, you did not prove as you didn't mention it. Read about this universal property anywhere in the web...even in Wikipedia
– DonAntonio
Nov 17 at 22:25
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to understand the basic theory of free abelian groups and let me ask you the following question. This is the excerpt of the book which I'm reading.
Lemma (Universal property of free abelian group): Let $X$- theset of generators of abelian group $F$, $|X|=n$. Then the following
statements are equivalent:
(i) $F=F^{ab}_n$ - free abelian groups and $X$- the set of its
generators.
(ii) Any mapping $varphi:Xto A$ where $A$ - abelian group
induces the homomorphism $tilde{varphi}:Fto A$.
Corollary: Any abelian group with $n$ generators is homomorphic image of some free abelian group $F^{ab}_n$.
Proof: Follows immediately from univeral property.
However, for me is not clear how it follows from lemma. But I have done in the following way: suppose that $(A,+)$ is abelian group with generators ${a_1,a_2,dots, a_n}$. Let $F^{ab}_n$ be some free abelian group with generators $X={x_1,dots,x_n}$. Consider the mapping $varphi:Xto A$ by $varphi: x_imapsto a_i$ then we can consider the homomorphism $tilde varphi: F^{ab}_nto A$ defined via $$tilde varphi:sum k_ix_i mapsto sum k_ia_i.$$
Then easy to see that $tilde varphi(F^{ab}_n)=A$.
Is my proof looks correct?
But I was not able to derive it immediately from Universal property.
Would be very grateful for help!
group-theory
I am trying to understand the basic theory of free abelian groups and let me ask you the following question. This is the excerpt of the book which I'm reading.
Lemma (Universal property of free abelian group): Let $X$- theset of generators of abelian group $F$, $|X|=n$. Then the following
statements are equivalent:
(i) $F=F^{ab}_n$ - free abelian groups and $X$- the set of its
generators.
(ii) Any mapping $varphi:Xto A$ where $A$ - abelian group
induces the homomorphism $tilde{varphi}:Fto A$.
Corollary: Any abelian group with $n$ generators is homomorphic image of some free abelian group $F^{ab}_n$.
Proof: Follows immediately from univeral property.
However, for me is not clear how it follows from lemma. But I have done in the following way: suppose that $(A,+)$ is abelian group with generators ${a_1,a_2,dots, a_n}$. Let $F^{ab}_n$ be some free abelian group with generators $X={x_1,dots,x_n}$. Consider the mapping $varphi:Xto A$ by $varphi: x_imapsto a_i$ then we can consider the homomorphism $tilde varphi: F^{ab}_nto A$ defined via $$tilde varphi:sum k_ix_i mapsto sum k_ia_i.$$
Then easy to see that $tilde varphi(F^{ab}_n)=A$.
Is my proof looks correct?
But I was not able to derive it immediately from Universal property.
Would be very grateful for help!
group-theory
group-theory
edited Nov 17 at 22:16
asked Nov 17 at 21:57
ZFR
4,90831337
4,90831337
1
The condition (ii) seems to be completely out of way. The usual (for me) universal property (ii) states that any map $;phi: Xto A;$ , with $;A;$ any abelian group, can be uniquely extended to a homomorphism of abelian groups $;overlinephi: F_n^{ab}to A;$ ...Nothing about bijection.
– DonAntonio
Nov 17 at 22:06
@DonAntonio, Is my proof looks correct?
– ZFR
Nov 17 at 22:13
Yes, it is. Extremely easy...and useful. Delete that "bijective" there. Completely nonsense.
– DonAntonio
Nov 17 at 22:15
@DonAntonio, thanks a lot! Firstly I thought that without bijection the induced homomorphism will not be surjective.
– ZFR
Nov 17 at 22:17
If you leave the word "bijective, it'd mean $;|A|=n;$ , which of course not necessarily is true. It'd be true if you'd ask that the map be from $;X;$ to a generating set with $;n;$ elements of $;A;$ ...! Observe you also teh very important property of uniqueness ...which, btw, you did not prove as you didn't mention it. Read about this universal property anywhere in the web...even in Wikipedia
– DonAntonio
Nov 17 at 22:25
add a comment |
1
The condition (ii) seems to be completely out of way. The usual (for me) universal property (ii) states that any map $;phi: Xto A;$ , with $;A;$ any abelian group, can be uniquely extended to a homomorphism of abelian groups $;overlinephi: F_n^{ab}to A;$ ...Nothing about bijection.
– DonAntonio
Nov 17 at 22:06
@DonAntonio, Is my proof looks correct?
– ZFR
Nov 17 at 22:13
Yes, it is. Extremely easy...and useful. Delete that "bijective" there. Completely nonsense.
– DonAntonio
Nov 17 at 22:15
@DonAntonio, thanks a lot! Firstly I thought that without bijection the induced homomorphism will not be surjective.
– ZFR
Nov 17 at 22:17
If you leave the word "bijective, it'd mean $;|A|=n;$ , which of course not necessarily is true. It'd be true if you'd ask that the map be from $;X;$ to a generating set with $;n;$ elements of $;A;$ ...! Observe you also teh very important property of uniqueness ...which, btw, you did not prove as you didn't mention it. Read about this universal property anywhere in the web...even in Wikipedia
– DonAntonio
Nov 17 at 22:25
1
1
The condition (ii) seems to be completely out of way. The usual (for me) universal property (ii) states that any map $;phi: Xto A;$ , with $;A;$ any abelian group, can be uniquely extended to a homomorphism of abelian groups $;overlinephi: F_n^{ab}to A;$ ...Nothing about bijection.
– DonAntonio
Nov 17 at 22:06
The condition (ii) seems to be completely out of way. The usual (for me) universal property (ii) states that any map $;phi: Xto A;$ , with $;A;$ any abelian group, can be uniquely extended to a homomorphism of abelian groups $;overlinephi: F_n^{ab}to A;$ ...Nothing about bijection.
– DonAntonio
Nov 17 at 22:06
@DonAntonio, Is my proof looks correct?
– ZFR
Nov 17 at 22:13
@DonAntonio, Is my proof looks correct?
– ZFR
Nov 17 at 22:13
Yes, it is. Extremely easy...and useful. Delete that "bijective" there. Completely nonsense.
– DonAntonio
Nov 17 at 22:15
Yes, it is. Extremely easy...and useful. Delete that "bijective" there. Completely nonsense.
– DonAntonio
Nov 17 at 22:15
@DonAntonio, thanks a lot! Firstly I thought that without bijection the induced homomorphism will not be surjective.
– ZFR
Nov 17 at 22:17
@DonAntonio, thanks a lot! Firstly I thought that without bijection the induced homomorphism will not be surjective.
– ZFR
Nov 17 at 22:17
If you leave the word "bijective, it'd mean $;|A|=n;$ , which of course not necessarily is true. It'd be true if you'd ask that the map be from $;X;$ to a generating set with $;n;$ elements of $;A;$ ...! Observe you also teh very important property of uniqueness ...which, btw, you did not prove as you didn't mention it. Read about this universal property anywhere in the web...even in Wikipedia
– DonAntonio
Nov 17 at 22:25
If you leave the word "bijective, it'd mean $;|A|=n;$ , which of course not necessarily is true. It'd be true if you'd ask that the map be from $;X;$ to a generating set with $;n;$ elements of $;A;$ ...! Observe you also teh very important property of uniqueness ...which, btw, you did not prove as you didn't mention it. Read about this universal property anywhere in the web...even in Wikipedia
– DonAntonio
Nov 17 at 22:25
add a comment |
1 Answer
1
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oldest
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up vote
1
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accepted
You are sort of partially reproving the universal property for this case, which is not needed.
The universal property ought to be stated as
For every function $phi: X to A$ where $A$ is an Abelian group, there is a unique homomorphism $f:F= F^{textrm{ab}}_n to A$ such that $f| X=phi$.
Now, if $A$ is Abelian group with your $n$ generators, indeed define $phi(x_i) =a_i$ when ${x_1,ldots,x_n} = X$. Then by the universal property, a homomorphism $f: F to A$ exists, extending $phi$.
As $f[X]=phi[X]$ is a set of generators for $A$, $f[F]=A$, as $f[F]$ is a subgroup containing the $a_i$, and the smallest subgroup containing the $a_i$ is $A$ (that is what it means to be a generating set). So surjectivity is almost automatic, and we're done proving the corollary.
Very nice and good answer! +1
– ZFR
Nov 17 at 23:59
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You are sort of partially reproving the universal property for this case, which is not needed.
The universal property ought to be stated as
For every function $phi: X to A$ where $A$ is an Abelian group, there is a unique homomorphism $f:F= F^{textrm{ab}}_n to A$ such that $f| X=phi$.
Now, if $A$ is Abelian group with your $n$ generators, indeed define $phi(x_i) =a_i$ when ${x_1,ldots,x_n} = X$. Then by the universal property, a homomorphism $f: F to A$ exists, extending $phi$.
As $f[X]=phi[X]$ is a set of generators for $A$, $f[F]=A$, as $f[F]$ is a subgroup containing the $a_i$, and the smallest subgroup containing the $a_i$ is $A$ (that is what it means to be a generating set). So surjectivity is almost automatic, and we're done proving the corollary.
Very nice and good answer! +1
– ZFR
Nov 17 at 23:59
add a comment |
up vote
1
down vote
accepted
You are sort of partially reproving the universal property for this case, which is not needed.
The universal property ought to be stated as
For every function $phi: X to A$ where $A$ is an Abelian group, there is a unique homomorphism $f:F= F^{textrm{ab}}_n to A$ such that $f| X=phi$.
Now, if $A$ is Abelian group with your $n$ generators, indeed define $phi(x_i) =a_i$ when ${x_1,ldots,x_n} = X$. Then by the universal property, a homomorphism $f: F to A$ exists, extending $phi$.
As $f[X]=phi[X]$ is a set of generators for $A$, $f[F]=A$, as $f[F]$ is a subgroup containing the $a_i$, and the smallest subgroup containing the $a_i$ is $A$ (that is what it means to be a generating set). So surjectivity is almost automatic, and we're done proving the corollary.
Very nice and good answer! +1
– ZFR
Nov 17 at 23:59
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You are sort of partially reproving the universal property for this case, which is not needed.
The universal property ought to be stated as
For every function $phi: X to A$ where $A$ is an Abelian group, there is a unique homomorphism $f:F= F^{textrm{ab}}_n to A$ such that $f| X=phi$.
Now, if $A$ is Abelian group with your $n$ generators, indeed define $phi(x_i) =a_i$ when ${x_1,ldots,x_n} = X$. Then by the universal property, a homomorphism $f: F to A$ exists, extending $phi$.
As $f[X]=phi[X]$ is a set of generators for $A$, $f[F]=A$, as $f[F]$ is a subgroup containing the $a_i$, and the smallest subgroup containing the $a_i$ is $A$ (that is what it means to be a generating set). So surjectivity is almost automatic, and we're done proving the corollary.
You are sort of partially reproving the universal property for this case, which is not needed.
The universal property ought to be stated as
For every function $phi: X to A$ where $A$ is an Abelian group, there is a unique homomorphism $f:F= F^{textrm{ab}}_n to A$ such that $f| X=phi$.
Now, if $A$ is Abelian group with your $n$ generators, indeed define $phi(x_i) =a_i$ when ${x_1,ldots,x_n} = X$. Then by the universal property, a homomorphism $f: F to A$ exists, extending $phi$.
As $f[X]=phi[X]$ is a set of generators for $A$, $f[F]=A$, as $f[F]$ is a subgroup containing the $a_i$, and the smallest subgroup containing the $a_i$ is $A$ (that is what it means to be a generating set). So surjectivity is almost automatic, and we're done proving the corollary.
answered Nov 17 at 23:27
Henno Brandsma
102k344108
102k344108
Very nice and good answer! +1
– ZFR
Nov 17 at 23:59
add a comment |
Very nice and good answer! +1
– ZFR
Nov 17 at 23:59
Very nice and good answer! +1
– ZFR
Nov 17 at 23:59
Very nice and good answer! +1
– ZFR
Nov 17 at 23:59
add a comment |
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The condition (ii) seems to be completely out of way. The usual (for me) universal property (ii) states that any map $;phi: Xto A;$ , with $;A;$ any abelian group, can be uniquely extended to a homomorphism of abelian groups $;overlinephi: F_n^{ab}to A;$ ...Nothing about bijection.
– DonAntonio
Nov 17 at 22:06
@DonAntonio, Is my proof looks correct?
– ZFR
Nov 17 at 22:13
Yes, it is. Extremely easy...and useful. Delete that "bijective" there. Completely nonsense.
– DonAntonio
Nov 17 at 22:15
@DonAntonio, thanks a lot! Firstly I thought that without bijection the induced homomorphism will not be surjective.
– ZFR
Nov 17 at 22:17
If you leave the word "bijective, it'd mean $;|A|=n;$ , which of course not necessarily is true. It'd be true if you'd ask that the map be from $;X;$ to a generating set with $;n;$ elements of $;A;$ ...! Observe you also teh very important property of uniqueness ...which, btw, you did not prove as you didn't mention it. Read about this universal property anywhere in the web...even in Wikipedia
– DonAntonio
Nov 17 at 22:25