Proof of a lemma for finitely generated abelian groups
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In the proof of the classification theorem for finitely generated abelian groups we used the following proposition: For every abelian group with generators $x_1,...,x_n$ and relation $a_1x_1+...+a_nx_n=0$ with integer coefficients there are generators $y_1,...,y_n$ with relation $gcd(a_1,...,a_n)y_1+gcd(a_1,...,a_n)y_2=0$.
I try to understand this proposition in an example and would like to generalize this to a proof of the statement. However, I got stuck in an (simple?) example. Could someone explain to me, which $y_1,...,y_n$ I would have to choose for generators $x_1,x_2,x_3$ and relation $4x_1+8x_2+10x_3=0$?
I was able to produce the desired relation but then my choice of $y_i$ did not generate the group anymore.
abstract-algebra abelian-groups
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
asked Nov 22 at 14:47
Hectorx
6
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Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
-1
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In the proof of the classification theorem for finitely generated abelian groups we used the following proposition: For every abelian group with generators $x_1,...,x_n$ and relation $a_1x_1+...+a_nx_n=0$ with integer coefficients there are generators $y_1,...,y_n$ with relation $gcd(a_1,...,a_n)y_1+gcd(a_1,...,a_n)y_2=0$.
I try to understand this proposition in an example and would like to generalize this to a proof of the statement. However, I got stuck in an (simple?) example. Could someone explain to me, which $y_1,...,y_n$ I would have to choose for generators $x_1,x_2,x_3$ and relation $4x_1+8x_2+10x_3=0$?
I was able to produce the desired relation but then my choice of $y_i$ did not generate the group anymore.
abstract-algebra abelian-groups
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
asked Nov 22 at 14:47
Hectorx
6
New contributor
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
My first question is, can you confirm there should be the same number of $x$ and $y$ (since both have index ending in $n$)? Furthermore, your relation only needs to hold for $y_1$ and $y_2$? And the 2 coefficients in the relation are the exact same?
– NazimJ
Nov 22 at 15:43
All three is exactly true according to our lecture.
– Hectorx
Nov 22 at 15:46
Ok please also share with us the desired relation you produced (Even though they no longer generated the group)
– NazimJ
Nov 22 at 15:55
$2(2x_1)=-2(4x_2+5x_3)$ choosing $y_1=2x_2$ and $y_2=4x_2+5x_3$, $y_3=x_3$, obviously not generating $x_1$.
– Hectorx
Nov 22 at 16:08
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In the proof of the classification theorem for finitely generated abelian groups we used the following proposition: For every abelian group with generators $x_1,...,x_n$ and relation $a_1x_1+...+a_nx_n=0$ with integer coefficients there are generators $y_1,...,y_n$ with relation $gcd(a_1,...,a_n)y_1+gcd(a_1,...,a_n)y_2=0$.
I try to understand this proposition in an example and would like to generalize this to a proof of the statement. However, I got stuck in an (simple?) example. Could someone explain to me, which $y_1,...,y_n$ I would have to choose for generators $x_1,x_2,x_3$ and relation $4x_1+8x_2+10x_3=0$?
I was able to produce the desired relation but then my choice of $y_i$ did not generate the group anymore.
abstract-algebra abelian-groups
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
asked Nov 22 at 14:47
Hectorx
6
New contributor
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
In the proof of the classification theorem for finitely generated abelian groups we used the following proposition: For every abelian group with generators $x_1,...,x_n$ and relation $a_1x_1+...+a_nx_n=0$ with integer coefficients there are generators $y_1,...,y_n$ with relation $gcd(a_1,...,a_n)y_1+gcd(a_1,...,a_n)y_2=0$.
I try to understand this proposition in an example and would like to generalize this to a proof of the statement. However, I got stuck in an (simple?) example. Could someone explain to me, which $y_1,...,y_n$ I would have to choose for generators $x_1,x_2,x_3$ and relation $4x_1+8x_2+10x_3=0$?
I was able to produce the desired relation but then my choice of $y_i$ did not generate the group anymore.
abstract-algebra abelian-groups
abstract-algebra abelian-groups
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
asked Nov 22 at 14:47
Hectorx
6
New contributor
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
asked Nov 22 at 14:47
Hectorx
6
New contributor
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
edited Nov 22 at 14:56
Davide Giraudo
124k16149254
124k16149254
asked Nov 22 at 14:47
Hectorx
6
New contributor
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 22 at 14:47
Hectorx
6
asked Nov 22 at 14:47
Hectorx
6
6
New contributor
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hectorx is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
My first question is, can you confirm there should be the same number of $x$ and $y$ (since both have index ending in $n$)? Furthermore, your relation only needs to hold for $y_1$ and $y_2$? And the 2 coefficients in the relation are the exact same?
– NazimJ
Nov 22 at 15:43
All three is exactly true according to our lecture.
– Hectorx
Nov 22 at 15:46
Ok please also share with us the desired relation you produced (Even though they no longer generated the group)
– NazimJ
Nov 22 at 15:55
$2(2x_1)=-2(4x_2+5x_3)$ choosing $y_1=2x_2$ and $y_2=4x_2+5x_3$, $y_3=x_3$, obviously not generating $x_1$.
– Hectorx
Nov 22 at 16:08
add a comment |
My first question is, can you confirm there should be the same number of $x$ and $y$ (since both have index ending in $n$)? Furthermore, your relation only needs to hold for $y_1$ and $y_2$? And the 2 coefficients in the relation are the exact same?
– NazimJ
Nov 22 at 15:43
All three is exactly true according to our lecture.
– Hectorx
Nov 22 at 15:46
Ok please also share with us the desired relation you produced (Even though they no longer generated the group)
– NazimJ
Nov 22 at 15:55
$2(2x_1)=-2(4x_2+5x_3)$ choosing $y_1=2x_2$ and $y_2=4x_2+5x_3$, $y_3=x_3$, obviously not generating $x_1$.
– Hectorx
Nov 22 at 16:08
My first question is, can you confirm there should be the same number of $x$ and $y$ (since both have index ending in $n$)? Furthermore, your relation only needs to hold for $y_1$ and $y_2$? And the 2 coefficients in the relation are the exact same?
– NazimJ
Nov 22 at 15:43
My first question is, can you confirm there should be the same number of $x$ and $y$ (since both have index ending in $n$)? Furthermore, your relation only needs to hold for $y_1$ and $y_2$? And the 2 coefficients in the relation are the exact same?
– NazimJ
Nov 22 at 15:43
All three is exactly true according to our lecture.
– Hectorx
Nov 22 at 15:46
All three is exactly true according to our lecture.
– Hectorx
Nov 22 at 15:46
Ok please also share with us the desired relation you produced (Even though they no longer generated the group)
– NazimJ
Nov 22 at 15:55
Ok please also share with us the desired relation you produced (Even though they no longer generated the group)
– NazimJ
Nov 22 at 15:55
$2(2x_1)=-2(4x_2+5x_3)$ choosing $y_1=2x_2$ and $y_2=4x_2+5x_3$, $y_3=x_3$, obviously not generating $x_1$.
– Hectorx
Nov 22 at 16:08
$2(2x_1)=-2(4x_2+5x_3)$ choosing $y_1=2x_2$ and $y_2=4x_2+5x_3$, $y_3=x_3$, obviously not generating $x_1$.
– Hectorx
Nov 22 at 16:08
add a comment |
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Hectorx is a new contributor. Be nice, and check out our Code of Conduct.
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My first question is, can you confirm there should be the same number of $x$ and $y$ (since both have index ending in $n$)? Furthermore, your relation only needs to hold for $y_1$ and $y_2$? And the 2 coefficients in the relation are the exact same?
– NazimJ
Nov 22 at 15:43
All three is exactly true according to our lecture.
– Hectorx
Nov 22 at 15:46
Ok please also share with us the desired relation you produced (Even though they no longer generated the group)
– NazimJ
Nov 22 at 15:55
$2(2x_1)=-2(4x_2+5x_3)$ choosing $y_1=2x_2$ and $y_2=4x_2+5x_3$, $y_3=x_3$, obviously not generating $x_1$.
– Hectorx
Nov 22 at 16:08