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
124k16150259
asked Nov 22 at 14:47
Hectorx
161
add a comment |
up vote
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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
124k16150259
asked Nov 22 at 14:47
Hectorx
161
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
124k16150259
asked Nov 22 at 14:47
Hectorx
161
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
124k16150259
asked Nov 22 at 14:47
Hectorx
161
edited Nov 22 at 14:56
Davide Giraudo
124k16150259
asked Nov 22 at 14:47
Hectorx
161
edited Nov 22 at 14:56
Davide Giraudo
124k16150259
edited Nov 22 at 14:56
Davide Giraudo
124k16150259
edited Nov 22 at 14:56
Davide Giraudo
124k16150259
124k16150259
asked Nov 22 at 14:47
Hectorx
161
asked Nov 22 at 14:47
Hectorx
161
asked Nov 22 at 14:47
Hectorx
161
161
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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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