Understanding rules on an Abelian group decomposition
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In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!
1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?
2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)
3- And if writing them as powers of primes so why following the below steps?
abstract-algebra group-theory abelian-groups
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In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!
1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?
2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)
3- And if writing them as powers of primes so why following the below steps?
abstract-algebra group-theory abelian-groups
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reputation from 72D ending in 6 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
add a comment |
up vote
1
down vote
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up vote
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In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!
1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?
2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)
3- And if writing them as powers of primes so why following the below steps?
abstract-algebra group-theory abelian-groups
In Aluffi's book Algebra, just from possibility of writing an abelian group G isomorphic to $langle grangle oplus G/langle grangle$ it concludes that by induction $G cong mathbb{Z}/d_1mathbb{Z} oplus dots mathbb{Z}/d_nmathbb{Z}$ with specific rules on $d_i$'s!
1- So why not $G cong mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_1mathbb{Z} dots mathbb{Z}/p_1mathbb{Z} oplus mathbb{Z}/p_2mathbb{Z} dots mathbb{Z}/p_2mathbb{Z} oplus dots dots dots mathbb{Z}/p_nmathbb{Z}$?
2- Why $d_i|d_{i+1}$? (a must/theorem or a choice/standard-rule?)
3- And if writing them as powers of primes so why following the below steps?
abstract-algebra group-theory abelian-groups
abstract-algebra group-theory abelian-groups
edited 2 days ago
Arturo Magidin
259k32581902
259k32581902
asked Nov 30 at 0:26
72D
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48716
This question has an open bounty worth +50
reputation from 72D ending in 6 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
This question has an open bounty worth +50
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I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.
1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.
2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds
- Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.
But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
– André 3000
17 hours ago
@André3000 Yes you're right. I'll edit.
– Joel Pereira
12 hours ago
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.
1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.
2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds
- Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.
But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
– André 3000
17 hours ago
@André3000 Yes you're right. I'll edit.
– Joel Pereira
12 hours ago
add a comment |
up vote
0
down vote
I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.
1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.
2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds
- Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.
But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
– André 3000
17 hours ago
@André3000 Yes you're right. I'll edit.
– Joel Pereira
12 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.
1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.
2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds
- Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.
I'm assuming you're talking about finite abelian groups, otherwise one would need a free group $mathbb{Z}^r$ in the decomposition.
1) We may not have necessarily have prime orders. Note $<g> cong mathbb{Z}/dmathbb{Z}$ if $g$ has order $d$ in $G$. As an example, $mathbb{Z}/4mathbb{Z} notcong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/2mathbb{Z}$.
2) The order just gives a standardized way of representing the group. One can show that the representation is unique up to order of the terms in the direct sum. If one writes the invariant factors in this way, there is an algorithm to compute the elementary divisors. Similarly, given the elementary divisors, the algorithm gives the invariant factors in such a way that the divisibility condition holds
- Elementary divisors and invariant factors are different things. The elementary divisors are a decomposition of the group into $primary$ ideals, that is power of prime ideals. One can determine these primary ideals given the invariant factors using the process described above.
edited 11 hours ago
the_fox
2,2791430
2,2791430
answered 20 hours ago
Joel Pereira
43618
43618
But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
– André 3000
17 hours ago
@André3000 Yes you're right. I'll edit.
– Joel Pereira
12 hours ago
add a comment |
But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
– André 3000
17 hours ago
@André3000 Yes you're right. I'll edit.
– Joel Pereira
12 hours ago
But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
– André 3000
17 hours ago
But $mathbb{Z}/6mathbb{Z} cong mathbb{Z}/2mathbb{Z} oplus mathbb{Z}/3mathbb{Z}$. A better example might be $mathbb{Z}/4mathbb{Z}$.
– André 3000
17 hours ago
@André3000 Yes you're right. I'll edit.
– Joel Pereira
12 hours ago
@André3000 Yes you're right. I'll edit.
– Joel Pereira
12 hours ago
add a comment |
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