Understanding rules on an Abelian group decomposition
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
edited Nov 30 at 17:46
Arturo Magidin
260k32584904
asked Nov 30 at 0:26
72D
545116
add a comment |
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
edited Nov 30 at 17:46
Arturo Magidin
260k32584904
asked Nov 30 at 0:26
72D
545116
add a comment |
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
edited Nov 30 at 17:46
Arturo Magidin
260k32584904
asked Nov 30 at 0:26
72D
545116
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 Nov 30 at 17:46
Arturo Magidin
260k32584904
asked Nov 30 at 0:26
72D
545116
edited Nov 30 at 17:46
Arturo Magidin
260k32584904
asked Nov 30 at 0:26
72D
545116
edited Nov 30 at 17:46
Arturo Magidin
260k32584904
edited Nov 30 at 17:46
Arturo Magidin
260k32584904
edited Nov 30 at 17:46
Arturo Magidin
260k32584904
260k32584904
asked Nov 30 at 0:26
72D
545116
asked Nov 30 at 0:26
72D
545116
asked Nov 30 at 0:26
72D
545116
545116
add a comment |
add a comment |
1 Answer
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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.
edited Dec 2 at 14:36
the_fox
2,41411431
answered Dec 2 at 5:42
Joel Pereira
65119
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
Dec 2 at 9:06
@André3000 Yes you're right. I'll edit.
– Joel Pereira
Dec 2 at 13:30
add a comment |
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1 Answer
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1 Answer
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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.
edited Dec 2 at 14:36
the_fox
2,41411431
answered Dec 2 at 5:42
Joel Pereira
65119
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
Dec 2 at 9:06
@André3000 Yes you're right. I'll edit.
– Joel Pereira
Dec 2 at 13:30
add a comment |
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 Dec 2 at 14:36
the_fox
2,41411431
answered Dec 2 at 5:42
Joel Pereira
65119
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
Dec 2 at 9:06
@André3000 Yes you're right. I'll edit.
– Joel Pereira
Dec 2 at 13:30
add a comment |
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 Dec 2 at 14:36
the_fox
2,41411431
answered Dec 2 at 5:42
Joel Pereira
65119
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 Dec 2 at 14:36
the_fox
2,41411431
answered Dec 2 at 5:42
Joel Pereira
65119
edited Dec 2 at 14:36
the_fox
2,41411431
edited Dec 2 at 14:36
the_fox
2,41411431
edited Dec 2 at 14:36
the_fox
2,41411431
2,41411431
answered Dec 2 at 5:42
Joel Pereira
65119
answered Dec 2 at 5:42
Joel Pereira
65119
answered Dec 2 at 5:42
Joel Pereira
65119
65119
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
Dec 2 at 9:06
@André3000 Yes you're right. I'll edit.
– Joel Pereira
Dec 2 at 13:30
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
Dec 2 at 9:06
@André3000 Yes you're right. I'll edit.
– Joel Pereira
Dec 2 at 13:30
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
Dec 2 at 9:06
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
Dec 2 at 9:06
@André3000 Yes you're right. I'll edit.
– Joel Pereira
Dec 2 at 13:30
@André3000 Yes you're right. I'll edit.
– Joel Pereira
Dec 2 at 13:30
add a comment |
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