What Does Up to Isomorphism Really Mean?

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    Definition Isomorphism
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Discussion Overview

The discussion revolves around the classification of abelian groups of order 12 up to isomorphism, exploring what it means to list groups in this context and the implications of isomorphism in group theory.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that classifying groups of order 12 up to isomorphism requires listing groups such that every group of that order is isomorphic to one of the listed groups.
  • It is proposed that \(\mathbb{Z}_{12}\) should be included in the list of abelian groups of order 12.
  • Participants mention other abelian groups of order 12, such as \(\mathbb{Z}_{3} \times \mathbb{Z}_{4}\) and \(\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3}\).
  • There is a discussion about the isomorphism \(\mathbb{Z}_{2} \times \mathbb{Z}_{6}\) and its relation to other groups, with a claim that it is isomorphic to \(\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3}\) under certain conditions.
  • Some participants introduce the concept of the Chinese Remainder Theorem (CRT) and its relevance in proving isomorphisms when \( \text{gcd}(a,b) = 1 \).
  • There is a clarification that the coprimality of \(a\) and \(b\) is necessary for the application of the CRT in this context.

Areas of Agreement / Disagreement

Participants generally agree on the inclusion of certain groups in the classification but express differing views on the implications and conditions under which isomorphisms hold, particularly regarding the use of the Chinese Remainder Theorem.

Contextual Notes

The discussion includes assumptions about the properties of groups and the necessity of coprimality for certain isomorphism proofs, which remain unresolved in terms of their broader implications.

Bachelier
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what does it really mean?

for instance if asked to list all abelian grps of order 12 up to iso, then do we include Z12 or not?
 
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When you are asked to classify all groups of order 12 up to isomorphism, this means two things:
  • You need to create a list of groups such that every group of order 12 is isomorphic to one of these groups.
  • None of you the groups you listed are isomorphic to each other.
 
Specifically: yes, you need to include [itex]\mathbb{Z}_{12}[/itex] or some group isomorphic to it. There are more groups of order 12 though...
 
Yes thanks I get it now.

Hence for ABELIAN groups of order 12 we have [itex]\mathbb{Z}_{12}[/itex] ≈ [itex]\mathbb{Z}_{3}[/itex] X [itex]\mathbb{Z}_{4}[/itex], and [itex]\mathbb{Z}_{2}[/itex] X [itex]\mathbb{Z}_{2}[/itex] X [itex]\mathbb{Z}_{3}[/itex]

what would be [itex]\mathbb{Z}_{2}[/itex] X [itex]\mathbb{Z}_{6}[/itex] isomorphic to?
 
Bachelier said:
Yes thanks I get it now.

Hence for ABELIAN groups of order 12 we have [itex]\mathbb{Z}_{12}[/itex] ≈ [itex]\mathbb{Z}_{3}[/itex] X [itex]\mathbb{Z}_{4}[/itex], and [itex]\mathbb{Z}_{2}[/itex] X [itex]\mathbb{Z}_{2}[/itex] X [itex]\mathbb{Z}_{3}[/itex]

what would be [itex]\mathbb{Z}_{2}[/itex] X [itex]\mathbb{Z}_{6}[/itex] isomorphic to?

In general: if gcd(a,b)=1, then [itex]\mathbb{Z}_{ab}\cong \mathbb{Z}_a\times \mathbb{Z}_b[/itex] (try to prove this!).

So we would have [itex]\mathbb{Z}_2\times \mathbb{Z}_6\cong \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3[/itex].
 
micromass said:
In general: if gcd(a,b)=1, then [itex]\mathbb{Z}_{ab}\cong \mathbb{Z}_a\times \mathbb{Z}_b[/itex] (try to prove this!).

So we would have [itex]\mathbb{Z}_2\times \mathbb{Z}_6\cong \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_3[/itex].

we can construct an isomorphism ψ: from [itex]\mathbb{Z}_{ab}[/itex] to [itex]\mathbb{Z}_a\times \mathbb{Z}_b[/itex]

such that ψ(x)= (x mod a, x mod b), it is a homomorphism, a surjection(for any y in the range, there exists an x congruent to z(mod ab) "the solution to the congruence system, and finally ψ is an injection because domain and codomain are finite sets with equal cardinality or order .
 
Yes, but where did you use that gcd(a,b)=1?? You do need this!
 
my computer just died.
we used the fact that a and b are coprime to show that the system of equations has a solution by CRT.
 
Bachelier said:
my computer just died.
we used the fact that a and b are coprime to show that the system of equations has a solution by CRT.

That seems alright! :smile:
 

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