Group Formation: Converting Sets into Groups with Binary Operations

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Discussion Overview

The discussion revolves around the concept of converting sets into groups using binary operations. Participants explore various methods and structures, including cyclic groups, vector spaces, and the use of the well-ordering theorem, with a focus on both countable and uncountable sets.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant suggests that any non-empty set can be turned into a group by defining a binary operation on it.
  • Another participant proposes that for countable sets, a cyclic group can be formed, while for uncountable sets, the well-ordering theorem can be used to define a group structure through finite subsets of an equivalent ordinal.
  • A different viewpoint introduces the idea of using a vector space over Q with a basis derived from the given set, or constructing a free group generated by the set.
  • One participant mentions the concept of a Hamel basis in relation to vector spaces and discusses the construction of a vector space from a given basis.
  • Another participant elaborates on the vector space formed by functions on the set, emphasizing the existence of a basis consisting of functions with a single non-zero value.

Areas of Agreement / Disagreement

Participants present multiple competing views on how to convert sets into groups, with no consensus reached on a single method or solution.

Contextual Notes

Some methods rely on specific assumptions, such as the countability of the set or the applicability of the well-ordering theorem. The discussion also touches on the distinction between constructing a vector space from a basis versus proving the existence of a basis using Zorn's lemma.

Kummer
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I have seen this problem a long time ago. It is really supprising, maybe you shall like it as well.

Given any non-empty set we can define a binary operation on this set to turn it into a group.
 
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If the set is countable, then this is no problem, because we can turn it into the appropriate cyclic group. I forgot the actual solution, but for uncountable sets I think we can use the well-ordering theorem to get an equivalent ordinal, and then we give the set of finite subsets of this ordinal the group operation of symmetric difference. There is a bijection between this latter set and our original one, so we simply copy the new-found group structure. I believe this even gives us an abelian group.

Is this the solution you saw?
 
if you want one less exotic, or non commutative, try the vector space over Q with basis your given set, or even the free group generated by your set.

again you can use the "same cardinality" trick of morphism, to replace the elements of your group by the original set elements.
 
Vector space over Q with basis given X

Just thought I'd mention a keyword: Hamel basis.
 
well, given a set S, i just meant the vector space which is the set of all functions on the set S, into Q, with value zero except at a finite number of elements of S.

so this space has an obvious basis, namely the functions with exactly one non zero value.

the phrase hamel basis to me usually means you have a vector space first, and then invoke the zorn lemma to prove a basis exists, in cases where an obvious one is not at hand.

so it is much easier and more constructive, to go from a basis to a vector space spanned by that basis, than the other way around.
 

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