Does AB Equal G When |A| + |B| Exceeds |G|?

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Discussion Overview

The discussion revolves around the proposition that if A and B are subsets of a group G and the sum of their cardinalities exceeds the cardinality of G, then the product set AB equals G. Participants explore the implications of this proposition, examining specific cases, subgroup properties, and the conditions under which the statement may hold true.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims that if |A| + |B| > |G|, then AB = G, using a specific case where A has more than half of G's elements and B is a subgroup.
  • Another participant questions the validity of having a subgroup of G with half the elements, citing Lagrange's theorem and the implications for groups of prime order.
  • A later reply clarifies that the argument does not imply all groups have a subgroup of half their elements but rather considers a hypothetical case to support the proposition.
  • Another participant argues that it is impossible to have |A| + |B| > |G| unless one of the subsets is the whole group, challenging the initial claim.
  • One participant emphasizes that A and B are merely subsets and not necessarily subgroups, suggesting that the union of the generated sets can still cover G.
  • Another participant asserts that having |G|/2 + 1 elements as a subset of G guarantees that the group they generate is G, regardless of subgroup status.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the proposition, with some supporting it under specific conditions while others challenge its general applicability. No consensus is reached regarding the truth of the proposition.

Contextual Notes

Participants highlight limitations related to subgroup properties, the implications of Lagrange's theorem, and the specific conditions under which the proposition may or may not hold true.

joeblow
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If A and B are mere SUBSETS of G and |A| + |B| > |G|, then AB = G.

My thought is that a "weakest case" would be where A has |G|/2 + 1 elements (if |A| + |B| > |G|, then one of A or B must have over half of G's elements) and |G|/2 of them form a subgroup of G. Then taking B= the subgroup, it is true that AB = G because that one stray element when multiplied with the subgroup gives the rest of G. So, since AB = G in this case, it must be true in general.

Is this valid, or should I be doing something else?
 
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I'm wondering exactly why |G|/2 of them form a subgroup of G. By Lagrange's theorem the order of any subgroup must divide the order of the group, but suppose your group has order p where p is a prime, then it clearly can't have a subgroup of order 2 unless p is in fact equal to 2 in which case your subgroup satisfying the stated property would simply be the identity which is trivial.
 
I don't mean to imply that any group will have a subgroup that has half the elements; I'm saying that *IF* G had a subgroup H with half of its elements, then choosing A = {H plus one element from G-H} and B = H, this proposition would hold. I think that such a case is the "weakest" case possible, so if it works for this case, it must be true in general. I think this because the closure property for subgroups allows for fewer elements to be introduced by elements from A being multiplied with elements of B.

Do you think that this is true?
 
When you add that stray element to H to form the set A, you actually got the whole group. Since the order of a group is always divisible by the order the subgroup, you cannot have a subgroup of G which has an order > |G|/2 unless this subgroup is the whole group.

This means that it is impossiple to have |A| + |B| > |G|, unless one of the subgoups is the whole group G.
 
A and B are not necessarily subgroups of G; they are just subsets of G. So when I multiply A and B, I get H U aH = G. (where a is the "stray" element)

Since it works for this case, *does* it work for all cases? Is what I have a valid proof of the proposition?
 
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You don't need a subset of A or B to be a group: If you just have |G|/2+1 elements as a subset of G, the group they generate needs to be G. It doesn't matter if they or some subset of them formed a subgroup to begin with or not
 

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