Group Theory: Is A a Left Coset of G?

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Discussion Overview

The discussion revolves around the question of whether a finite subset A of a group G, satisfying the condition |A²|=|A|², can be classified as a left coset of G. Participants explore various implications of this condition, particularly in the context of group theory, including both Abelian and non-Abelian groups.

Discussion Character

  • Debate/contested
  • Exploratory
  • Technical explanation

Main Points Raised

  • One participant suggests that if |A²|=|A|² holds for A, then A might be a left coset of G, having proven this for subsets with two elements.
  • Another participant questions whether A needs to be a subgroup rather than just a subset to be considered a coset.
  • It is proposed that the condition holds true for Abelian groups with |A| > 1, as the inequality |A²| ≤ Choose(|A|, 2) implies that the assumption cannot be satisfied.
  • Counterarguments are presented, asserting that in Abelian groups, the condition |G|²=|G²| can indeed be satisfied, challenging the earlier claim.
  • A participant mentions a potential solution involving the set K = {g ∈ G | gA = A}, suggesting that if |x^{-1} A| can be shown to be greater than or equal to |K|, a contradiction could be reached.
  • Another participant expresses confusion about the problem statement, indicating that it may not align with the original question posed.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the condition |A²|=|A|², with some supporting the idea that it leads to A being a left coset, while others challenge this notion, particularly in the context of Abelian groups. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations regarding the assumptions made about the nature of A and G, particularly concerning whether A must be a subgroup or if the properties of the group (Abelian vs. non-Abelian) affect the validity of the claims made.

micromass
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This is not a homework problem. I was just wondering.

Let G be a group and let A be a finite subset of G. If |A²|=|A|² (where A^2=\{a_1a_2~\vert~a_1,a_2\in A\} ). Is it true that A is a left coset of G?

If A has two elements, then I have proven that this is true. But for greater elements, it soon becomes very complicated. I do think this is true...

Anybody got a proof/counterexample or maybe some hints?
 
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I thought A would have to be a subgroup rather than just a subset?

whereas a coset is a subset of the group G.
 
It is true for Abelian groups where |A| > 1. Cause

|A2| ≤ Choose(|A|, 2) < |A|2 (so unequal)

So your assumption will always be false so the proposition will always be true.

When the group is not Abelian the key is to look at elements which do not commute (I would say).
 
Last edited:
Uh, no. Take G an abelian group, then |G|²=|G²|. The assumption can certainly be satisfied, even in abelian groups...
 
micromass said:
Uh, no. Take G an abelian group, then |G|²=|G²|.

If G is a group, then G^2 = G, so

|G^2| = |G| &lt; |G|^2

assuming G is nontrivial.
 
Damn... It appears that I have posted the wrong problem. Never mind this thread...
 
Huh the problem statement seems fine. It can be found on page 5 of this http://www.mfo.de/programme/schedule/2010/27/OWR_2010_29.pdf" (under Question: (1)), as my teacher received this problem from the author of that particular paper at a conference in India.

I do think I have a solution, thanks to some comments from my TA, and some good ideas from micromass.

As micromass suggested, the first step is to show xA = A^2, which we can of course write as A = x^{-1} A^2. The key insight is to consider the set K = \{g \in G \,|\, gA = A\}. It is straightforward to show that this is a subgroup of G. Unraveling the definition of K, we see that x^{-1} A \subset K, so |x^{-1} A| \leq |K| . If we can show |x^{-1} A|\geq |K|, we're done. Now |x^{-1} A| = |A| (exhibit a bijection), so it remains to prove |A| \geq |K|. Contradiction seemed to be the easiest route here.

And that's really it.

*EDIT* Obviously I left out a lot of details, mainly to allow those who might have been interested in the problem in the first place to figure out various individual arguments. I'll be happy to clarify any particular part of the proof, and please point out any oversights I might have made.
 
Last edited by a moderator:
thanks snipez. I was really breaking my head at this one :smile: Now I can be at rest...
 
snipez, the problem you are solving is not the one posed in the OP.
 
  • #10
Ah yes, finally I notice the |A| term is squared, but in fact the problem I posted (with the condition |A^2| = |A|) is the one micromass meant to refer to. That thread is somewhere lost in the homework help forums. Thanks.
 

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