Group Theory: Is A a Left Coset of G?

[micromass]

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?

 

[Beeraa]

I thought A would have to be a subgroup rather than just a subset?

whereas a coset is a subset of the group G.

 

[Outlined]

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).

 

[micromass]

Uh, no. Take G an abelian group, then |G|²=|G²|. The assumption can certainly be satisfied, even in abelian groups...

 

[jbunniii]

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.

 

[micromass]

Damn... It appears that I have posted the wrong problem. Never mind this thread...

 

[snipez90]

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.

 

[micromass]

thanks snipez. I was really breaking my head at this one Now I can be at rest...

 

[Office_Shredder]

snipez, the problem you are solving is not the one posed in the OP.

 

[snipez90]

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.