one may even take S^1= multiplicative group of all complex numbers
with modulus 1.every element in it is a square!  In
harmonicanalysis@yahoogroups.com, Michael Cowling <michaelc@m...>
wrote:
>
> In some compact groups, you have $G = G^2$, so that you can get
> $\lambda(G) = \lambda(G^2)$.
>
> Michael
>
>
>
> On Sat, 4 Dec 2004, ali akbar arefigamal wrote:
>
> > Hi, all
> >
> > Let G be a LCA group with the Haar measure $\lambda$. Can we
find
> > a measurable subset A (having finite measure )in G such that
> > $\lambda(A)=\lambda(A^{2})$, in which $A^{2}=\{a^{2}, a\in A\}$ .
> >
> > I think this is false, but i cann't prove it. It is possible
hint me.
> >
> > Thanks,
> >
> >
> > 
> > Do you Yahoo!?
> > Yahoo! Mail  now with 250MB free storage. Learn more.
>
> 
>
_____________________________________________________________________
__
> 

>  Professor Michael Cowling School of Mathematics

>  Telephone: +61 2 9385 7101 University of New
South Wales 
>  Fax: +61 2 9385 7123 UNSW Sydney NSW
2052 
>  Mobile: +61 4 0936 0678 AUSTRALIA

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>
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