Given a group $G$ we may consider its group ring $\mathbb C[G]$ consisting of all finitely supported functions $f\colon G\to\mathbb C$ with pointwise addition and convolution. Take $f,g\in\mathbb C[G]$ such that $f\*g=1$. Does this imply that $g\*f=1$?


If $G$ is abelian, its group ring is commutative, so the assertion holds. In the non-abelian case we have $f\*g(x)=\sum\_y f(xy^{-1})g(y)$, while $g\*f(x)=\sum\_y f(y^{-1}x)g(y)$, and this doesn't seem very helpful.


If $G$ is finite, $\dim\_{\mathbb C} \mathbb C[G]= |G|<\infty$, and we may consider a linear operator $T\colon \mathbb C[G]\to\mathbb C[G]$ defined by $T(h) = f\*h$. It is obviously surjective, and hence also injective. Now, the assertion follows from $T(g\*f)=f=T(1)$.


What about infinite non-abelian groups? Is a general proof or a counterexample known?



Is a left invertible element of a group ring also right invertible?
A ring is called Dedekind-finite if that property holds. Semisimple rings are Dedekind finite, so this covers $\mathbb CG$ for a finite group $G$; this is easy to do by hand. It is a theorem of Kaplansky that this also holds $KG$ for arbitrary groups $G$ and arbitrary fields $K$ of characteristic zero. See [Kaplansky, Irving. Fields and rings. The University of Chicago Press, Chicago, Ill.-London 1969 ix+198 pp. [MR0269449](http://www.ams.org/mathscinet-getitem?mr=MR0269449)] It is open, I think, for general fields.


