ra.rings and algebras - Is a left invertible element of a group ring also right invertible?

Given a group $G$ we may consider its group ring $mathbb C[G]$ consisting of all finitely supported functions $fcolon Gtomathbb C$ with pointwise addition and convolution. Take $f,ginmathbb C[G]$ 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 $Tcolon mathbb C[G]tomathbb C[G]$ 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?

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