Consider a finite group $G$ and complex group algebra $\mathbb{C}(G)$, i.e. formal sums $$ \sum\_{g \in G} a\_gg, \ a\_g \in \mathbb{C},$$ with algebra structure: $$ \sum\_g a\_gg+\sum\_gb\_gg=\sum\_g(a\_g+b\_g)g,$$
$$ \left( \sum\_ga\_gg \right) \left( \sum\_g b\_gg \right) =\sum\_{g,h}a\_gb\_h(gh).$$


Does there group $G$ and $x\in \mathbb{C}(G), x \not = 0$, such that $x^2=0$?



Square of non-zero element in group algebra is always non-zero?
${\mathbb C}S\_3 \cong {\mathbb C}\oplus {\mathbb C} \oplus M\_2({\mathbb C})$ and so the answer to your original question is yes -- take $x$ to be the element which corresponds in this decomposition to $(0,0,a)$ where
$$a= \pmatrix{0 & 1 \\ 0 & 0 }.$$


The same kind of idea shows that for any finite non-abelian group, ${\mathbb C}G$ contains a non-trivial square-zero element; as Loren Spice has observed, if G is an abelian group then ${\mathbb C}G$ cannot contain any non-trivial square-zero elements.




---


While I am here (and ignoring my pile of marking, ahem), I can't resist mentioning two old results which generalize the result stated above. Both were mentioned somewhere before on MathOverflow but I can't remember where at the moment.


**Theorem** (Kaplansky, 1948). Let $A$ be a noncommutative ${\rm C}^\*$ algebra. Then $A$ has a non-trivial, square-zero element.


**Theorem** (Behncke , 1971) Let $G$ be a noncommutative, locally compact group. Then the convolution algebra $L^1(G)$ has a non-trivial, square-zero element.


