Square of non-zero element in group algebra is always non-zero? 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 $x^2=0$?
 A: EDIT: I did not realize that it was asked for a finite group. Let me remark that, in the following example, $G$ is not finite.
Take two cyclic non-trivial finite groups $C_n$ and $C_m$ with generators $a$ and $b$, of order $n>1$ and $m>1$ respectively. Let $G$ be the free product of $C_n$ and $C_m$. 
Let $x=(1-a)b(1+a+a^2+\ldots+a^{n-1})$. Then,
$$x^2=(1-a)b(1+a+a^2+\ldots+a^{n-1})(1-a)b(1+a+a^2+\ldots+a^{n-1})$$
$$=(1-a)b 0 b(1+a+a^2+\ldots+a^{n-1})=0$$
But clearly $x\neq 0$.
A: ${\mathbb C}S_3 \cong {\mathbb C}\oplus {\mathbb C} \oplus M_2({\mathbb C})$ 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.
A: The question is equivalent to asking whether $\mathbb{C}G$ contains non-zero nilpotent elements. As Yemon Choi's answer implicitly demonstrates, the answer is yes if and only if $G$ is non-Abelian.
This is a consequence of the Wedderburn structure theorems applied to the semisimple algebra $\mathbb{C}G$, as has been noted in comments.
Here is an elementary proof that the group algebra $\mathbb{C}G$ contains no non-zero nilpotent right ideal, which is equivalent to the semisimplicity of $\mathbb{C}G$ but the proof ( which has appeared in the book on character theory by D. Goldschmidt) requires little machinery.
Recall that a right ideal $I$ is nilpotent if every sufficiently long product of elements of $I$ is zero. In particular, every element of $I$ must be nilpotent.
Suppose that $I$ is a nilpotent right ideal of $\mathbb{C}G$ and that $r \in I$. Then $rg$ is nilpotent for each $g \in G$.
Consider the right regular matrix representation of $\mathbb{C}G$, and let $t$ be the trace it affords. Then $t(1_{G}) = |G|$ and $t(g) = 0 $ for all $g \neq 1_{G} \in G$.
Hence for every $g \in G$, we have that $t(rg^{-1})$ is $|G|$ times the coefficient of $g$ in $r$ ( when $r$ is expressed as a linear combination of elements of $G$).
However $rg^{-1}$ is nilpotent for all $g \in G$ as $I$ is a right ideal, and nilpotent matrices always have trace zero, so $t(rg^{-1}) = 0$ for all $g \in G$, and thus $r = 0$. 
