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 $x^2=0$?

• For an Abelian group, taking the Fourier transform shows that this cannot happen. Feb 17, 2016 at 17:23
• Why the votes to close? The question is natural, and for infinite groups questions about zero divisors in the group algebra are actually quite difficult. As it happens, the answer is yes, but I don't think this would be obvious without knowing some representation theory Feb 17, 2016 at 17:49
• @YemonChoi To be fair, Maschke's Theorem and Artin-Wedderburn is not exactly research level. And to ask that question for a matrix algebra would also not be research level. For the record: I did not vote to close, because I cannot be bothered, but I can easily see why someone would. Feb 17, 2016 at 18:27
• It should be noted that for infinite, torsion-free groups, this question is wide open (it is a conjecture of Kaplansky that there are no nonzero nilpotent elements in that case). Feb 17, 2016 at 18:39
• @AlexB. Fair enough. If the question has been asked for a matrix algebra I would have voted to close. But Maschke+AW is, I think, something that not every mathematician would have seen, so that was the criterion I used for myself Feb 17, 2016 at 18:43

${\mathbb C}S_3 \cong {\mathbb C}\oplus {\mathbb C} \oplus M_2({\mathbb C})$ 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.

• YemonChoi, stop updating this post! I can only upvote it once! :-) Feb 17, 2016 at 18:14
• To elaborate on YemonChoi's nice answer, and the comments above, this is a representation-theoretic question in disguise. The examples for non-Abelian groups, resp. the non-existence for Abelian groups, correspond to the absence (resp. presence) of representations of dim. $>1$. The nilpotent element here is a matrix coefficient of two orthogonal vectors, e.g., $(1,0,-1)$ and $(1,-2,1)$, in the 2-dimensional permutation representation of $\mathrm S_3$ on the trace-0 vectors in 3-space. It is $-3$ at $s_1$ and $s_1s_2$, $3$ at $s_2$ and $s_2s_1$, and $0$ else ($s_1 = (1\ 2)$, $s_2 = (2\ 3)$). Feb 17, 2016 at 21:03
• Nice, but what would be a square zero element in $\ell^1(F_2)$? Feb 17, 2016 at 23:34
• @AndreasThom, if $\mathrm F_2$ is free on generators $a$ and $b$, then wouldn't the modification $(1 - a/2)b(1 + a/2 + (a/2)^2 + \dotsb)$ of SimoneVirili's answer work? EDIT: No, it wouldn't. Feb 18, 2016 at 1:40
• @AndreasThom Behncke's proof, when specialized to $F_2=\langle a,b\rangle$, starts as follows: the subalgebra $\ell^1(\langle a\rangle)$ is, once, you apply the Gelfand transform, regular, so we can certainly cook up non-zero elements $u$ and $v$ in this subalgebra with $uv=vu=0$. Now we just need to find $y\in \ell^1(F_2)$ such that $uyv\neq 0$. I can't remember right now how Behncke does this in the general case, but for this example, since we can cook up explicit $u$ and $v$ easily with Fourier analysis, I suspect one could show with some direct work that $y=\delta_b$ (or similar) works. Feb 18, 2016 at 10:15

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 $x\neq 0$.

• But $G$ isn't finite …. Feb 17, 2016 at 17:52
• The OP wanted the group G to be finite, although your example is a nice one otherwise Feb 17, 2016 at 17:53
• Since the free product is free, one can specialize this example to any finite group generated by two noncommuting elements $a$ and $b$. For example, $a=(1,2)$ and $b=(1,2,3) \in S_3$ yield the concrete $x = (1,2,3) - (1,3,2) + (2,3) - (1,3)$. Feb 17, 2016 at 18:19
• (Err, on reflection, that's inevitable, since there's a unique conjugacy class of non-0 nilpotents in $\mathbb C\mathrm S_3$ ….) Feb 17, 2016 at 21:43
• My earlier assertion that we can specialize this example to any finite group $G = \langle a, b \rangle$, is wrong, by the way: We might get $x = 0$. Indeed, this happens exactly, when the commutator $[b,a] \in \langle a \rangle$. A concrete example would be the quaternion group $Q_8$. In fact, the group algebra of $Q_8$ over the rationals contains no nilpotent elements, since it is a direct product of division rings. Feb 18, 2016 at 13:51

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 $r = 0$.