Are there a finite group $G$ and a field $\mathbb{F}$ such that $\gcd(3,|G|)=1$ and the group algebra $\mathbb{F}[G]$ contains a zero divisor whose support is of size $3$?

Recall that the support of an element $\alpha$ of $\mathbb{F}[G]$ is the set $\{x\in G \;|\; \alpha(x)\neq 0\}$.

This question is related to the following one:

Zero divisors of the form $1+x+y$ in the rational group algebra

One motivation to propose the question is the following well-known observation: if $a$ is an element of order $3$ in a group $G$ then $1+a+a^2$ is a zero divisor over any group algebra of $G$ (whose support is of size $3$). So the hypothesis $\gcd(3,|G|)=1$.

  • $\begingroup$ Why isn't the support of every element all of $G$? Aren't elements of $G$ units in $\mathbb{F}[G]$? Do I misunderstand the notation $\alpha(x)$? $\endgroup$ Commented Aug 30, 2015 at 8:32
  • $\begingroup$ @GabrielC.Drummond-Cole: I'd think the support of an element of $\mathbb{F}[G]$ is meant to be the set of group elements whose coefficient is $\neq 0$ -- but the notation $\alpha(x) \neq 0$ looks also confusing to me. $\endgroup$
    – Stefan Kohl
    Commented Aug 30, 2015 at 8:56
  • $\begingroup$ One of the definitions of a group algebra regards its elements as functions from $G$ to $\mathbb F$. $\endgroup$ Commented Aug 30, 2015 at 10:12
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    $\begingroup$ This is a very legitimate question, however, you should assume that the other factor (in the factorization of zero) also has support-size coprime to the order of the group. $\endgroup$ Commented Aug 31, 2015 at 9:02
  • $\begingroup$ @AndreasThom: Another motivation to propose the question is the question of the existence of zero divisors with support of size 3 in the group algebra of torsion-free groups which are residually finite. The number ``3" is the first unsettled. $\endgroup$ Commented Aug 31, 2015 at 13:51

2 Answers 2


Multiplication by $\sum_{g\in G}g$ annihilates any element whose sum of coefficients is zero. It remains to choose such element with support of size 3, which is possible whenever $\mathbb F\neq GF(2)$.

  • $\begingroup$ Many thanks. This perfectly and very simply answer my question. Actually the answer shows all elements in the augmentation ideal of the group algebra whose support is of size 3 can be chosen as a candidate for the answer over any field with more than 2 elements. The construction may interpret as canonical as it is valid over any finite group and any field with more than 2 elements. Can one find a zero divisor with support of size $3$ over $GF(2)$? anyway again many thanks for your answer. $\endgroup$ Commented Aug 31, 2015 at 5:49
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    $\begingroup$ Yes: let $G = \langle g \mid g^7 = 1 \rangle$ and $\alpha = g + g^2 + g^4$; then $\alpha (\alpha+1) = 0$. $\endgroup$ Commented Sep 1, 2015 at 18:43
  • $\begingroup$ @NoamD.Elkies: Have you any idea to generalize this example? $\endgroup$ Commented Sep 5, 2015 at 9:47
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    $\begingroup$ One way to explain this example is starting from an identity $g+g^2+g^4=0$ in the field of $2^3$ elements, which contains an element $g$ of multiplicative order $7$. There are similar identities from other finite fields or characteristic $2$; for example, $x^5 + x^2 + 1$ is irreducible mod $2$ so $g^5 + g^2 + 1$ is a zero-divisor when $g$ has exponent $31$. $\endgroup$ Commented Sep 5, 2015 at 13:13

It is not really an answer to the question, but Ilya Bogdanov's answer puts quite strong restrictions on zero divisors of $\mathbb{F}G$ when $G$ is finite. Let $I(G)$ denote the augmentation ideal of $\mathbb{F}G$, that is $\{ \sum_{g \in G} \lambda_{g} g : \sum_{g \in G} \lambda_{g} = 0 \}$, which is indeed a two-sided ideal of $\mathbb{F}G$. Then every element of $I(G)$ is annihilated by $\sum_{ g \in G} g$, as Ilya observed, so $I(G)$ consists of zero divisors.

In the opposite direction, we can draw the conclusion that whenever $ab = 0$ for $a,b \in \mathbb{F}G$, at least one of $a$ or $b$ is in $I(G)$. For if $a$ is not in $ I(G)$, we can write $ a = \lambda 1_{G} + i$ where $0 \neq \lambda \in \mathbb{F}$ and $i \in I(G)$. Then $0 = ab = \lambda b +ib$. However, $ib \in I(G)$, so that $\lambda b \in I(G)$, and hence $b \in I(G)$. Similarly if $b$ is outside $I(G)$, we find that $a \in I(G)$, since $I(G)$ is a two-sided ideal. In particular, note that all nilpotent elements of $\mathbb{F}G$ lie in $I(G)$ by repeated applications of this.

(Later edit: In fact, note that a zero divisor $b \in \mathbb{F}G$ which annihilates an element $a$ lying outside $I(G)$ must itself lie in $\cap_{n = 1}^{\infty}I(G)^{n}$, since we see above that there is some $i \in I(G)$ and non-zero $\lambda \in \mathbb{F}$ such that $ib = -\lambda b$ (and a similar argument if we had $ba = 0$).

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    $\begingroup$ One may simplify the opposite direction argument by observing that $\mathbb F[G]/I(G)\cong \mathbb F$ via the epimorphism $\sum_{g\in G}\alpha_gg\mapsto \sum_{g\in G}\alpha_g$. This applies to all groups, not only finite ones. $\endgroup$ Commented Sep 1, 2015 at 11:51
  • $\begingroup$ Yes, thanks, that's true for $I(G)$ itself- not sure if that makes it clear that the element lies in the intersection of the powers of $I(G)$. I had noticed that that direction works (with my argument) for infinite groups too since $I(G)$ is still an ideal ( if we think of the group algebra as consisting of finitely supported elements) $\endgroup$ Commented Sep 1, 2015 at 12:36

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