Let $K$ be a finite field and $G$ be a discrete group.

Is it true that for every $a,b\in K[G]$ the condition $ab=0$ implies $ba=0$?

It does not seem to be related to zero divisor problem, any ideas if this can be true and for which fields?

  • $\begingroup$ Can you be more specific for what you mean by $K[G]$? Is this the group ring, or the ring of functions on $G$? $\endgroup$ Mar 12, 2011 at 15:10
  • $\begingroup$ $K[G]$ is the group ring. $\endgroup$ Mar 12, 2011 at 15:26
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    $\begingroup$ I don't believe this even for finite $G$. After all a group ring of a finite group $G$ over a field of characteristic coprime to $\left|G\right|$ is semisimple and therefore a product of matrix rings (after corresponding field extension!), which generally are not 1x1 matrix rings. It is easy to find two 2x2 matrices with product $0$ in one order but not $0$ in the other order. $\endgroup$ Mar 12, 2011 at 15:52
  • $\begingroup$ unfortunately, this does not give an example where the property is not satisfied. $\endgroup$ Mar 12, 2011 at 16:05
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    $\begingroup$ Dear Kate, here is an explicit example of what Darij says: in ${\mathbb F_5}[S_3]$, $A=1+(23)$ and $B=(12)-(123)$ satisfy $AB=0$ and $BA\ne 0$. $\endgroup$ Mar 12, 2011 at 16:48

5 Answers 5


Let G be non-abelian of order 6, with x of order 2 and y of order 3. In such a group yxy = x, since both x and xy have order 2. Let K be a field with 2 elements. Then (x+y)⋅(1+xy) = x+y + y+yxy = x+y + y+x = 0, but (1+xy)⋅(x+y) = x+y + xyx+xyy = x+y + yy + xyy ≠ 0.

You may be thinking of the property: if a⋅b = 0 then there is some non-zero c such that c⋅a = 0. This holds in all (two-sided) Artinian rings (because elements are either units or zero-divisors). I believe this is true for two-sided self-injective rings as well. I don't know if it is possessed by group rings of infinite groups over finite fields.

(Thanks to Greg Marks:) The classification of finite group rings over fields where ab=0 implies ba=0 is given in:

Gutan, Marin; Kisielewicz, Andrzej. "Reversible group rings." J. Algebra 279 (2004), no. 1, 280–291. MR2078399 DOI:j.jalgebra.2004.02.011.

In particular, K is a field of order 22n-1 and G is the quaternion group of order 8, or G is abelian. Li and Parmenter (2007) extend this to finite group rings over commutative rings with 1 in MR2372321.

  • $\begingroup$ also there are some answers in comments $\endgroup$ Mar 12, 2011 at 17:14
  • $\begingroup$ G=Q8 over K=GF(2) has the ab=0 implies ba=0 property, but G=D8 over K=GF(2) does not. It seems the characterization might be a little strange. $\endgroup$ Mar 12, 2011 at 17:21
  • $\begingroup$ For finite groups $G$, I can at least show that $[G,[G,G]]=1$ if $G$ satisfies the property in question. (Characteristic can be arbitrary.) $\endgroup$ Mar 12, 2011 at 17:24
  • $\begingroup$ The field definitely matters in general, as G=Q8 over K=GF(3) does not have the property. I would be curious to hear a counter-example to: if KG has the ab=0 implies ba=0 property then either G is abelian, G is infinite, or (K has characteristic 2 and G=Q8). $\endgroup$ Mar 12, 2011 at 18:13
  • $\begingroup$ You are very kind to credit me, but I doubt I deserve it. I am embarrassed to say that the very interesting paper of Gutan and Kisielewicz had slipped my mind. Gutan and Kisielewicz completely resolve the group algebra reversibility question for the case of torsion groups. They remark that the torsion-free case is connected with the zero divisor problem, in the sense that it will be very difficult to produce a non-reversible group algebra with a torsion-free group, since this will of course resolve the zero divisor problem in the negative. $\endgroup$
    – Greg Marks
    Mar 14, 2011 at 2:08

The condition $ab=0 \Rightarrow ba=0$ defines what are often called reversible rings, which, for example, have the property that the set of nilpotent elements is an ideal that coincides with the prime radical.  A full matrix ring can't have this property, so you can construct counterexamples by taking any finite field $K$ and nonabelian group $G$ to which Maschke's theorem applies.  An alternative example of a non-reversible group algebra is $K[G]$ where $K$ is the field of two elements and $G$ is the dihedral group of order $8$.  Here the set of nilpotent elements does coincide with the prime radical (the ring is local artinian), but one can find elements $a,b \in K[G]$ with $ab=0$ but $ba \neq 0$.


Using the miniscule amounts of representation theory I know, you can construct an example as follows.

Let $G=S_3=D_6=\left< r,s\mid r^3=s^2=srsr=e \right>$ (rotation $r$ and flip $s$), and let $K=\mathbb F_5[\omega]$ where $\omega^3=1$ (finite field!). Then an irreducible representation of $G$ is given by the $2$-dimensional vector space $V$ with basis $\{v, sv\}$ and action of $G$ on $V$ $\rho\colon G\to Aut(V)$ generated by by $\rho(s)(v)=sv$, $\rho(s)(sv)=v$, $\rho(r)v=\omega v$ and $\rho(r)(sv)=\omega^2 sv$.

Hence, we have $\rho(e)=\left[\begin{matrix}1&0\newline 0&1\end{matrix}\right]$, $\rho(r)=\left[\begin{matrix}\omega&0\newline 0&\omega^2\end{matrix}\right]$, $\rho(r^2)=\left[\begin{matrix}\omega^2&0\newline 0&\omega\end{matrix}\right]$, $\rho(s)=\left[\begin{matrix}0&1\newline 1&0\end{matrix}\right]$, $\rho(sr)=\left[\begin{matrix}0&\omega^2\newline \omega&0\end{matrix}\right]$, $\rho(sr)=\left[\begin{matrix}0&\omega\newline \omega^2&0\end{matrix}\right]$.

The first four matrices are linearly independent over $\mathbb K$ and hence generate $End_{\mathbb K}(V)$ which is $4$-dimensional. Now we know that if $\bar K$ is the algebraic closure of $K$, then $\bar K[G]=\bigoplus_W End_{\bar{\mathbb K}}(W)$ where $W$ are irreducible representations over $\bar K$. Character theory tells us that the projection of $\bar K[G]$ onto $End_{\bar {\mathbb K}}(W)$ is given by left-multiplication by the idempotent $\frac{\dim W}{|G|}\sum_{g\in G}\chi_W(g^{-1})g$ where $\chi_W$ is the trace of $\rho_W\colon G\to Aut(W)$.

So in particular we have the idempotent $\phi=\frac13(2e-r-r^2)\in D_6$, which projects $\bar K[G]$ onto $End_{\bar{\mathbb K}}(V)$, and hence must project $K[G]$ into $End_{\mathbb K}(V)$. But we see that $\phi(e)=\frac13(2e-r-r^2)$, $\phi(r)=\frac13(-e+2r-r^2)$, $\phi(r^2)=\frac13(-e-r+2r^2)$, and $\phi(s)=\frac13(2s-rs-r^2s)$ are four linearly independent elements in the image of $\psi$ and thus span $End_{\mathbb K}(V)$.

So take your favorite $2\times 2$ matrices $A$ and $B$ (over $K)$ such that $AB=0$ but $BA\neq 0$, find $a_g$ and $b_g$ such that $A=\sum_{g\in \{e,r,r^2,s\}} a_g\rho(g)$ and $B=\sum_{g\in \{e,r,r^2,s\}} b_g\rho(g)$, and then $a=\sum_{g\in \{e,r,r^2,s\}}a_g\phi(g)$ and $b=\sum_{g\in \{e,r,r^2,s\}}b_g\phi(g)$ will be such that $ab=0$ but $ba\neq 0$.


If $G$ is torsion-free then the question of reversibility of $K[G]$ (that is, does $ab = 0$ imply $ba = 0$) is in fact equivalent to the zero divisor conjecture, for any field $K$.

Connell showed that such a $K[G]$ is prime, so given non-zero $a, b \in K[G]$ there exists $c \in K[G]$ such that $bca \neq 0$. If it were the case that $ab = 0$ then we have $a (bc) = 0$ but $(bc) a \neq 0$ and $K[G]$ is not reversible. This trick is how you show that the unit conjecture implies the zero divisor conjecture.

(In the other direction, if a ring has no zero divisors then reversibility holds trivially.)

Connell, I. G., On the group ring, Can. J. Math. 15, 650-685 (1963). ZBL0121.03502.


Somewhat belatedly, here is a class of examples. Let $G$ be a nontrivial free product of two groups (nontrivial means excluding $G = Z_2 * Z_2$). Suppose that $KG$ has a zero divisor (e.g., if at least one of the groups in $G = G_2 * G_2$ contains a torsion element). Then $KG$ has a right zero divisor that is not a left zero divisor. This is a consequence of Left and right zero divisors in group algebras by D Handelman (me) and J Lawrence, Bulletin of the Australian Mathematical Society (1976) 15 (3), 453-454. This is independent of the choice of field.


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