Directly finite group algebras

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

If a square matrix has a right inverse, that inverse is also a left inverse. A group algebra \(K[G]\) is usually infinite-dimensional, but each of its elements is a finite sum. Kaplansky asked whether this finiteness suffices: does \(ab=1\) imply \(ba=1\) for all \(a,b\in K[G]\)? The answer is yes over every field of characteristic zero, and over every field for large classes of groups. In 2026 OpenAI showed that the answer is no in general: there are a finite field \(K\) of characteristic two, a finitely presented group \(G\) and \(a,b\in K[G]\) with \(ab=1\) and \(ba\ne1\) [OpenAI-DF2]. The same group is not sofic, and it carries a cellular automaton that is injective but not surjective. This course proves these statements. This lesson sets up the question and proves the classical positive results.

1. Directly and stably finite rings

All rings have an identity \(1\), and ring homomorphisms preserve it. A ring \(R\) is directly finite if \(ab=1\) implies \(ba=1\) for all \(a,b\in R\). It is stably finite if every matrix ring \(\operatorname{Mat}_n(R)\), \(n\ge1\), is directly finite.

A field, a commutative ring and a finite-dimensional algebra over a field are directly finite. For a finite-dimensional algebra \(R\) over a field \(K\), the left multiplications \(\lambda_a(x)=ax\) are \(K\)-linear maps of the finite-dimensional space \(R\), with \(\lambda_a\lambda_b=\lambda_{ab}\). If \(ab=1\), then \(\lambda_a\lambda_b=\mathrm{id}\), so \(\lambda_b\) is injective, hence bijective, and \(\lambda_b\lambda_a=\mathrm{id}\) as well. Applying this to \(1\) gives \(ba=\lambda_b\lambda_a(1)=1\). Since \(\operatorname{Mat}_n(R)\) is again finite-dimensional, \(R\) is stably finite.

The ring of all linear maps of an infinite-dimensional vector space is not directly finite (Exercise 8.1). If \(ab=1\) in any ring, then \(e=ba\) satisfies \[ e^2=b(ab)a=e,\quad ae=a,\quad eb=b , \] so \(e\) is an idempotent, and \(ba=1\) exactly when the idempotent \(1-e\) is zero.

Lemma 1.1. Let \(ab=1\) in a ring \(R\). If \(a\) has a left inverse \(c\), that is \(ca=1\), then \(c=b\) and \(ba=1\).

Proof. \(c=c(ab)=(ca)b=b\), so \(ba=ca=1\). \(\square\)

2. Group algebras, finite groups and finite supports

Let \(K\) be a field and \(G\) a group. The group algebra \(K[G]\) is the \(K\)-vector space with basis \(G\): its elements are finite sums \(c=\sum_{g\in G}c_gg\) with \(c_g\in K\), almost all zero. The support of \(c\) is \(\operatorname{supp}(c)=\{g:c_g\ne0\}\). The product is \[ (cd)_w=\sum_{uv=w}c_ud_v , \] a finite sum, and \(K[G]\) is an associative \(K\)-algebra with identity \(1=1_G\). For a subgroup \(H\le G\), the elements supported in \(H\) form the subalgebra \(K[H]\subseteq K[G]\), and every homomorphism of groups \(G\to G'\) extends linearly to a homomorphism of algebras \(K[G]\to K[G']\), injective when the group homomorphism is.

Lemma 2.1. Let \(G\) be finite and \(K\) any field. Then \(K[G]\) is stably finite.

Proof. \(K[G]\) has dimension \(|G|\) over \(K\), so Section 1 applies. \(\square\)

This holds also when the characteristic of \(K\) divides \(|G|\), where \(K[G]\) is not semisimple.

Lemma 2.2 (supports). Let \(a,b\in K[G]\) and let \(H\) be the subgroup generated by \(\operatorname{supp}(a)\cup\operatorname{supp}(b)\). Then \(a,b\in K[H]\), the products \(ab\) and \(ba\) are the same in \(K[H]\) and in \(K[G]\), and \(H\) is finitely generated.

Proof. The products are computed by the same finite sums of products of elements of \(H\). The generating set is finite. \(\square\)

So a pair with \(ab=1\ne ba\) lives on a finitely generated group. Being finitely generated is automatic; the group of the theorem below is moreover finitely presented, which needs a separate argument. By Lemma 2.1, such a group is infinite.

3. Characteristic zero

Theorem 3.1 (Kaplansky). Let \(K\) be a field of characteristic zero and \(G\) any group. Then \(K[G]\) is stably finite.

The proof uses the trace of The regular trace and spectral phase, Section 1, with the group \(G\) in place of \(A\). There, \(L(g)\delta_x=\delta_{gx}\) and \(R(g)\delta_x=\delta_{xg^{-1}}\) on \(\ell^2(G)\), \(\mathcal M\) is the algebra of bounded operators commuting with every \(R(g)\), the algebra \(\operatorname{Mat}_n(\mathcal M)\) acts on \(H_n=\mathbb C^n\otimes\ell^2(G)\) and is exactly the set of bounded operators commuting with every \(I\otimes R(g)\), and \[ \tau_n(T)=\sum_{j=1}^n\langle T(e_j\otimes\delta_1),e_j\otimes\delta_1\rangle . \] By Lemma 1.3 there, \(\tau_n(ST)=\tau_n(TS)\) for \(S,T\in\operatorname{Mat}_n(\mathcal M)\), and \(\tau_n(T)\ge0\) for \(T\ge0\), with equality only for \(T=0\). Also \(\tau_n(I)=n\).

The map \(\lambda:\mathbb C[G]\to\mathcal M\), \(\lambda\bigl(\sum c_gg\bigr)=\sum c_gL(g)\), is a homomorphism of algebras, since \(L(g)L(h)=L(gh)\), and it is injective, since \(\lambda(c)\delta_1=\sum_gc_g\delta_g\). Applied entrywise, it gives an injective homomorphism \(\operatorname{Mat}_n(\mathbb C[G])\to\operatorname{Mat}_n(\mathcal M)\).

Lemma 3.2 (range projections). Let \(E\in\operatorname{Mat}_n(\mathcal M)\) be an idempotent and let \(P\) be the orthogonal projection of \(H_n\) onto the image of \(E\). Then \(P\in\operatorname{Mat}_n(\mathcal M)\) and \(\tau_n(P)=\tau_n(E)\).

Proof. The image of \(E\) is the kernel of \(I-E\): if \(y=Ex\), then \((I-E)y=Ex-E^2x=0\), and if \((I-E)y=0\), then \(y=Ey\). So it is a closed subspace. Each \(V_g=I\otimes R(g)\) commutes with \(E\), so it maps the image of \(E\) into itself; the same holds for \(V_g^*=V_{g^{-1}}\), so the orthogonal complement is also invariant, and \(PV_g=V_gP\). Hence \(P\in\operatorname{Mat}_n(\mathcal M)\).

Since \(P\) is the identity on the image of \(E\), we have \(PE=E\); since \(E\) is the identity on its own image, \(EP=P\). Put \(X=E-P\). Then \(PX=PE-P=X\) and \(XP=EP-P=0\). By the trace property, \[ \tau_n(X)=\tau_n(PX)=\tau_n(XP)=0 , \] so \(\tau_n(E)=\tau_n(P)\). \(\square\)

Proof of Theorem 3.1 for \(K=\mathbb C\). Let \(A,B\in\operatorname{Mat}_n(\mathbb C[G])\) with \(AB=I\). Write \(A'\), \(B'\) for their images in \(\operatorname{Mat}_n(\mathcal M)\). Then \(A'B'=I\), so \(E=B'A'\) is an idempotent, and \[ \tau_n(E)=\tau_n(B'A')=\tau_n(A'B')=n . \] Let \(P\) be the range projection of Lemma 3.2. Then \(\tau_n(I-P)=n-\tau_n(E)=0\), and \(I-P\ge0\), so \(P=I\). The image of \(E\) is then all of \(H_n\), and \(E\) is the identity on its image, so \(E=I\). Since the homomorphism to \(\operatorname{Mat}_n(\mathcal M)\) is injective, \(BA=I\). \(\square\)

For a subfield \(K\subseteq\mathbb C\), the inclusion \(\operatorname{Mat}_n(K[G])\subseteq\operatorname{Mat}_n(\mathbb C[G])\) gives the result. For a general field of characteristic zero we use the following lemma.

Lemma 3.3. Every finitely generated field \(F=\mathbb Q(a_1,\ldots,a_r)\) of characteristic zero has an injective homomorphism into \(\mathbb C\).

Proof. Put \(F_0=\mathbb Q\) and \(F_i=F_{i-1}(a_i)\), and suppose that \(\varphi:F_{i-1}\to\mathbb C\) is an injective homomorphism. If \(a_i\) is algebraic over \(F_{i-1}\) with minimal polynomial \(\mu\), then \(F_i\cong F_{i-1}[x]/(\mu)\), and the polynomial \(\varphi(\mu)\) has a root \(z\in\mathbb C\) by the fundamental theorem of algebra (Complex Analysis: Lebl, Guide to Cultivating Complex Analysis, Theorem 3.3.11). Sending \(x\) to \(z\) extends \(\varphi\) to \(F_i\); the extension is injective because \(F_i\) is a field. If \(a_i\) is transcendental over \(F_{i-1}\), then \(F_i\) is the field of rational functions in \(a_i\). The field \(\varphi(F_{i-1})\) is countable, so it has countably many polynomials, each with finitely many roots; since \(\mathbb C\) is uncountable, some \(z\in\mathbb C\) is not a root of any nonzero polynomial with coefficients in \(\varphi(F_{i-1})\). Sending \(a_i\) to \(z\) extends \(\varphi\) to \(F_i\), because no nonzero polynomial in \(a_i\) is sent to zero. \(\square\)

Proof of Theorem 3.1. Let \(A,B\in\operatorname{Mat}_n(K[G])\) with \(AB=I\). The finitely many coefficients of the entries of \(A\) and \(B\) generate a subfield \(F\) of \(K\), and \(A,B\in\operatorname{Mat}_n(F[G])\). By Lemma 3.3 there is an injective homomorphism \(F\to\mathbb C\); applied to coefficients, it gives an injective homomorphism \(\operatorname{Mat}_n(F[G])\to\operatorname{Mat}_n(\mathbb C[G])\). The case \(K=\mathbb C\) gives \(BA=I\) in the image, hence in \(\operatorname{Mat}_n(F[G])\). \(\square\)

The argument uses only a positive, faithful trace with \(\tau(1)\ne0\); this is the "finite tracial setting" that Kaplansky used, and Montgomery gave a short proof with the operator-norm closure of the regular representation [Montgomery 1969]. Exercise 8.5 extracts a rank inequality from the same argument.

4. What was known in positive characteristic

In positive characteristic no such trace is available, and the question stayed open. Positive results came from approximation. Ara, O'Meara and Perera proved stable finiteness of \(D[G]\) over every division ring \(D\) for residually amenable groups \(G\), as recorded in [Elek–Szabó, Section 1]. Elek and Szabó proved it for all sofic groups [Elek–Szabó, Corollary 4.7]: groups whose multiplication can be imitated by maps of finite sets in which nonidentity elements move almost every point (Sofic groups and surjunctivity, Section 1). Whether every group is sofic was a central open question [Pestov, Open question 3.8].

Dykema, Heister and Juschenko showed that direct finiteness of \(K[G\times H]\) for all finite groups \(H\) implies stable finiteness of \(K[G]\), and studied finitely presented groups that are universal for one-sided inverses with small supports [Dykema–Heister–Juschenko, abstract]. A counterexample to stable finiteness therefore gives a counterexample to direct finiteness; Packing and the reverse defect, Section 4, carries out this transfer for the matrices that occur in this course.

A second line of work connects the question to symbolic dynamics. Gottschalk conjectured that every group is surjunctive: every injective cellular automaton over it is surjective. Gromov and Weiss proved this for sofic groups [Pestov, Theorem 6.3; Capraro–Lupini, Section II.13]. A surjunctive group satisfies Kaplansky's conjecture over finite fields [Capraro–Lupini, Section I.5], and Phung showed that Gottschalk's conjecture implies stable finiteness over every field [Phung, abstract]; Ceccherini-Silberstein, Coornaert and Phung extended this to monoid algebras [CCP, abstract]. Sofic groups and surjunctivity proves the two implications that this course needs.

5. The theorem and its consequences

Theorem 5.1 ([OpenAI-DF2, Theorem 1.1]). There are a finite field \(K\) of characteristic two, a finitely presented group \(G\) containing an element of odd prime order, and \(a,b\in K[G]\) such that \[ ab=1,\qquad ba\ne1 . \]

The field, the group and the two elements are defined by a terminating finite search together with explicit formulas; the search need not be practical, and the inequality \(ba\ne1\) is proved by an algebraic certificate, without solving the word problem of \(G\).

Corollary 5.2. The group \(G\) of Theorem 5.1 is not sofic, and the cellular automaton \(x\mapsto\bigl(g\mapsto\sum_ub_ux(gu)\bigr)\) on \(K^G\) is injective but not surjective. In particular \(G\) is a finitely presented group that is neither sofic nor surjunctive.

Theorem 5.1 is proved in Line partitions, girth and the counterexample, and Corollary 5.2 follows there from Sofic groups and surjunctivity. By Theorem 3.1, the field must have positive characteristic; by Lemma 2.1, the group must be infinite.

The OpenAI release contains three further manuscripts in the same family.

In characteristic zero the trace of Section 3 goes further. OpenAI proved in 2026 that the Hattori–Stallings trace of a projective module over \(\mathbb C[G]\) vanishes at the classes of elements of infinite order; for torsion-free \(G\) this makes the trace of an idempotent an integer, and the range projections of Lemma 3.2 then show that the only idempotents of \(R[G]\) are \(0\) and \(1\), for every commutative domain \(R\) of characteristic zero. This is Kaplansky's idempotent conjecture in characteristic zero (Idempotents and integral traces, Theorem 2.1). The completion behaves differently: OpenAI's torsion-free group with a projection of trace strictly between \(0\) and \(\frac12\) in its reduced group \(C^*\)-algebra is the subject of A torsion-free counterexample to the Kadison–Kaplansky conjecture.

Later work builds on the existence of nonsofic groups. Gelander constructs invariant random subgroups that are not co-sofic in rank-one simple Lie groups and in automorphism groups of trees, giving counterexamples to a continuous analogue of the Aldous–Lyons conjecture; the construction starts from a finitely generated nonsofic group [Gelander 2026, abstract]. On the positive side, every one-relator group is sofic [Son 2026, abstract], so the group of Theorem 5.1 has no one-relator presentation.

7. The plan of the course

8. Exercises

8.1 (easy). Let \(V\) be the space of finitely supported sequences \((x_0,x_1,\ldots)\) over a field. Let \(r(x)=(0,x_0,x_1,\ldots)\) and \(l(x)=(x_1,x_2,\ldots)\). Show that \(lr=\mathrm{id}\) and \(rl\ne\mathrm{id}\).

8.2 (easy). Show that every commutative ring is directly finite, and that \(K[G]\) is directly finite when \(G\) is abelian.

8.3 (easy). Let \(K\subseteq L\) be fields. Show that if \(L[G]\) is directly finite, so is \(K[G]\).

8.4 (medium). Let \(R\) be directly finite and \(f\in R\) an idempotent. Show that the ring \(fRf\), with identity \(f\), is directly finite.

8.5 (medium). Let \(A\) be an \(n\times m\) and \(B\) an \(m\times n\) matrix over \(\mathbb C[G]\) with \(AB=I_n\). Show that \(n\le m\), and that \(BA=I_m\) when \(n=m\).

9. Solutions

8.1. \(lr(x)=l(0,x_0,x_1,\ldots)\), which is \((x_0,x_1,\ldots)=x\). For \(x=(1,0,0,\ldots)\), \(rl(x)=r(0,0,\ldots)=0\ne x\).

8.2. If \(ab=1\) and \(R\) is commutative, then \(ba=ab=1\). If \(G\) is abelian, the product formula \((cd)_w=\sum_{uv=w}c_ud_v\) is symmetric in \(c\) and \(d\), so \(K[G]\) is commutative.

8.3. The inclusion \(K\subseteq L\) gives an injective homomorphism \(K[G]\to L[G]\). If \(ab=1\) in \(K[G]\), the same holds in \(L[G]\), so \(ba=1\) there, and by injectivity \(ba=1\) in \(K[G]\).

8.4. Let \(x,y\in fRf\) with \(xy=f\). Since \(xf=x\), \(fy=y\) and \(f(1-f)=(1-f)f=0\), \[ \begin{aligned} (x+1-f)(y+1-f)&=xy+(1-f)\\ &=1 . \end{aligned} \] By direct finiteness of \(R\), \((y+1-f)(x+1-f)=1\), that is \(yx+(1-f)=1\), so \(yx=f\).

8.5. Let \(A'\), \(B'\) be the images in rectangular matrices over \(\mathcal M\), and let \(\tau\) denote the sum of the diagonal coefficients \(\langle T_{jj}\delta_1,\delta_1\rangle\) of a square matrix. The computation in the proof of Lemma 1.3 of the trace lesson, \(\tau(ST)=\sum_{j,l}\tau(S_{jl}T_{lj})=\tau(TS)\), applies to rectangular \(S\) and \(T\) of compatible sizes, so \(\tau_m(B'A')=\tau_n(A'B')=n\). The square matrix \(E=B'A'\) is an idempotent in \(\operatorname{Mat}_m(\mathcal M)\). With \(P\) its range projection, Lemma 3.2 gives \(\tau_m(P)=n\), and \(I-P\ge0\) gives \(m-n=\tau_m(I-P)\ge0\). If \(m=n\), then \(\tau_m(I-P)=0\), so \(P=I\) and \(E=I\) as in the proof of Theorem 3.1.

References