Isaacs's fake degree conjecture for algebra groups

About 11 years old · traced to 2015

Let $J$ be a finite-dimensional nilpotent algebra over a finite field $\operatorname{F}$ , let $G=1+J$ be the associated algebra group, and let $J^*=\operatorname{Hom}_{\operatorname{F}}(J,\operatorname{F})$ , with $G$ acting on $J^*$ by conjugation. Isaacs's fake degree conjecture. In every algebra group $G=1+J$ , the character degrees coincide, counting multiplicities, with the square roots of the cardinalities of the $G$ -orbits in $J^*$ . This extends the known correspondence when $J^p=0$ ; the paper's abstract states that algebra groups arising from finite $p$ -groups with nontrivial Bogomolov multiplier provide counterexamples, so the conjecture is refuted.

References

Primary source

Javier Garcia-Rodriguez, Andrei Jaikin-Zapirain and Urban Jezernik, “Units of group rings, the Bogomolov multiplier, and the fake degree conjecture”, arXiv:1502.03242 (2015).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2021 paper gives counterexamples in every characteristic, refuting the conjecture.

The conjecture predicts that character degrees of every algebra group $G=1+J$ match the square roots of orbit sizes in $J^*$ . The correspondence is known when $J^p=0$ .

May 28, 2021 counterexamples

The article Units of group rings, the Bogomolov multiplier, and the fake degree conjecture derives

\left|(1+I_{\mathbb{F}_q})^{\mathrm{ab}}\right|=q^{k(\pi)-1}\left|B_0(\pi)\right|.

For finite $p$ -groups with nontrivial Bogomolov multiplier, this contradicts the equality implied by the conjecture; Corollary 4 states that the conjecture is invalid in every characteristic.

Current status (as of September 2026): The conjecture is reported refuted in every characteristic by counterexamples from finite $p$ -groups with nontrivial Bogomolov multiplier; no unresolved case is identified in the retrieved sources.

Sources

No solutions have been posted yet.