Nonabelian Erdős–Ginzburg–Ziv conjecture

For every finite nonabelian group GG, let exp⁡(G)\exp(G) be the exponent of GG. Define s(G)s(G) to be the least integer ℓ\ell such that every sequence over GG of length at least ℓ\ell has a product-one subsequence of length exactly exp⁡(G)\exp(G), and define η(G)\eta(G) to be the least integer ℓ\ell such that every sequence over GG of length at least ℓ\ell has a nonempty product-one subsequence of length at most exp⁡(G)\exp(G). Here a subsequence g1,…,gkg_1,\ldots,g_k is product-one if its terms can be ordered so that gπ(1)⋯gπ(k)=1Gg_{\pi(1)}\cdots g_{\pi(k)}=1_G. The conjecture asserts that s(G)=η(G)+exp⁡(G)−1s(G)=\eta(G)+\exp(G)-1.

References

Primary source

arXiv

Progress summary

Refreshed
Claimed progress

A new preprint proves the conjectured equality for a broad but restricted family of nonabelian groups, while the full conjecture remains open.

Gao and Li formulated the conjecture in 2010: for every finite noncyclic group GG, the Erdős–Ginzburg–Ziv constant should satisfy E(G)≤3∣G∣/2\mathsf{E}(G)\leq 3|G|/2. The general finite nonabelian case is not settled.

Known results

  • Gao, Li, and Qu (2023) proved the conjectured bound for all groups of odd order.
  • Earlier work handled additional special families, including groups of order 2pt2p^t under an abelian-subgroup hypothesis.
  • For finite noncyclic groups with 4∤∣G∣4\nmid |G|, E(G)≤3∣G∣/2\mathsf{E}(G)\leq 3|G|/2, with equality exactly when GG has a cyclic subgroup of index 22.

August 16, 2026 restricted-family proof

Qu, Wang, and Li prove the equality s(G)=η(G)+exp⁡(G)−1=d(G)+exp⁡(G)s(G)=\eta(G)+\exp(G)-1=\mathsf{d}(G)+\exp(G) when GG has a cyclic subgroup HH of index pp, with pp the smallest prime divisor of ∣G∣|G|. They also determine smexp⁡(G)(G)=η(G)+mexp⁡(G)−1s_{m\exp(G)}(G)=\eta(G)+m\exp(G)-1 for every m≥1m\geq1. This advances the conjecture substantially but leaves other finite nonabelian groups open.

Current status (as of August 2026): A substantial restricted family is settled by a new preprint, but the conjecture for all finite nonabelian groups remains open.

Sources

Solutions 0

No solutions have been posted yet.