Gao's theorem for nonabelian groups

Definitions and mathematical statement

Let $G$ be a finite group, written multiplicatively with identity $1$. A sequence is a finite indexed list of elements, with repetitions allowed. A nonempty selection of its positions is product-one if the selected elements can be ordered so that their product is $1$. This ordering may differ from their order in the original list.

Define $s(G)$ to be the least integer $m$ such that every length-$m$ sequence has a product-one selection, and define $s'(G)$ in the same way but requiring exactly $|G|$ selected positions. Both thresholds exist: a sufficiently long sequence contains $|G|$ copies of one element, whose product is $1$. The conjecture is \[ s'(G)=s(G)+|G|-1 \qquad\text{for every finite group }G. \] These are thresholds for sequences, not sets of distinct elements. The integer $s(G)-1$ is the largest length of a product-one-free sequence. Appending $|G|-1$ identities to such a sequence shows $s'(G)\ge s(G)+|G|-1$: a product-one selection of size $|G|$ would have to use a nonempty selection of the original terms. Thus the matching upper bound is the substantive universal claim. Noncommuting elements are allowed throughout, and order may be chosen separately for each candidate selection.

Short English statement

Does the usual relation between the two product-one subsequence thresholds hold for every finite group, including noncommutative groups?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3378 (OPG-337), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, Gao's theorem for nonabelian groups. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/gaos_theorem_for_nonabelian_groups.