The small Davenport constant $\mathsf{d}(G)$ of a finite group $G$ is the maximal length of a product-one-free sequence over $G$. For the exponent-$p$ Heisenberg group $H_{p^3}$ of order $p^3$, Godara and Sarkar proved $\mathsf{d}(H_{27})=6$ and posed $\mathsf{d}(H_{p^3})=3p-3$ for every odd prime $p$, leaving $p\ge5$ open. We settle the first open case: $\mathsf{d}(H_{125})=12$. The lower bound is the explicit product-one-free sequence $x^4y^4v^4$. For the upper bound we record a product-one criterion that reduces the non-commutative problem to additive combinatorics over $\mathbb{F}_5^2$, and then reduce"every length-13 sequence has a product-one subsequence"to a single finite statement -- a spread bound on quotient multisets -- which we verify by an exhaustive, memory-flat search in C, its verdict independently reproduced by a second search with a different pruning strategy. Every auxiliary lemma is machine-checked. The argument is genuinely $p$-specific: we identify the exact step that fails for $p\ge7$ (a Chevalley-Warning shortcut whose forced block need not be wide), exhibit the obstructing multiset for $p=7$, and leave only $18\le\mathsf{d}(H_{343})\le24$. The techniques -- the Cauchy-Davenport theorem, Chevalley-Warning, and Olson's value of the Davenport constant of $C_p^2$ -- are standard; the contribution is their assembly against a new non-abelian target and the finite verification that closes it.
For a finite group $G$, let $\mathsf{d}(G)$ denote the maximum length of a sequence having no nonempty subsequence whose terms can be ordered to have product one. For an odd prime $p$, let $H_{p^3}=\operatorname{UT}*3(\mathbb{F}*p)$. Godara and Sarkar proved $\mathsf{d}(H*{27})=6$ and conjectured $\mathsf{d}(H*{p^3})=3p-3$; in a recent preprint, White proved the next case $\mathsf{d}(H_{125})=12$ and left $18\leq\mathsf{d}(H_{343})\leq24$. We prove $\mathsf{d}(H_{343})=18$. We adopt White's product-one criterion and spread framework and develop a $p=7$-specific direction stratification. An explicit product-one-free sequence gives the lower bound. For the upper bound, we stratify a hypothetical product-one-free sequence of length $19$ by the number of central terms and by the occupied projective directions of its quotient multiset. Supports on at most two directions are excluded by a theoretical argument whose finite auxiliary statements are exhaustively checked; the three-direction case and the case of five central terms are settled by exact finite computations. The remaining thirty strata are encoded by a counterexample-guided SAT procedure. A separately implemented checker verifies all $9{,}920{,}815$ seed cuts and all $27{,}207$ learned cuts, and each final unsatisfiable instance is accompanied by a checked LRAT certificate. A separate implementation-level audit verifies the master encoding, the proof archives, and the lower-bound witness.
Let $p$ be an odd prime and let $H_{p^3}=\operatorname{UT}_3(\mathbb{F}_p)$ be the Heisenberg group of order $p^3$ and exponent $p$. We prove $\mathsf{d}(H_{p^3})=3p-3$. The main ingredient of the proof is an order-value growth theorem. If $B$ is a noncollinear zero-sum sequence of $n$ nonzero vectors in $\mathbb{F}_p^2$, then the alternating areas obtained by ordering $B$ assume at least $\min(p,n-1)$ distinct values. Its proof is a short contraction induction: contract a suitable independent pair, replace the contracted vector in both orders, and apply Cauchy--Davenport. A polynomial relative-subsum theorem and a sharp representation-rigidity lemma then turn this local growth into a uniform spread bound. Combined with the standard product-one criterion for $H_{p^3}$, the spread bound yields the upper bound; the usual sequence $x^{p-1}y^{p-1}v^{p-1}$ gives the lower bound.
Let $G$ be a finite nonabelian group. The small Davenport constant $\mathsf d(G)$ of $G$ is the largest integer $\ell$ such that there exists a product-one free sequence over $G$ of length $\ell$, while the Gao constant $E(G)$ of $G$ is the least integer $\ell$ such that every sequence over $G$ of length at least $\ell$ contains a product-one subsequence of length exactly $|G|$. A long-standing conjecture of Gao and Zhuang \cite{ZG2005} asserts that $E(G)=\mathsf d(G)+|G|$ for every finite nonabelian group $G$. Let $p$ be an odd prime and let $H_{p^3}=\operatorname{UT}_3(\mathbb F_p)$ be the finite Heisenberg group over $\mathbb F_p$. Godara and Sarkar proved the Gao-Zhuang equality for $H_{27}=\operatorname{UT}_3(\mathbb F_3)$ and asked whether the same equality holds for $H_{p^3}$ for every odd prime $p$. Recently, Volkmann proved that $\mathsf d(H_{p^3})=3p-3$. In this paper, we determine the Gao constant of $H_{p^3}$ and prove that $E(H_{p^3})=\mathsf d(H_{p^3})+|H_{p^3}|=p^3+3p-3$. Together with the known abelian and cyclic-index cases, this completes the verification of the Gao-Zhuang equality for all groups of order $p^3$, for every prime $p$.
For a finite abelian group $G$ and $k \geq 1$, the generalized Davenport constant $D_k(G)$ is the least $\ell$ such that every sequence over $G$ of length at least $\ell$ has $k$ pairwise disjoint nonempty zero-sum subsequences. A theorem of Freeze and Schmid gives $D_k(C_5^3) \geq 5k+10$ for every $k \geq 2$. We prove the matching upper bound: $D_4(C_5^3)=30$, and hence $D_k(C_5^3)=5k+10$ for every $k \geq 2$, so the Freeze--Schmid bound is attained by $C_5^3$ from $k=2$ onward, as it is by $C_2^3$ and unlike $C_3^3$. The proof is finite and computer-assisted. The remaining case reduces to showing that every zero-sum sequence of length $31$ over $C_5^3$ contains a nonempty zero-sum subsequence of length at most five. A saturation argument confines the multiplicities of a hypothetical counterexample to $\{1,2,4\}$, its support pattern to one of $60$ solutions of two linear equations, and its geometry to one of $78$ rank/plane branches normalized to a standard basis; an exhaustive search exhausts every branch with no survivor. The search was carried out by three independently written implementations, and the branch cover was regenerated by separate programs from the lemmas alone; two further machine-verified values, $D_3(C_5^3)=25$ and $s_{\leq 6}(C_5^3)=24$, enter the second statement, and their records accompany the paper.
We show that von Neumann algebras of fundamental groups of closed orientable surfaces of genus $g\geq2$ are free group factors on $2g-1$generators. The key technical ingredient involves a proof that the element $w=ABA^{-1}B^{-1}$ of the free group $\mathbb{F}_{2}=\langle A,B\rangle$ is freely complemented in the group factor: $L(\mathbb{F}_{2})=W^{*}(w)*W^{*}(v)$ for some Haar unitary $v\in L(\mathbb{F}_{2})$ that is freely independent from $w$. Combined with previous results, we conclude that for an arbitrary finitely generated torsion-free non-elementary discrete subgroup $\Gamma\subset PSL_{2}(\mathbb{R})$, $L(\Gamma)$ is a free group factor, settling a conjecture of de la Harpe and Voiculescu. This result was obtained using OpenAI's ChatGPT Pro 6.0.
Fix an integer $K\ge2$, and let $C_K^n =\{(e^{\frac{2\pi ij}{K}})_{j=0}^{K-1}\}^n$ be the product of cyclic groups of order $K$. For a Fourier character $\chi_\alpha$, let $s(\alpha)$ be the number of active coordinates. We give a self-contained proposed proof that the dimension-free Bohnenblust--Hille constants governed by interaction order grow polynomially: if $p_d=2d/(d+1)$ and \[ \gamma_2=\frac12, \qquad \gamma_K=\frac{K\log(K-1)}{4(K-2)}\quad(K\ge3), \] then the $\ell^{p_d}$ norm of Fourier coefficients $\{\hat f(\alpha)\}$ is bounded by $L^\infty$ norm of $f$ multiplied by $C(K) d^{4\gamma_K+5}$. The constant $C(K)$ is actually at most of the order $K^{5/2}$.
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