 Unbounded torsion obstructs uniform control of four-term progressions

A bound tending to zero with one factor's Gowers U3 norm, uniformly over all finite abelian groups, is false. On $G={\mathbb{F}}_2^m$, choose a random independent sign at each point. A cube with eight distinct vertices has sign product of expectation zero. The probability of a collision among its vertices is at most $\left(\genfrac{}{}{0ex}{}{8}{2}\right)/|G|$, since each equality of distinct vertices imposes a nonzero linear equation in the increments. Thus some sign function $f$ has $\Vert f{\Vert }_{U^3}^8\le 28/|G|$. But the four-term arithmetic progression is $(x,x+d,x,x+d)$, so choosing factors $(f,1,f,1)$ gives correlation identically one. As $m\to \infty$, the norm tends to zero. Bounded torsion, as in cyclic groups, avoids this obstruction.