Each cell stands for all groups of the given order and rank (number of generators). The colour says what settles the
question for all of them: a computation, a proof, or both. Hover over a cell for the statement of the draft it rests on; a click opens that statement.
$2$-groups of class $3$: is $D_5(Q)=1$?
This is Problem 20.84 itself: by the reduction (Proposition 2.1), $D_5\subseteq\gamma_4$ for all groups if and only if
$D_5(Q)=1$ for every finite $2$-group of class $3$. Class $3$ forces $d\le n-2$ at order $2^n$.
The open cells at order $2^{11}$, ranks $5$–$7$, are outlined.
C computer C+P computer + paper P paper ? open. Empty cells: no such groups.
Up to $2^8$: the lattice $\Delta^n$ for all groups; $2^9$: the Jennings sieve and the lattice.
$2^{10}$: the minimal-counterexample lemma reduces to step-$1$ descendants of groups of order $2^9$ (ANUPQ), $3400$ candidates.
$2^{11}$: the same, with whole subtrees cut; rank $8$ by quadratic forms and Lemma 5.6, rank $9$ impossible.
$3$-groups: is $D_5(Q)=\gamma_5(Q)$?
This is the question of $3$-torsion in $D_5/\gamma_5$ (Tahara: the exponent divides $6$). The “paper” cells are closed thus:
if $Q/\gamma_5$ is metabelian, then $D_5=\gamma_5$ by Gupta–Tahara, and a non-metabelian group of class $4$ and rank $d$
has order $\ge3^{d+5}$ (Proposition 6.5); for $d=2$ the factor $\gamma_2/\gamma_3$ is cyclic. The rest is the search up to $3^9$.
Open cells are hatched: there the case of $Q/\gamma_2$ with at most one invariant $>3$ is proved (Theorem 6.8).
C computer C+P computer + paper P paper ? open; the case of $Q/\gamma_2$ with at most one invariant $>3$ proved (Thm. 6.8). Empty cells: no such groups.