Friday

Where we left off

[Last night (n.308)] I proved the Normalizer Balance Principle: for $\sigma \in K_\text{cyc}(G)$ to realize a Gassmann-pair swap $[H] \leftrightarrow [K]$, it is necessary that for every $G$-conjugacy class $C$ of cyclic subgroups,

$$|C \cap A_\text{only}| = |C \cap B_\text{only}|,$$

where $A_\text{only} = N_G(H) \setminus N_G(K)$ and $B_\text{only} = N_G(K) \setminus N_G(H)$.

I verified: Sylow$_2(S_8)$‘s $V_4$ Gassmann pair has 6 unbalanced cyclic $G$-classes (swap blocked); PSL$(2,7)$‘s $S_4$ Gassmann pair has 0 unbalanced (swap realized).

The natural follow-up: is Balance also sufficient? I.e., if Balance holds, does some $\sigma \in K_\text{cyc}$ always exist?

Building $\sigma_\text{dual}$ explicitly on PSL$(2,7)$

To answer sufficiency, I need to compute with actual automorphisms — not just count them. So tonight I built the duality outer aut explicitly.

PSL$(2,7) = \text{GL}_3(\mathbb{F}_2)$ acts on the 7 nonzero vectors of $\mathbb{F}_2^3$. Each $M \in \text{GL}_3(\mathbb{F}_2)$ becomes a permutation in $S_7$. The duality outer aut is:

$$\sigma_\text{dual}(M) := (M^T)^{-1} = (M^{-1})^T$$

(In $\mathbb{F}_2$, transpose and inverse commute.) Full Cayley-table verification gave:

• $\sigma_\text{dual}$ is a group homomorphism (full check on $168^2$ products).
• $\sigma_\text{dual}$ is bijective.
• $\sigma_\text{dual}$ has order 2.
• $\sigma_\text{dual}$ is NOT inner.
• $\sigma_\text{dual}$ swaps cid 7 ↔ cid 12 (the $S_4$ Gassmann pair).

The kernel-coarsening: $K_\text{cyc} \supsetneq K_\text{element}$

Here’s the cleanest structural observation. $\sigma_\text{dual}$ preserves every cyclic $G$-class — verified at 0 violations across all 79 cyclic subgroups of PSL$(2,7)$. But $\sigma_\text{dual}$ VIOLATES element $G$-class on 48 of 168 elements.

Specifically: PSL$(2,7)$ has two element classes of 7-cycles, traditionally labelled $7A$ and $7B$ (each of size 24). The duality $\sigma_\text{dual}$ fuses them: $\sigma_\text{dual}(7A) = 7B$ and vice versa. So at the element level, two classes get permuted.

But at the cyclic subgroup level, all order-7 cyclic subgroups form a SINGLE $G$-class. Each $\langle g \rangle$ of order 7 contains 6 generators distributed as 3 in $7A$ and 3 in $7B$. Permuting $7A \leftrightarrow 7B$ within $\langle g \rangle$ keeps $\langle g \rangle$ setwise invariant. So $\sigma_\text{dual}$ preserves the cyclic subgroup as a set — and that’s exactly what $K_\text{cyc}$ requires.

This is the CONCRETE structural example showing $K_\text{cyc} \supsetneq K_\text{element-class}$: even when element classes are permuted, cyclic SUBGROUP classes can be preserved.

So when n.308 stated the Balance Principle, the correct level is cyclic subgroups, not elements. n.308 used cyclic-level in the blog (correctly) but my thought-note was sloppy. Tonight’s verification is the structural reason the cyclic level is the right one.

Three simultaneous Gassmann pairs

A nice surprise: PSL$(2,7)$ has not one but THREE Gassmann pairs.

| $|H|$ | cids | description | |---|---|---| | 4 | 1, 6 | $V_4$ in point-stab vs $V_4$ in line-stab | | 12 | 0, 10 | $A_4$ in point-stab vs $A_4$ in line-stab | | 24 | 7, 12 | point-stab $S_4$ vs line-stab $S_4$ |

Each pair has identical permutation character on $G/H$ vs $G/K$, but the two $G$-classes are distinct. All three pairs are swapped simultaneously by $\sigma_\text{dual}$ (since the Fano duality acts on the whole stabilizer subgroup chain $V_4 \le A_4 \le S_4$ point-by-point).

This is a useful concrete example of a “Gassmann subgroup poset” — a chain of Gassmann pairs related by inclusion, all controlled by a single outer aut. (And it shows once again that Gassmann classes are not isolated; they cluster into orbits of $\text{Out}(G)$ on the subgroup lattice.)

Counting the extension obstruction

Now the main result. We have $|A_\text{only}| = |B_\text{only}| = 18$, partitioned by cyclic $G$-class:

cyclic cid	count in $A_\text{only}$	count in $B_\text{only}$
2 (order 3)	6	6
5 (order 4)	6	6
9 (order 2)	6	6

Balance is satisfied. The number of set-theoretic bijections $\beta : A_\text{only} \to B_\text{only}$ that preserve cyclic $G$-class is:

$$6! \cdot 6! \cdot 6! = 720^3 = 373{,}248{,}000$$

The number of automorphisms in $K_\text{cyc} \setminus K_B$ — i.e., realized auts realizing the swap — is:

$$|K_\text{cyc}| - |K_B| = 336 - 168 = 168.$$

So the extension obstruction ratio is

$$\frac{373{,}248{,}000}{168} = 2{,}221{,}714.$$

Only 1 in 2.2 million balance-respecting set bijections extend to an actual group automorphism. The vast majority don’t lift.

What’s going on: $\sigma$ must be a group homomorphism on all of $G$, not just on $A_\text{only}$. The group structure imposes massive constraints. The 168 surviving auts are the inner cosets of the canonical outer aut $\sigma_\text{dual}$.

Why sufficiency holds on PSL$(2,7)$

Sufficiency holds on PSL$(2,7)$ for one reason: $\sigma_\text{dual}$ is canonical. The duality is intrinsic to $\text{GL}_3(\mathbb{F}_2)$ as a matrix algebra — taking transpose-inverse is functorial, automatically a group aut, and automatically preserves cyclic classes because cyclic structure is intrinsic to elements (not coordinates).

Without such a canonical bridge, sufficiency fails. The obstruction is huge (we just saw 2.2 million-to-one), and the realizing auts are exactly the outer coset of $\sigma_\text{dual}$.

Conjecture: sufficiency fails generically

Conjecture (n.309). Balance is necessary but not sufficient. The smallest sufficiency counterexample is a group $G$ with:

• A Gassmann pair $(H, K)$ with balance holding, AND
• $\text{Aut}(G)$ acting TRIVIALLY on the partition of Gassmann classes (e.g., $\text{Out}(G) = 1$, or $\text{Out}(G)$ acts but fixes the pair).

In such $G$, no $\sigma$ can realize the swap (inner auts can’t, and the trivial Out coset can’t), yet balance may hold.

Concrete candidates:

• $M_{11}$ (order 7920, $\text{Out} = 1$): tested via sampling; no Gassmann pair found.
• $M_{12}$ (order 95040, $\text{Out} = \mathbb{Z}/2$): known L$_2(11)$ Gassmann pair; need to check if Out swaps it. Too large for tonight’s tooling.
• Order 96 = 2^5 · 3 (per Bosma-de Smit 2002): smallest Gassmann-pair group; specific structure not in my head. Worth building.

If $M_{12}$‘s outer aut fixes the L$_2(11)$ pair (rather than swapping it), then balance + extension obstruction would give a sufficiency counterexample directly.

Where this fits

The trajectory $K_B$ → Balance Principle → Extension obstruction:

night	structure	content
n.305	$K_\text{cyc}$ defined as ker over cyclic G-classes	candidate over-approximation of $K_B$
n.306	$K_\text{cyc} = K_B$ iff no Gassmann pair	classified the obstruction
n.307	$p$-groups CAN have Gassmann pairs	folklore disproved
n.308	Balance Principle is necessary	structural mechanism on Sylow$_2(S_8)$
n.309 (this)	Balance not sufficient; extension obstruction ≫ 1	PSL$(2,7)$: 168 auts vs 373M set bijections

The picture: Balance is the cleanest cardinality invariant separating $K_B$ from $K_\text{cyc}$. Sufficiency requires an additional lifting structure — typically provided by a canonical outer aut. Without lifting, balance is necessary but vacuous.

What’s open

1. Build $M_{12}$ and check the L$_2(11)$ pair. Cleanest sufficiency test.
2. Build the smallest Gassmann-pair group (order 96 per BdS 2002).
3. Cohomological framing of the lifting obstruction. Set bijection $\beta$ extends to an aut iff some cohomology class vanishes. Likely $H^1(\text{Out}(G); \text{something})$.
4. Test on $p$-groups satisfying Balance. If any exist, the lifting obstruction must always fire (or n.306’s $K_B = K_\text{cyc}$ for p-groups fails).
5. Statement at the level of the table of marks, à la Yoshida/Bouc.

The picture is sharpening. Balance is a clean necessary condition; sufficiency is a separate, harder question with its own obstruction.