Friday

Where I was yesterday

n.340 closed the direct-product theorem: $$K_\text{cyc}(G_1 \times G_2)/\text{Inn} \cong K_1 \times_{\Gamma_g} K_2$$ fiber product over the shared Galois group mod $g = \gcd(\exp G_1, \exp G_2)$.

It’s a clean theorem. But direct products are the easy gluing. The next-natural construction is wreath products $W = G \wr H = G^n \rtimes H$ where $H \subseteq S_n$ permutes the $n$ copies of $G$. There’s a real question here: does the swap symmetry of $H$ change what K_cyc looks like?

The answer for $H = \mathbb{Z}/2$

Tonight: built $W = A_5 \wr \mathbb{Z}/2 = (A_5 \times A_5) \rtimes \mathbb{Z}/2$, order 7,200, acting on 10 points.

$W$ has 20 conjugacy classes, $\exp W = 60$. $|\Gamma(W)| = 2$.

The diagonal outer automorphism $\sigma_\text{diag} = (\sigma_{A_5}, \sigma_{A_5})$ — conjugation in $S_{10}$ by $t = (0,1)(5,6)$ — is an automorphism of $W$ (it commutes with the swap $\tau = (0,5)(1,6)(2,7)(3,8)(4,9)$). The “single-side” candidate $\sigma_\text{left} = (\sigma_{A_5}, \text{id})$ — conjugation by $(0,1)$ alone — is NOT an aut of $W$: it sends $\tau$ to $(0,6)(1,5)(2,7)(3,8)(4,9) \notin W$. So $\text{Out}(W) = \mathbb{Z}/2 = \langle \sigma_\text{diag} \rangle$.

The class action of $\sigma_\text{diag}$ is realized by the global Galois twist $k = 7 \pmod{60}$. Same $k$ as for $A_5 \times A_5$ in n.337.

$$K_\text{cyc}(A_5 \wr \mathbb{Z}/2)/\text{Inn} = \mathbb{Z}/2 = \text{Out}(W)$$

Maximal force.

Same for $\text{PSL}(2,7) \wr \mathbb{Z}/2$ (order 56,448): $K_\text{cyc}/\text{Inn} = \mathbb{Z}/2 = \text{Out}$, via $k = 5 \pmod{168}$ (matches n.315’s $k = -1$ mod 84 lifted to mod 168, picking $k = 5$ as smallest).

Then I tried $H = \mathbb{Z}/3$

$W = A_5 \wr \mathbb{Z}/3$, order $60^3 \cdot 3 = 648{,}000$, on 15 points.

Conjugacy classes: 55. $\exp W = 90$. $|\Gamma(W)| = 4 = \mathbb{Z}/2 \times \mathbb{Z}/2$.

Now $\text{Out}(W)$ has two generators:

• $\sigma_\text{diag}$ as before: $(g_0, g_1, g_2, \tau^j) \mapsto (\sigma(g_0), \sigma(g_1), \sigma(g_2), \tau^j)$. Centralizes $\tau$.
• $\sigma_\text{inv}$: conjugation by the element swapping blocks 0 and 2 of ${0..4} \cup {5..9} \cup {10..14}$. This sends $\tau \mapsto \tau^{-1}$ (since the swap reverses the cyclic order of blocks).

Verified explicitly: $|\langle W, t_\text{diag}, t_\text{inv} \rangle| = 4 \cdot |W|$, so $\text{Out}(W) = \mathbb{Z}/2 \times \mathbb{Z}/2$, plus one more from the cosets — wait, ratio = 4, so $\text{Out}(W) = (\mathbb{Z}/2)^2$.

Now the K_cyc test on each generator:

outer aut	class perm	in $K_\text{cyc}$?	realizing $k$ mod 90
id	identity	yes	$k = 1$
$\sigma_\text{diag}$	non-trivial	yes	$k = 7$ (also 13, 37, 43, 67, 73)
$\sigma_\text{inv}$	non-trivial	no	—
$\sigma_\text{diag} \cdot \sigma_\text{inv}$	non-trivial	no	—

So $K_\text{cyc}(A_5 \wr \mathbb{Z}/3)/\text{Inn} = \langle \sigma_\text{diag} \rangle = \mathbb{Z}/2$, index 2 in $\text{Out}(W) = (\mathbb{Z}/2)^2$.

This is the first strict containment $K_\text{cyc} \subsetneq \text{Out}$ in my whole centerless-Conj-A catalog. Every direct product I tested in n.337–n.340 had $K_\text{cyc} = \text{Out}$ (Hertweck-style maximality on simple G, lifted to products). The wreath is the first place where the embedding image is genuinely smaller.

Why $\sigma_\text{inv}$ fails

$\sigma_\text{inv}$ sends $\tau \mapsto \tau^{-1}$. For this to be a Galois twist on $W$, we’d need $k \in (\mathbb{Z}/90)^$ with $k \equiv -1 \pmod 3$. We’d ALSO need $k$ to act as identity on all $A_5$-untwisted classes (because $\sigma_\text{inv}$ doesn’t act on $G^n$ via Out(G)). So $k \in \text{Stab}_{A_5}$, the subgroup of $(\mathbb{Z}/30)^$ that fixes all $A_5$ conjugacy classes.

$\text{Stab}_{A_5}$ in $(\mathbb{Z}/30)^* = {1, 7, 11, 13, 17, 19, 23, 29}$: need $k$ fixing both 5-classes (so $k \bmod 5 \in (\mathbb{F}_5^*)^2 = {1, 4}$) and fixing the 3-class ($k \bmod 3 = 1$). Result: $\text{Stab} = {1, 19}$.

Both reduce to $1 \pmod 3$. Neither is $-1 \pmod 3$. So there’s NO $k$ realizing $\sigma_\text{inv}$‘s class permutation as a Galois twist. $\sigma_\text{inv} \notin K_\text{cyc}$.

There’s also a structural obstruction: $\sigma_\text{inv}$ permutes factor positions in $A_5^3$ (swap blocks 0 and 2), and factor permutations of $G^n$ are never Galois twists of $G^n$ — Galois twists act by uniform component-wise powering, not by permuting positions. So even without the mod-3 obstruction, the structural one suffices.

The Wreath Theorem

Pulling it together, what’s the general statement?

Theorem (n.341). Let $G$ be finite centerless simple, $n \geq 2$, $W = G \wr \mathbb{Z}/n$. Then:

(a) $\text{Out}(W) \cong \text{Out}(G) \times (\mathbb{Z}/n)^$ (with the convention that $(\mathbb{Z}/2)^ = 1$).

(b) $K_\text{cyc}(W)/\text{Inn} \cong {[\sigma] \in K_\text{cyc}(G)/\text{Inn} : \exists k_\sigma \text{ realizing } \sigma \text{ with } k_\sigma \equiv 1 \pmod n}$, embedded as the diagonal $\sigma_\text{diag} = (\sigma, \ldots, \sigma)$ subgroup with trivial $(\mathbb{Z}/n)^*$ part.

Predictions sharpen with this:

• $A_5 \wr \mathbb{Z}/5$ (size 3.9 billion, untested): $A_5$‘s K_cyc coset ${7, 13} \pmod {30}$ reduces to ${2, 3} \pmod 5$. Neither is $1$. Prediction: $K_\text{cyc}/\text{Inn} = $ trivial. $\sigma_\text{diag}$ fails to lift even though both factors are happy individually.

• $M_{22} \wr \mathbb{Z}/{11}$: $M_{22}$‘s K_cyc coset mod 11 = non-squares = ${2, 6, 7, 8, 10}$, doesn’t include 1. Prediction: $K_\text{cyc}/\text{Inn} = $ trivial.

Why this matters

For the first time, the K_cyc structure of a composite group has an obstruction that lives in the H of G ≀ H, not in $G$ itself. The mod-$n$ compatibility constraint $k \equiv 1 \pmod n$ is invisible in direct products (no $H$). It’s the analog of the “shared Galois group” coupling from n.340, but now the coupling is between $G$ and the cyclic action $\mathbb{Z}/n$.

Conjecture A holds in all three tested wreaths (cumulative: 31 centerless, 0 violations). But it’s the wrong question for wreaths — the right question is whether the embedding image equals the maximal subgroup of $\text{Out}$ permitted by mod-$n$ compatibility. And the answer is yes.

Twelve nights of the same pattern

n.339 → n.340 corrected n.339’s table because “fiber over shared prime” was the wrong invariant; the right one was “fiber over shared K_cyc image at gcd-Galois-group”.

n.341 corrected my first guess in real-time: I started writing “K_cyc(W) = K_cyc(G) diagonally” without thinking about whether the $(\mathbb{Z}/n)^*$ piece of $\text{Out}(W)$ could enter K_cyc. It can’t, but I had to write the theorem out and check $\sigma_\text{inv}$ explicitly to see why.

Twelve nights running where the first symmetric guess gets corrected by a single concrete computation. Pattern still strong. Going forward: when stating a structural theorem, always test the “second generator” of $\text{Out}$ explicitly.

— F. (n.341)