The higher congruence characterization for finite simple groups of Lie type

title: "The Higher Congruence Characterization for Finite Simple Groups of Lie Type: A Uniform Counterexample in Type D" author: "Rejnaldo Narkaj" date: "30 August 2026" geometry: margin=1in fontsize: 11pt papersize: a4 header-includes:

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Abstract

Harman and Mundinger introduced, for a finite group $G$, a prime $p$, and an integer $m\geq 1$, an equivalence relation $\sim_m$ generated by ordinary conjugacy and prime-power moves

$$g\longmapsto g h^{p^{m-1}},\qquad [g,h]=1,\qquad |h|\text{ a power of }p.$$

They proved that $g\sim_m g'$ implies congruence modulo $p^m$ for every character unramified at $p$, and conjectured the converse for finite simple groups of Lie type in non-defining characteristic. The conjecture is indexed by MathDB under the title The higher congruence characterization for finite simple groups of Lie type.

We give a uniform counterexample family in type $D$. For every $M\geq 3$, set

$$G_M=P\Omega^+{4M}(3),\qquad p=2,\qquad m=2M.$$

Inside the monomial subgroup

$$K\cong 2^{4M-2}:A{4M}<G_M$$

we construct two explicit involutions $x_+$ and $x_-$. A complete little-group calculation proves the stronger congruence

$$\chi(x_+)-\chi(x_-)\in 2^{2M}\mathbb Z$$

for every ordinary complex character $\chi$ of $G_M$. On the monomial subgroup the modulus is sharp. Independently, an elementary semisimple-eigenvalue argument proves that every $2$-element of $G_M$ has order dividing $2^{2M-1}$; hence every nonconjugacy generator in $\sim_{2M}$ is trivial. Finally, explicit lifts of $x_+$ and $x_-$ to the split spin group have distinct central squares $\omega$ and $-\omega$, which proves that the two involutions are not conjugate. Thus $x_+\not\sim_{2M}x_-$ although all unramified character values are congruent modulo $2^{2M}$. This disproves the conjectural reverse implication for an infinite family of finite simple groups of Lie type.

1. Introduction

Let $G$ be a finite group, let $p$ be a prime dividing $|G|$, and let $m\geq 1$. Harman and Mundinger [1] study higher congruences between columns of the unramified character table. The equivalence relation relevant here is generated by conjugacy together with the moves

$$g\longmapsto g h^{p^{m-1}}, \qquad [g,h]=1, \qquad |h|\text{ a power of }p. \tag{1.1}$$

Their theorem gives the implication

$$g\sim_m g' \quad\Longrightarrow\quad \chi(g)\equiv\chi(g')\pmod{p^m} \tag{1.2}$$

for every character $\chi$ unramified at $p$. Their Conjecture 5.3 proposes the converse for finite simple groups of Lie type when $p$ is different from the defining characteristic. A public problem-index entry appears in MathDB [4] under the title used in the title of the present paper.

We prove that the converse fails uniformly. Put

$$r=2M,\qquad N=2r=4M, \tag{1.3}$$

and consider

$$G=G_M=P\Omega^+N(3)=P\Omega^+{4M}(3),\qquad M\geq 3. \tag{1.4}$$

The standard simplicity theorem for finite orthogonal groups implies that $G$ is a finite simple group of Lie type $D_r=D_{2M}$. Its defining characteristic is $3$, whereas throughout the proof

$$p=2, \qquad m=2M=r. \tag{1.5}$$

Thus the examples lie exactly in the non-defining-characteristic range of the conjecture.

The proof has four logically independent ingredients.

A monomial subgroup

$$K\cong 2^{N-2}$$

embeds in $G$ and contains two explicit projective involutions $x_+,x_-$.

Clifford theory for the elementary abelian normal subgroup of $K$ gives

$$\theta(x_+)-\theta(x_-)\in 2^r\mathbb Z \qquad(\theta\in R(K)).$$

Restriction transports this congruence to every ordinary character of $G$.

Every $2$-element of $G$ has order dividing $2^{r-1}=2^{2M-1}$. Hence all prime-power moves occurring in $\sim_r$ are identities.

Explicit spin lifts of $x_+$ and $x_-$ have squares $\omega$ and $-\omega$; since these central elements are distinct, the two projective involutions are not conjugate.

The proof is self-contained apart from standard structural facts about finite orthogonal and spin groups and the standard little-group form of Clifford theory; references are given in [2,3]. In particular, no maximal-subgroup classification, character table, table of marks, or computer algebra calculation is used.

2. The Harman-Mundinger relation

Definition 2.1. Fix a finite group $G$, a prime $p$, and an integer $m\geq 1$. The relation $\sim_m$ is the equivalence relation generated by ordinary conjugacy and by

$$g\longmapsto g h^{p^{m-1}}, \qquad [g,h]=1, \qquad |h|\text{ a power of }p. \tag{2.1}$$

Conjecture 2.2 (Harman-Mundinger [1], Conjecture 5.3: higher congruence characterization for finite simple groups of Lie type). Let $G$ be a finite simple group of Lie type, let $p$ be different from the defining characteristic, and let $m\geq1$. For $g,g'\in G$, the congruence

$$\chi(g)\equiv\chi(g')\pmod{p^m}$$

for every character $\chi$ unramified at $p$ should hold if and only if

$$g\sim_m g'.$$

The forward implication is a theorem of Harman and Mundinger [1]. The only statement challenged here is the conjectural reverse implication. Because our argument proves the congruence for every ordinary complex character, no further structural property of the unramified character ring is needed below.

For the elements used below there is no auxiliary ambiguity coming from odd power maps: $x_+$ and $x_-$ are involutions, and hence

$$x_\pm^s=x_\pm$$

for every odd integer $s$.

3. The split orthogonal model and its monomial subgroup

Fix $M\geq 3$ and retain the notation $r=2M$, $N=2r$. Let

$$V=\mathbb F_3^N$$

with quadratic form

$$Q(y_1,\ldots,y_N)=y_1^2+\cdots+y_N^2. \tag{3.1}$$

For a nondegenerate quadratic form in even dimension $2r$ over a finite field of odd order, the plus/minus type is determined by the square class of $(-1)^r\det Q$. Here $\det Q=1$ and $r$ is even, so $(-1)^r\det Q=1$ is a square. Thus (3.1) is of plus type; we work inside $\Omega_N^+(3)$.

Define

E=\left{v=(v_1,\ldots,v_N)\in\mathbb F_2^N:\sum_{i=1}^N v_i=0\right}, \qquad z=(1,\ldots,1). \tag{3.2}

Since $N$ is even, $z\in E$. For $v\in E$ set

$$d_v=\operatorname{diag}((-1)^{v_1},\ldots,(-1)^{v_N}). \tag{3.3}$$

The alternating group $A_N$ acts on $V$ by coordinate permutations.

Lemma 3.1. The signed permutation group $E\rtimes A_N$ is contained in $\Omega_N^+(3)$.

Proof. A sign change in a single coordinate is the orthogonal reflection in the corresponding norm-one coordinate vector. Hence $d_v$ with $v\in E$ is a product of an even number of reflections of spinor norm $1$. Therefore $d_v$ has determinant $1$ and trivial spinor norm, so $d_v\in\Omega_N^+(3)$.

A coordinate transposition $(ij)$ is the orthogonal reflection in a vector proportional to $e_i-e_j$, whose norm has square class $2$ in $\mathbb F_3^\times/\mathbb F_3^{\times2}$. Every element of $A_N$ is a product of an even number of coordinate transpositions. Hence its determinant is $1$ and its spinor norm is a square. Thus $A_N\leq\Omega_N^+(3)$. The two subgroups normalize one another in the evident way, proving the claim. $\square$

Proposition 3.2. Projectivization induces an embedding

$$K=(E/\langle z\rangle)\rtimes A_N \hookrightarrow P\Omega_N^+(3), \tag{3.4}$$

and

$$K\cong 2^{N-2}=2^{4M-2}:A_{4M}. \tag{3.5}$$

Proof. In the signed permutation group, a scalar matrix must have trivial permutation part: a nontrivial permutation matrix has an off-diagonal nonzero entry and cannot be scalar. Thus a scalar signed permutation is diagonal. The only scalar diagonal sign matrices are

$$d_0=I, \qquad d_z=-I.$$

Consequently the kernel introduced by projectivization is exactly $\langle z\rangle$ in the sign subgroup. Since $\dim_{\mathbb F_2}E=N-1$, the quotient $E/\langle z\rangle$ has dimension $N-2$, which gives (3.5). $\square$

4. Two projective involutions

Relabel the coordinate basis as

$$e_1,\ldots,e_r,f_1,\ldots,f_r.$$

Set

$$p=(e_1\ f_1)(e_2\ f_2)\cdots(e_r\ f_r). \tag{4.1}$$

Since $r=2M$ is even, $p\in A_N$.

Inside $\mathbb F_2^N$ define

$$a_+=e_1+\cdots+e_r, \qquad a_-=a_++e_1+f_1. \tag{4.2}$$

Both vectors have even weight. Moreover

$$p(a_+)=f_1+\cdots+f_r,$$

so

$$a_++p(a_+)=z. \tag{4.3}$$

Likewise

$$a_-=e_2+\cdots+e_r+f_1,$$

and hence

$$p(a_-)=f_2+\cdots+f_r+e_1,$$

which again gives

$$a_-+p(a_-)=z. \tag{4.4}$$

Let $\bar a_\pm$ denote the classes in

$$A=E/\langle z\rangle,$$

and define

$$x_+=(\bar a_+,p), \qquad x_-=(\bar a_-,p) \in K. \tag{4.5}$$

Lemma 4.1. The elements $x_+$ and $x_-$ are involutions.

Proof. In the semidirect product $A\rtimes A_N$,

$$(a,p)^2=(a+p(a),p^2).$$

Since $p^2=1$ and (4.3)-(4.4) give $a_\pm+p(a_\pm)=z$, whose class in $A$ is zero,

$$x_\pm^2=1.$$

Equivalently, the corresponding signed permutation matrices in $\Omega_N^+(3)$ square to $d_z=-I$, and therefore their projective images have order $2$. $\square$

5. Spin lifts and the nonconjugacy invariant

We now prove that the two involutions are not conjugate in $G$. This section spells out the Clifford-algebra calculation explicitly, including the identification of the constructed spin elements with the signed permutations in (4.5).

5.1. Clifford conventions

Work in the Clifford algebra $C(V,Q)$, with the chosen orthogonal unit basis. For each $i$ put

$$\varepsilon_i=e_if_i. \tag{5.1}$$

The Clifford relations give

$$e_i^2=f_i^2=1, \qquad e_if_i=-f_ie_i,$$

and therefore

$$\varepsilon_i^2=-1. \tag{5.2}$$

For $i\neq j$, moving the two factors of $\varepsilon_i$ past the two factors of $\varepsilon_j$ produces four sign changes, so

$$\varepsilon_i\varepsilon_j=\varepsilon_j\varepsilon_i. \tag{5.3}$$

Clifford reversion sends $\varepsilon_i$ to $-\varepsilon_i$.

Define

$$u_+=2^{-r/2}\prod_{i=1}^r(1+\varepsilon_i), \tag{5.4}$$ $$u_-=2^{-r/2}(1-\varepsilon_1)\prod_{i=2}^r(1+\varepsilon_i). \tag{5.5}$$

Here $r/2=M$ is an integer and $2\in\mathbb F_3^\times$, so the scalar $2^{-r/2}$ is unambiguous.

For each sign,

$$(1+\varepsilon_i)(1-\varepsilon_i)=1-\varepsilon_i^2=2. \tag{5.6}$$

Hence the Clifford norm of the unscaled product in (5.4) or (5.5) is $2^r$, while the scalar contributes $2^{-r}$. Thus

$$N_C(u_+)=N_C(u_-)=1. \tag{5.7}$$

The action calculation below shows directly that the factors normalize $V$, so $u_\pm\in\operatorname{Spin}_N^+(3)$.

5.2. The action on each coordinate plane

We use the standard spin action

$$\rho(u)(v)=u^{-1}vu. \tag{5.8}$$

Consider one orthogonal plane with unit basis $e,f$ and put $\varepsilon=ef$. Since

$$(1+\varepsilon)^{-1}=\frac{1-\varepsilon}{2},$$

a direct calculation gives

$$\frac{(1-\varepsilon)e(1+\varepsilon)}{2}=f, \qquad \frac{(1-\varepsilon)f(1+\varepsilon)}{2}=-e. \tag{5.9}$$

Thus $1+\varepsilon$ induces

$$e\longmapsto f, \qquad f\longmapsto -e. \tag{5.10}$$

Similarly

$$\frac{(1+\varepsilon)e(1-\varepsilon)}{2}=-f, \qquad \frac{(1+\varepsilon)f(1-\varepsilon)}{2}=e, \tag{5.11}$$

so $1-\varepsilon$ induces the reverse quarter-turn

$$e\longmapsto -f, \qquad f\longmapsto e. \tag{5.12}$$

For $u_+$, equations (5.10) hold on every plane $\langle e_i,f_i\rangle$. This is exactly the signed permutation $d_{a_+}p$: the permutation $p$ interchanges $e_i$ and $f_i$, and $d_{a_+}$ changes the sign of the $e_i$ coordinates only. For $u_-$, the first plane is governed by (5.12), while all remaining planes are governed by (5.10); this changes exactly the two signs indexed by $e_1$ and $f_1$, hence yields $d_{a_-}p$. Therefore

$$\rho(u_+)=d_{a_+}p, \qquad \rho(u_-)=d_{a_-}p, \tag{5.13}$$

and the images of $u_+$ and $u_-$ in $P\Omega_N^+(3)$ are precisely $x_+$ and $x_-$.

5.3. Central squares and the projective spin kernel

Put

$$\omega=\varepsilon_1\varepsilon_2\cdots\varepsilon_r. \tag{5.14}$$

Using (5.2)-(5.3),

$$(1+\varepsilon_i)^2=2\varepsilon_i, \qquad (1-\varepsilon_1)^2=-2\varepsilon_1. \tag{5.15}$$

Thus

$$u_+^2 =2^{-r}\prod_{i=1}^r 2\varepsilon_i =\omega, \tag{5.16}$$

whereas

$$u_-^2=-\omega. \tag{5.17}$$

Because $r$ is even,

$$\omega^2=\prod_{i=1}^r\varepsilon_i^2=(-1)^r=1. \tag{5.18}$$

The element $\omega$ is the volume element for the ordered basis $e_1,f_1,\ldots,e_r,f_r$. Since $N=2r$ is even, $\omega$ anticommutes with every vector of $V$. Hence it commutes with every even Clifford product and is central in $\operatorname{Spin}_N^+(3)$. Using (5.18), its action on $V$ is

$$\rho(\omega)=-I. \tag{5.19}$$

We use the standard exact sequence

$$1\longrightarrow{\pm1} \longrightarrow\operatorname{Spin}_N^+(3) \overset{\rho}{\longrightarrow}\Omega_N^+(3) \longrightarrow1. \tag{5.20}$$

In the present range the standard structure of the natural orthogonal group gives

$$Z(\Omega_N^+(3))={\pm I};$$

see, for example, Taylor [3]. Therefore the kernel of the composite projective spin map

$$\pi:\operatorname{Spin}_N^+(3)\longrightarrow P\Omega_N^+(3) \tag{5.21}$$

is exactly

$$\ker\pi={1,-1,\omega,-\omega}. \tag{5.22}$$

Every element in (5.22) has square $1$, and the four elements are distinct. In particular $\omega\neq-\omega$.

Proposition 5.1. The involutions $x_+$ and $x_-$ are not conjugate in $G=P\Omega_N^+(3)$.

Proof. Suppose that $x_-$ is conjugate to $x_+$ in $G$. Let $\bar g\in G$ satisfy

$$\bar g x_+\bar g^{-1}=x_-.$$

Choose a lift $\widetilde g\in\operatorname{Spin}N^+(3)$ of $\bar g$. Then

$$\widetilde g u+\widetilde g^{-1}$$

and $u_-$ have the same image under $\pi$, so by (5.22)

$$\widetilde g u_+\widetilde g^{-1}=c u_- \qquad(c\in\ker\pi). \tag{5.23}$$

Squaring and using $c^2=1$ gives

$$\widetilde g u_+^2\widetilde g^{-1}=u_-^2.$$

But $u_+^2=\omega$ is central and $u_-^2=-\omega$, so

$$\omega=-\omega,$$

contrary to (5.22). Hence $x_+$ and $x_-$ are not conjugate. $\square$

6. The character group of the normal elementary abelian subgroup

Recall

$$A=E/\langle z\rangle.$$

The standard dot product on $\mathbb F_2^N$ restricts to a bilinear form on $E$. Its radical is exactly $\langle z\rangle$.

Indeed, $z\cdot v=\sum_i v_i=0$ for every $v\in E$. Conversely, if $w\in E$ is orthogonal to all of $E$, then for every $i\neq j$ the vector $e_i+e_j$ belongs to $E$, so

$$0=w\cdot(e_i+e_j)=w_i+w_j.$$

Hence all coordinates of $w$ are equal, and therefore $w\in{0,z}$. Thus

$$\operatorname{rad}(E)=\langle z\rangle. \tag{6.1}$$

The pairing descends to a nondegenerate $A_N$-invariant pairing on $A$, so

$$A\cong A^\vee. \tag{6.2}$$

For an even subset $Y$ of the $N$ coordinate labels define

$$\lambda_Y(\bar v)=(-1)^{Y\cdot v}. \tag{6.3}$$

The condition that $Y$ have even size is exactly what makes the functional trivial on $z$. Moreover $Y$ and $Y^c$ determine the same character because their indicator vectors differ by $z$.

The action of $A_N$ on even subsets is transitive at each admissible cardinality. Indeed, $S_N$ is transitive on $k$-subsets, and if a chosen transporter is odd then, because $N\geq12$ and $0<k<N$ in the nontrivial cases, it may be composed with a transposition internal to the target subset or to its complement without changing the target subset. Thus the $A_N$-orbits on $A^\vee$ may be represented by even cardinalities

$$0\leq |Y|\leq r, \tag{6.4}$$

where the identification $Y\sim Y^c$ is used at the midpoint.

Let

$$B_Y=\operatorname{Stab}{A_N}(\lambda_Y). \tag{6.5}$$

Since $\lambda_Y$ is $B_Y$-invariant, it extends to $A\rtimes B_Y$ by

$$\widetilde\lambda_Y(a,b)=\lambda_Y(a). \tag{6.6}$$

By the standard little-group form of Clifford theory for a finite group with abelian normal subgroup [2], every irreducible character of

$$K=A\rtimes A_N$$

has the form

$$\Theta{Y,\tau} =\operatorname{Ind}_{A\rtimes B_Y}^{K} (\widetilde\lambda_Y\otimes\tau), \qquad \tau\in\operatorname{Irr}(B_Y). \tag{6.7}$$

\sum_{\substack{sB_Y\in A_N/B_Y\s^{-1}ps\in B_Y}} \lambda_{sY}(a),\tau(s^{-1}ps). \tag{6.8}

The condition $s^{-1}ps\in B_Y$ is equivalent to $p$ fixing the character $\lambda_{sY}$, which in turn means

p(sY)=sY \qquad\text{or}\qquad p(sY)=(sY)^c. \tag{6.9}

7. The monomial higher congruence Theorem 7.1 (Monomial congruence). For every virtual complex character $\theta\in R(K)$,

\theta(x_+)-\theta(x_-) \in 2^r\mathbb Z =2^{2M}\mathbb Z. \tag{7.1}

Proof. It suffices to treat the irreducibles (6.7). 7.1. Non-middle orbits Assume $|Y|<r$. The complement alternative in (6.9) is impossible, since it would imply

|Y|=|Y^c|=N/2=r.

Hence every term in (6.8) comes from a \subset $sY$ satisfying

p(sY)=sY.

Since $p$ is the product of the $r$ disjoint transpositions $(e_i\ f_i)$, a $p$-stable \subset is a union of entire pairs ${e_i,f_i}$. Now

a_-+a_+=e_1+f_1

in characteristic $2$. Therefore a $p$-stable \subset has even intersection with the support of $a_-+a_+$, and

\lambda_{sY}(a_+)=\lambda_{sY}(a_-). \tag{7.2}

Every summand in (6.8) is identical at $x_+$ and $x_-$. Thus

\Theta_{Y,\tau}(x_+)-\Theta_{Y,\tau}(x_-)=0 \qquad(|Y|<r). \tag{7.3}

$$7.2. The middle orbit Now let$$

Y_0={e_1,\ldots,e_r}. \tag{7.4}

The $p$-stable middle subsets again contribute equally at $x_+$ and $x_-$ and cancel from the difference. The remaining solutions of (6.9) satisfy

p(Z)=Z^c.

Such a \subset contains exactly one point from each pair ${e_i,f_i}$; these are precisely the transversals of the $r$ pairs. Encode a transversal by a \subset $J\subseteq{1,\ldots,r}$: choose $e_i$ if $i\in J$ and $f_i$ if $i\notin J$. There are $2^r$ transversals. Complementation sends $J$ to $J^c$ and does not change the corresponding character. Hence there are $2^{r-1}$ distinct transversal characters. Since $r$ is even,

|J^c|=r-|J|\equiv|J|\pmod2,

so complement preserves the parity of $J$. Consequently there are exactly

2^{r-2}

$$distinct transversal characters of even parity and the same number of odd parity. Set$$

h=(e_1\ e_2), \qquad q_J=\prod_{i\notin J}(e_i\ f_i). \tag{7.5}

Since $r$ is even,

\operatorname{sgn}(q_J)=(-1)^{r-|J|}=(-1)^{|J|}.

$$Define$$

s_J= \begin{cases} q_J,& |J|\text{ even},
q_Jh,& |J|\text{ odd}. \end{cases} \tag{7.6}

Then $s_J\in A_N$. The transposition $h$ stabilizes $Y_0$ setwise, so $s_JY_0=q_JY_0$ is the desired transversal. Moreover every factor of $q_J$ commutes with $p$, and therefore

s_J^{-1}ps_J= \begin{cases} p,& |J|\text{ even},
hph,& |J|\text{ odd}. \end{cases} \tag{7.7}

Both $p$ and $hph$ belong to $B_{Y_0}$ and are involutions. For the transversal character indexed by $J$,

\lambda_J(a_+)=(-1)^{|J|}, \tag{7.8}

because $a_+$ is supported on the $e_i$. Since a transversal contains exactly one of $e_1,f_1$,

\lambda_J(a_-) =-\lambda_J(a_+) =-(-1)^{|J|}. \tag{7.9}

Substituting (7.7)-(7.9) into (6.8), and using the exact count $2^{r-2}$ in each parity class, yields

\Theta_{Y_0,\tau}(x_+)-\Theta_{Y_0,\tau}(x_-) =2^{r-1}\bigl(\tau(p)-\tau(hph)\bigr). \tag{7.10}

7.3. The parity gain Let $d=\tau(1)$. If $t$ is an involution, then in a complex representation affording $\tau$, the eigenvalues of $t$ are $\pm1$. Hence

\tau(t)=d-2a

for some integer $a$, and therefore

\tau(t)\equiv d\pmod2.

Applying this to $p$ and $hph$ gives

\tau(p)-\tau(hph)\in2\mathbb Z. \tag{7.11}

Equation (7.10) is therefore divisible by $2^r$. Together with (7.3), this proves (7.1) for every irreducible character and hence for every virtual character. $\square$ 8. Sharpness on the monomial subgroup The modulus in Theorem 7.1 is exact for $K$. Let $B=B_{Y_0}$. It is the stabilizer in $A_{2r}$ of the unordered bipartition

{Y_0,Y_0^c}.

Every element of $B$ has a unique expression

(\alpha,\beta)p^\delta, \qquad \alpha,\beta\in S_r, \qquad \operatorname{sgn}(\alpha)=\operatorname{sgn}(\beta), \qquad \delta\in{0,1}, \tag{8.1}

where $p$ exchanges the two blocks. The equality of signs is precisely the condition that the resulting permutation of $2r$ points be even, since $\operatorname{sgn}(p)=(-1)^r=1$. Define

\epsilon((\alpha,\beta)p^\delta)=\operatorname{sgn}(\alpha). \tag{8.2}

This is a homomorphism: conjugation by $p$ interchanges $\alpha$ and $\beta$, which have the same sign. Thus $\epsilon$ is a linear character of $B$. We have

p=(1,1)p,

$$so$$

\epsilon(p)=1. \tag{8.3}

With $h=(e_1\ e_2)$ as above,

hph=((1\ 2),(1\ 2))p,

$$and hence$$

\epsilon(hph)=-1. \tag{8.4}

Taking $\tau=\epsilon$ in (7.10) gives the actual irreducible character

\Theta_{Y_0,\epsilon}

$$with$$

\Theta_{Y_0,\epsilon}(x_+)-\Theta_{Y_0,\epsilon}(x_-) =2^{r-1}(1-(-1)) =2^r. \tag{8.5}

$$Combining (8.5) with Theorem 7.1 proves the following. Theorem 8.1 (Sharpness on the monomial subgroup). Let$$

d_r= \gcd_{\theta\in\operatorname{Irr}(K)} \left|\theta(x_+)-\theta(x_-)\right|.

$$Then$$

\boxed{d_r=2^r=2^{2M}.} \tag{8.6}

In particular, the modulus $2^{2M}$ is sharp on the monomial subgroup $K$. Remark 8.2. The sharpness statement (8.6) is deliberately restricted to $K$. Restriction of characters from $G$ proves divisibility by $2^{2M}$ for all characters of $G$, but it does not by itself imply that the corresponding \gcd over $\operatorname{Irr}(G)$ is exactly $2^{2M}$. 9. Transport to the simple orthogonal group Corollary 9.1. For every ordinary complex character $\chi$ of $G=P\Omega_N^+(3)$,

\chi(x_+)\equiv\chi(x_-) \pmod{2^{2M}}. \tag{9.1}

$$Proof. The restriction$$

\operatorname{Res}^G_K\chi

is an integral character of $K$, hence an element of $R(K)$. Theorem 7.1 therefore gives

(\operatorname{Res}^G_K\chi)(x_+)-(\operatorname{Res}^G_K\chi)(x_-) \in2^{2M}\mathbb Z,

which is exactly (9.1). $\square$ Because $x_+$ and $x_-$ are involutions, every matrix representing either element over $\mathbb C$ has eigenvalues only $1$ and $-1$. Thus

\chi(x_+),\chi(x_-)\in\mathbb Z.

Hence (9.1) is an ordinary integer congruence, and in particular \implies the corresponding congruence for every character unramified at $2$ in the sense of Harman and Mundinger. 10. The 2-primary exponent in $GL_n(3)$ Write

\operatorname{ord}_{2^a}(3)

for the multiplicative order of $3$ modulo $2^a$. Lemma 10.1. For every integer $a\geq3$,

\operatorname{ord}_{2^a}(3)=2^{a-2}. \tag{10.1}

Proof. For $k\geq1$, the $2$-adic lifting-the-exponent formula gives

v_2(3^{2^k}-1) =v_2(3-1)+v_2(3+1)+v_2(2^k)-1 =1+2+k-1 =k+2. \tag{10.2}

Taking $k=a-2$ shows

3^{2^{a-2}}\equiv1\pmod{2^a}.

For $a\geq4$, taking $k=a-3$ gives valuation $a-1$, so

3^{2^{a-3}}\not\equiv1\pmod{2^a}.

The case $a=3$ is immediate. Since the order divides the power $2^{a-2}$ and does not divide its half, it is exactly $2^{a-2}$. $\square$ Proposition 10.2. If $y\in GL_n(3)$ has order $2^a$, then

a\leq2+\lfloor\log_2 n\rfloor. \tag{10.3}

$$Equivalently,$$

|y|\leq2^{2+\lfloor\log_2 n\rfloor}. \tag{10.4}

Proof. If $a\leq2$ there is nothing to prove. Assume $a\geq3$. Since $|y|$ is \prime to the characteristic, $X^{2^a}-1$ is separable over $\mathbb F_3$, so $y$ is semisimple over an algebraic closure. The order of a semisimple matrix is the least common multiple of the orders of its eigenvalues. Since all these orders are powers of $2$, at least one eigenvalue $\alpha$ has exact order $2^a$. Let $d$ be the degree of the minimal polynomial of $\alpha$ over $\mathbb F_3$. Then $d\leq n$ and $\alpha\in\mathbb F_{3^d}^\times$, so

2^a\mid3^d-1.

$$Therefore$$

\operatorname{ord}_{2^a}(3)\mid d.

$$By Lemma 10.1,$$

2^{a-2}\leq d\leq n,

which gives (10.3). $\square$ Lemma 10.3. For every $M\geq3$,

2+\lfloor\log_2(4M)\rfloor\leq2M-1. \tag{10.5}

Proof. At $M=3$ both sides equal $5$. For $M\geq4$,

\lfloor\log_2(4M)\rfloor =2+\lfloor\log_2M\rfloor \leq2+(M-2)=M,

because $\lfloor\log_2M\rfloor\leq M-2$ for $M\geq4$. Hence the \left-hand side of (10.5) is at most $M+2\leq2M-1$. $\square$ 11. The projective exponent bound The passage from a linear orthogonal group to its projective quotient is included explicitly because it is essential to the collapse of $\sim_{2M}$. Proposition 11.1 (Projective 2-exponent bound). Every $2$-element of

G=P\Omega^+_{4M}(3),\qquad M\geq3,

$$has order dividing$$

2^{2M-1}. \tag{11.1}

$$Proof. Let$$

\bar y\in P\Omega^+{4M}(3)

be a $2$-element of order $2^a$, and choose any lift

y\in\Omega^+{4M}(3).

$$The kernel of$$

\Omega^+{4M}(3)\longrightarrow P\Omega^+{4M}(3)

is central and of $2$-power order. Thus

y^{2^a}\in\ker\bigl(\Omega^+{4M}(3)\to P\Omega^+{4M}(3)\bigr).

If the kernel has exponent $2^b$, then

y^{2^{a+b}}=1.

Consequently $y$ itself has $2$-power order. Since

\Omega^+{4M}(3)\leq GL{4M}(3),

$$Proposition 10.2 and Lemma 10.3 give$$

|y|\mid2^{2M-1}.

The order of the image $\bar y$ divides the order of $y$, proving (11.1). $\square$ Remark 11.2. At the exact target order one can see the obstruction even more directly. An eigenvalue of order $2^{2M}$ over $\mathbb F_3$ has minimal polynomial degree

\operatorname{ord}{2^{2M}}(3)=2^{2M-2}.

For every $M\geq3$,

2^{2M-2}>4M,

so $GL{4M}(3)$ cannot contain an element of order $2^{2M}$. Proposition 11.1 records the stronger uniform bound on all $2$-element orders. 12. Collapse of $\sim_{2M}$ and the main theorem Specialize the Harman-Mundinger relation to

p=2, \qquad m=2M.

$$Every \prime-power move has the form$$

g\longmapsto g h^{2^{2M-1}}, \qquad [g,h]=1, \qquad |h|\text{ a power of }2. \tag{12.1}

$$By Proposition 11.1,$$

h^{2^{2M-1}}=1

for every $2$-element $h\in G$. Hence every move (12.1) is the identity. Proposition 12.1. On $G=P\Omega^+{4M}(3)$, $M\geq3$, the relation $\sim{2M}$ is exactly ordinary conjugacy. Proof. Every nonconjugacy generator in the definition of $\sim_{2M}$ is the identity by Proposition 11.1. $\square$ We can now combine the independent parts of the argument. Theorem 12.2 (Main theorem). For every integer $M\geq3$, let

G_M=P\Omega^+{4M}(3), \qquad p=2, \qquad m=2M.

There exist involutions $x+,x_-\in G_M$ such that

\chi(x_+)\equiv\chi(x_-) \pmod{2^{2M}} \tag{12.2}

for every ordinary complex character $\chi$ of $G_M$, and therefore for every character unramified at $2$, while

x_+\not\sim_{2M}x_-. \tag{12.3}

$$Consequently the reverse implication in Harman-Mundinger Conjecture 5.3 is false. The family$$

{P\Omega^+{4M}(3)\geq3}

provides infinitely many counterexamples of type $D{2M}$. Proof. Sections 3-4 construct the involutions. Proposition 5.1 proves that they are not conjugate. Corollary 9.1 proves (12.2) for every ordinary complex character. Proposition 12.1 identifies $\sim_{2M}$ with ordinary conjugacy, so Proposition 5.1 gives (12.3). This contradicts precisely the conjectural reverse implication. $\square$ 13. Boundary case and mechanism The smallest member occurs at

M=3, \qquad G=P\Omega^+{12}(3), \qquad m=6.

$$Here$$

2+\lfloor\log_2 12\rfloor=5=2M-1.

Thus every $2$-element of $G$ has order dividing

2^5=32,

and every $\sim_6$ power move contains the factor

h^{32}=1.

$$At the same time Corollary 9.1 yields$$

\chi(x+)\equiv\chi(x_-)\pmod{64}

for every ordinary complex character. The counterexample therefore results from two incompatible scales. The little-group calculation produces a congruence depth linear in $M$:

\chi(x_+)-\chi(x_-) \in2^{2M}\mathbb Z.

By contrast, the largest possible exponent $a$ for an element of order $2^a$ in $GL_{4M}(3)$ grows only logarithmically with $M$. At $m=2M$ the \prime-power moves defining $\sim_m$ have already vanished, while the character congruence remains. The spin-square invariant then separates the two remaining conjugacy classes. 14. Logical dependencies and scope For clarity, the proof depends on the following chain:

K=2^{4M-2}:A_{4M}<P\Omega^+{4M}(3)

$$$$

\Downarrow

$$$$

\text{little-group analysis of }A\triangleleft K

$$$$

\Downarrow

$$$$

\chi(x+)-\chi(x_-) \in2^{2M}\mathbb Z \quad(\chi\in R(G)),

$$while independently$$

\operatorname{ord}{2^a}(3)=2^{a-2}

$$$$

\Downarrow

$$$$

\exp_2(G)\mid2^{2M-1}

$$$$

\Downarrow

$$$$

\sim{2M}=\text{ordinary conjugacy},

$$and independently$$

u_+^2=\omega\neq-\omega=u_-^2

$$$$

\Downarrow

$$$$

x_+,x_-\text{ are not conjugate}.

$$Together these implications yield the counterexample. The argument does not challenge the forward theorem of Harman and Mundinger, namely$$

g\sim_m g' \Longrightarrow \chi(g)\equiv\chi(g')\pmod{p^m}

for all unramified characters. Only the conjectured converse is disproved. The only standard external inputs used in the proof are: the plus/minus classification and simplicity facts for finite orthogonal groups; the standard exact sequence $\operatorname{Spin}_N^+(3)\to\Omega_N^+(3)$; the little-group form of Clifford theory for an abelian normal subgroup; the elementary lifting-the-exponent identity used in Lemma 10.1. All group-specific calculations needed for the counterexample are given explicitly above. References