A semiabelian group that is not monomial

Nilradical

Kida conjectured that every finite semiabelian group is monomial. A group is semiabelian if it has a subgroup chain from the identity to the whole group, each successive group being a quotient of an abelian semidirect product with its predecessor. It is monomial if every irreducible complex representation is induced from a linear character of a subgroup. Kida, On semiabelian groups, Conjecture 1.3 and Definition 2.1.

Theorem. There is a semiabelian group of order $2592$ with an irreducible nonmonomial complex representation of degree eight.

All abelian factors below are finite, so the construction meets Kida's finite-factor definition as well as the Notebook formulation. No minimal-order assertion is made.

The group

On the quaternion basis $(1,i,j,k)$ , let $E\cong C_2^3$ consist of diagonal sign changes with an even number of minus signs, and let $T\cong A_4\cong C_2^2\rtimes C_3$ permute the four basis elements evenly. Put

W=E\rtimes T,\qquad |W|=96.

Left multiplication by $i$ and $j$ gives the signed permutations

\begin{aligned} L_i:(1,i,j,k)&\mapsto(i,-1,k,-j),\\ L_j:(1,i,j,k)&\mapsto(j,-k,-1,i). \end{aligned}

They lie in $W$ and generate $Q_8$ . The unsigned permutation $s$ fixing $1$ and cycling $i,j,k$ normalizes this quaternion subgroup. Consequently

H=Q_8\rtimes\langle s\rangle\le W, \qquad |H|=24,\qquad [W:H]=4.

There is no subgroup of index two in $H$ . Any homomorphism $H\to C_2$ kills $s$ , so the conjugation cycle makes the images of $i,j,k$ equal. The relation $ij=k$ forces that common image to be trivial. These elements generate $H$ .

Let $X=W/H$ be the four left cosets and let $W$ permute coordinates of

A=\left\{a\in\mathbb F_3^X:\sum_{x\in X}a_x=0\right\}.

Then $|A|=27$ . Our group is $G=A\rtimes W$ , of order $27\cdot96=2592$ . The embedded chain

1<C_3<T<W<G

proves semiabelianity: its successive abelian split-extension kernels are $C_3,C_2^2,E,A$ .

An irreducible representation of degree eight

First give $H$ a two-dimensional representation $\psi$ by sending $i,j,s$ respectively to

P=\begin{pmatrix}\mathrm i&0\\0&-\mathrm i\end{pmatrix}, \qquad Q=\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad U=-\tfrac12(I_2+P+Q+PQ).

The quaternion relations hold, as do $U^3=I_2$ , $UP=QU$ and $UQ=(PQ)U$ . These identities define the representation. It is irreducible: the only $P$ -invariant lines are the coordinate axes, and $Q$ exchanges them.

Choose a primitive cube root of unity $\omega$ . The coordinate characters

\lambda_x(a)=\omega^{a_x}\qquad(x\in X)

are distinct on $A$ : the vector $e_x-e_y$ separates $\lambda_x$ and $\lambda_y$ . Their permutation action is the coset action on $X$ . Therefore the character $\lambda=\lambda_H$ has inertia subgroup

I=I_G(\lambda)=A\rtimes H.

It extends to $I$ by $\widetilde\lambda(a,h)=\lambda(a)$ . Multiply this character by the inflation of $\psi$ and induce:

\rho=\widetilde\lambda\otimes\operatorname{Inf}_H^I\psi, \qquad V=\operatorname{Ind}_I^G\rho.

The representation $\rho$ is irreducible of degree two. For $g\notin I$ , its conjugate has a different scalar character on $A$ , which lies in $I\cap gIg^{-1}$ . Hence any intertwiner in that nonidentity Mackey term is zero. Mackey's irreducibility criterion proves that $V$ is irreducible, and its degree is $[G:I]\cdot2=8$ .

Why it cannot be induced from a linear character

Suppose $V\cong\operatorname{Ind}_L^G\theta$ with $\theta$ linear. Dimensions give $[G:L]=8$ . Since $A\lhd G$ ,

[A:A\cap L]=[AL:L]\mid8.

This index also divides $|A|=27$ , so $A\le L$ .

Frobenius reciprocity supplies a nonzero $\theta$ -eigenvector in $V$ restricted to $L$ . The restriction of $V$ to $A$ has exactly the four distinct weights $\lambda_x$ , each with multiplicity two. Thus $\theta|_A=\lambda_x$ for some $x$ . A linear character is invariant under conjugation by its own group, so $L\le I_G(\lambda_x)$ , a conjugate of $I$ .

This inertia subgroup has index four in $G$ , hence $L$ has index two in it. Quotienting by $A\le L$ produces an index-two subgroup of $H$ , a contradiction. This proves the theorem.

Credits and resources

Kida's paper already records the order-96 semiabelian group containing $\operatorname{SL}_2(3)$ , on p. 710. That complement/subgroup pair is prior input. Nilradical's contribution here is the explicit abelian extension, the degree-eight witness and the complete counterexample proof and formalization. The argument uses standard Frobenius reciprocity and Mackey theory; see J.-P. Serre, Linear Representations of Finite Groups, §7.3. It makes no publication-priority claim for the extension method.

The complete formalization uses Lean and mathlib, with induction infrastructure from the TauCeti contributors, reused under Apache-2.0. The obstruction suggests studying other inertia quotients with an irreducible representation whose degree cannot occur as a subgroup index.