An irreducible nonmonomial representation of degree eight in a semiabelian group of order 2592

Problem

Is every finite semiabelian group monomial?

G\text{ semiabelian},\quad\chi\in\operatorname{Irr}(G) \quad\stackrel{?}{\Longrightarrow}\quad \chi=\operatorname{Ind}_H^G\lambda

for some $H\le G$ and linear character $\lambda$ ? Semiabelian groups are generated from the trivial group by quotients and semidirect products with finite abelian kernels.

Statement

The class of finite semiabelian groups is the smallest class containing the trivial group and closed under quotients and semidirect products $A\rtimes H$ with $A$ finite abelian. A finite group is monomial if every irreducible complex representation is induced from a one-dimensional representation of a subgroup. Kida’s Conjecture 1.3 (2), also recorded as Kourovka Problem 21.68 (1), asserts that semiabelian groups are monomial.

Theorem 1. There is a semiabelian group of order $2592$ having an irreducible complex representation of degree eight which is not induced from any one-dimensional representation of a subgroup.

A signed permutation group and a quaternionic subgroup

Let $M$ be the group of signed permutation matrices on an ordered basis $(1,i,j,k)$ such that the underlying permutation is even and the product of the four signs is $+1$ . The diagonal subgroup has order eight and the even permutations form a complement. Hence

\tag{1} M\cong C_2^3\rtimes A_4,\qquad |M|=96.

The symbols $1,i,j,k$ initially denote basis vectors. Identify them also with the standard real quaternion basis. Encode a signed permutation by its signed images, and put

I=(2,-1,4,-3),\qquad J=(3,-4,-1,2),\qquad T=(1,3,4,2).

Each belongs to $M$ . The maps $I,J$ are left multiplication by $i,j$ , while $T$ fixes $1$ and cycles $i,j,k$ . Quaternion multiplication gives

\tag{3} \begin{gathered} I^2=J^2=-1,\qquad JI=-IJ,\qquad T^3=1,\\ TIT^{-1}=J,\qquad TJT^{-1}=IJ. \end{gathered}

Thus $K=\langle I,J,T\rangle$ satisfies

\tag{4} K=\langle I,J\rangle\rtimes\langle T\rangle\cong Q_8\rtimes C_3, \qquad |K|=24,\qquad [M:K]=4.

Indeed, the eight left multiplications by $\{\pm1,\pm i,\pm j,\pm k\}$ are distinct, $T$ normalizes them, and their intersection with $\langle T\rangle$ is trivial.

Lemma 2. The group $K$ has no subgroup of index two.

Proof. An index-two subgroup would give a nontrivial homomorphism $\phi:K\to C_2$ . Since $T^3=1$ , one has $\phi(T)=1$ . The conjugation relations give

\phi(I)=\phi(J)=\phi(IJ)=\phi(I)\phi(J).

It follows that $\phi(I)=\phi(J)=1$ . The three generators therefore have trivial image, a contradiction. ◻

Lemma 3. The group $K$ has an irreducible complex representation $\rho$ of degree two.

Proof. Embed the real quaternion algebra into $M_2(\mathbb C)$ by

i\longmapsto\begin{pmatrix}\mathrm i&0\\0&-\mathrm i\end{pmatrix},\qquad j\longmapsto\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad k\longmapsto ij.

Set $q=(-1-i-j-k)/2$ . Writing $v=i+j+k$ , one has $v^2=-3$ and hence $q^2=(-1+v)/2$ and $q^3=1$ . Multiplication also gives

qiq^{-1}=j,\qquad qjq^{-1}=ij.

Consequently $I\mapsto i$ , $J\mapsto j$ , $T\mapsto q$ defines a representation of the semidirect product (4). The two eigenlines of $\rho(I)$ are interchanged by $\rho(J)$ , so there is no common invariant line. Thus even the restriction to $Q_8$ is irreducible. ◻

The semiabelian extension

Let $\Omega=M/K$ , with distinguished point $\omega_0=K$ , and let $M$ act on $\mathbb F_3^{\Omega}$ by permutation of coordinates. Put

\tag{8} A=\left\{a\in\mathbb F_3^{\Omega}:\sum_{\omega\in\Omega}a_\omega=0\right\}, \qquad G=A\rtimes M.

Then $A$ is an $M$ -invariant elementary abelian group of order $3^3=27$ . The successive split extensions

C_3,\qquad C_2^2\rtimes C_3=A_4,\qquad C_2^3\rtimes A_4=M,\qquad A\rtimes M=G

show directly that $G$ is semiabelian. Its order is $27\cdot96=2592$ .

Let $\zeta=\exp(2\pi\mathrm i/3)$ and define $\lambda_\omega(a)=\zeta^{a_\omega}$ , with exponents interpreted modulo three. Write $\lambda=\lambda_{\omega_0}$ .

Lemma 4. The four characters $\lambda_\omega$ on $A$ are distinct. The stabilizer of $\lambda$ in $G$ is $B=A\rtimes K$ .

Proof. For distinct $\omega,\nu$ , choose a third point $\xi$ and put $a_\omega=1$ , $a_\xi=-1$ , and all other coordinates zero. Then $a\in A$ and $\lambda_\omega(a)=\zeta\neq1=\lambda_\nu(a)$ . Since $M$ permutes the coordinate characters just as it permutes $\Omega$ , the stabilizer of $\lambda$ in $M$ is $K$ . Conjugation by $A$ acts trivially on $A$ , giving the asserted stabilizer in $G$ . ◻

Because $K$ fixes $\lambda$ , the formula

\tau(a,k)=\lambda(a)\rho(k)\qquad(a\in A,\ k\in K)

defines an irreducible representation of $B$ . Define

\tag{11} \sigma=\mathop{\mathrm{Ind}}_B^G\tau.

Its degree is $[G:B]\dim\tau=4\cdot2=8$ .

Proposition 5. The representation $\sigma$ is irreducible.

Proof. The four cosets of $B$ in $G$ give the decomposition

\sigma|_A=\bigoplus_{\omega\in\Omega}\lambda_\omega^{\oplus2}.

Write $V_\omega$ for the corresponding two-dimensional weight spaces. For any $A$ -invariant subspace $W$ , the projections

e_\omega=\frac1{|A|}\sum_{a\in A}\lambda_\omega(a)^{-1}\sigma(a)

preserve $W$ . Orthogonality of the distinct characters gives $W=\bigoplus_\omega(W\cap V_\omega)$ .

Now assume $W$ is $G$ -invariant and nonzero. Some intersection is nonzero. Transitivity of $G$ on the four weight spaces gives $W\cap V_{\omega_0}\neq0$ . The stabilizer $B$ acts on $V_{\omega_0}$ as $\tau$ , which is irreducible by Lemma 3. Hence $V_{\omega_0}\subseteq W$ . Translating by $G$ gives every $V_\omega\subseteq W$ , so $W$ is the entire space. ◻

Excluding induction from linear characters

Proof of Theorem 1. It remains to prove that $\sigma$ is not monomial. Suppose

\sigma\cong\mathop{\mathrm{Ind}}_L^G\eta

for $L\leq G$ and a one-dimensional representation $\eta$ of $L$ . Equality of dimensions gives $[G:L]=8$ . Since $A\lhd G$ ,

[A:A\cap L]=[AL:L]\mid[G:L]=8.

This index also divides $|A|=27$ . It is therefore one, and $A\leq L$ .

In the induced representation, the identity-coset line is $A$ -invariant and affords $\eta|_A$ . Thus $\eta|_A$ is one of the four characters in $\sigma|_A$ . Conjugating $L$ and $\eta$ inside $G$ , we may assume $\eta|_A=\lambda$ . For $l\in L$ and $a\in A$ , one-dimensionality gives

\lambda(lal^{-1})=\eta(l)\eta(a)\eta(l)^{-1}=\lambda(a).

Hence $L\leq B$ by Lemma 4, and

[B:L]=\frac{[G:L]}{[G:B]}=2.

Since $A\leq L$ , the subgroup $L/A$ has index two in $B/A\cong K$ . This contradicts Lemma 2. Together with Proposition 5 and (8), this proves the theorem. ◻

References

Preprint · Lean (GitHub)

E. I. Khukhro and V. D. Mazurov (eds.), Unsolved Problems in Group Theory: The Kourovka Notebook, Problem 21.68. arXiv:1401.0300.

M. Kida, On semiabelian groups, J. Group Theory 28 (2025), 697–712. doi:10.1515/jgth-2024-0010.