Let $G$ be a finite group and $\phi\colon G\to \mathrm{GL}_d(\mathbb C)$ be an irreducible representation, with character $\chi$. Recall that

  • $\phi$ is complex type if $\chi$ is not real-valued,
  • $\phi$ is real type if $\phi$ is the complexification of a representation $G\to\mathrm{GL}_d(\mathbb R)$,
  • $\phi$ is quaternionic type otherwise, i.e. $\chi$ is real-valued but $\phi$ is not real.

(Equivalently, $\phi$ is quaternionic if $\mathbb{C}^d$ has a $G$-invariant symplectic form, or if there is a $G$-equivariant, conjugate-linear operator on $\mathbb{C}^d$ whose square is $-\mathrm{id}$.)

Question: Is there an example of a finite group $G$ of order $4k+2$ for some $k$, such that $G$ has a quaternionic irreducible representation?

Some thoughts that may be relevant:

  • The Frobenius--Schur indicator $\frac{1}{|G|}\sum_{g\in G}\chi(g^2)$ is $-1$ iff $\phi$ is quaternionic, $+1$ iff $\phi$ is real, and $0$ iff $\phi$ is of complex type.
  • A group of order $4k+2$ must decompose as the semidirect product of a cyclic group of order two acting on a group of order $2k+1$.
  • Every nontrivial irreducible representation of a group of odd order is automatically of complex type (maybe this is a theorem of Burnside).
  • The degree $d$ has to be even, since $\mathbb{C}^d$ must have a symplectic form.
  • A counting argument using the Frobenius--Schur indicator, and a count of the the number of elements of order 2, shows that the degree $d$ must be strictly greater than $2$. So we need an irreducible representation of degree at least $6$.
  • Searching on GAP didn't give any examples for $|G|\leq 150$.

This isn't really a research question, it's just something curious. I'm asking for a friend.


3 Answers 3


I'm not sure how acceptable it is to give answers referring to theorems in textbooks. If you'd like me to reproduce the relevant argument, then please let me know.

Chapter III.11 of Simon, Representations of finite and compact groups, shows the following (in the notation there): if $H \le G$ with $H$ finite and $[G : H] = 2$, then (Theorem 1) irreducible representations of $G$ are in bijection with orbits of irreducible representations of $H$ under the action of a conjugator $y \in G \setminus H$; and (Theorems 2 and 3), if (either) representation $V$ of $H$ in an orbit is complex, then the corresponding representation of $G$ is quaternionic only if $V(y^2) = -1$. However, this is impossible if $H$ has odd order.


A more general result is true: If a Sylow $2$-subgroup of $G$ is elementary abelian, then $G$ has no irreducible character of quaternionic type. Recall that $\chi$ is quaternionic iff the Schur index $m_R(\chi) = 2$, where $R$ is the real number field. Theorem 10.9 of my character theory book asserts that for any prime $p$ and any subfield of F of the complex numbers, $p$ cannot divide $m_F(\chi)$ if the Sylow $p$-subgroups of $G$ are elementary abelian.

This does $not$ generalize further to groups with abelian Sylow $2$-subgroups. There is a group of order $12$ with a cyclic Sylow $2$-subgroup that has a quaternionic irreducible character.

  • 1
    $\begingroup$ Dear Marty, to place your example into a larger context, Rod Gow's 1976 paper `Real-valued characters and the Schur index' includes the result that if T is a 2-group which is not elementary abelian, then there is finite group G with Sylow 2-subgroup isomorphic to T which possesses a quaternionic irreducible character. $\endgroup$ Nov 25, 2015 at 21:10

Let $G$ be a finite group of twice odd order, and let $N$ be its normal $2$-complement. Since only the trivial irreducible character of $N$ is real-valued, it follows from Clifford's theorem that the only real-valued irreducible characters of $G$ are the two linear representations of $G$ with $N$ in their kernel, and the irreducible characters of $G$ which are induced from irreducible characters of $N$ which are in the same orbit as their complex conjugate under the natural action of $G$ (on irreducible characters of $N).$ For a real-valued irreducible character $\theta$ of the second type (ie not with $N$ in its kernel), a Lemma in a paper of mine on the Frobenius Schur indicator shows that the Frobenius -Schur indicator $\nu(\theta)$ is equal to $\langle {\rm Res}^{G}_{C_{G}(t)}(\theta), 1 \rangle$ where $t$ is an involution of $G$. This is non-negative, while on the other hand, since $\theta$ is real-valued, we know that $\nu(\theta) = \pm 1.$ Hence we must have $\nu(\theta) = 1$ and $\theta$ is realizable over $\mathbb{R}.$ Hence the associated representation is not of quaternionic type. In fact, I think this can be checked directly by considering the matrices of the representation induced from $N.$

There are rather more general results of this nature in my paper, and a substantial improvement in a later paper of John Murray. Murray shows that if we consider the characteristic $2$- permutation module ( for general finite group $G$ of even order) obtained from the conjugation action of $G$ on its involutions, then the projective indecomposable summands (over an algebraically closed field) are precisely the (reductions (mod $2$) of) the real-valued $2$-blocks of defect zero of $G$, and furthermore, these all occur with multiplicity $1$. Standard Schur index theory then implies, in particular, that a real-valued $2$-block of defect zero of any finite group $G$ has Schur index $1$, a particular case which I believe was already known to P. Fong some time before. A real-valued irreducible character of Schur index $1$ is realizable over $\mathbb{R},$ and is not quaternionic.


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