TITLE: Can groups of twice-odd order have quaternionic representations?
QUESTION [5 upvotes]: 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.

REPLY [9 votes]: 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.