= Solution
The printed wording calls $\Sigma_3$ cyclic, but its meaning in question 9 is the <symmetric group> $S_3$. Under that interpretation, take its planar <dihedral group> representation, with
$$
D(r)=\begin{pmatrix}-1/2&-\sqrt3/2\\\sqrt3/2&-1/2\end{pmatrix},\qquad
D(s)=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.
$$
Here $r^3=s^2=1$ and $srs=r^{-1}$. This <representation> is <irreducible> over the complex numbers: the two distinct rotation <eigenspaces> are exchanged by the reflection, so neither is invariant under the whole group. Its <character> is $2$ on the identity, $-1$ on the two 3-cycles and $0$ on the three transpositions. Squaring maps each transposition to the identity and permutes the 3-cycles. Its <Frobenius-Schur indicator> is therefore
$$
\boxed{\nu=\frac{2+3(2)+2(-1)}6=1.}
$$
If “cyclic” were intended literally, no two-dimensional complex <irreducible representation> would exist, since an <irreducible representation of a finite abelian group> is one-dimensional.
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