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Embed the order-eight dihedral group as $H=⟨(1234),(12)(34)⟩$ in the symmetric group $S_4$, and let $W$ be its two-dimensional complex representation obtained from its square action. Let $C$ be the three-dimensional complex representation from the rotational symmetry group of a cube. In the class order $1,(12),(12)(34),(123),(1234)$, the induced character is $(6,0,-2,0,0)$: only the identity and the square's half-turn have nonzero character values. This equals the character of $C\otimes (1\oplus \mathrm{sgn})$, since ${\chi }_C=(3,-1,-1,0,1)$. Characters distinguish finite-dimensional complex semisimple representations, giving the displayed isomorphism. Its trivial character multiplicity is $(6-6)/24=0$.

In the order of conjugacy classes used in part (c), the elements of $H$ have the following types:By part (d), only $1$ and $r^2$ have nonzero character values. Formula (b), or equivalently the element-sum version proved there, therefore giveswith zero on the other three conjugacy classes. Here the centralizer order is $24/3=8$. Thus ${\chi }_V=(6,0,-2,0,0)$.

Let $\epsilon$ denote the sign representation and take the explicit two-dimensional $G$-moduleIts character is ${\chi }_U=(2,0,2,2,0)$. The character of a tensor product of group representations is the pointwise product of their characters, because $\mathrm{tr}(M\otimes N)=\mathrm{tr}(M)\mathrm{tr}(N)$. ConsequentlyBy Maschke's theorem both complex representations are semisimple representations, and characters distinguish finite-dimensional complex semisimple representations. ThereforeThis is the induction of the planar square representation to the symmetric group on four letters.

The dimension of the fixed-point subspace of a group action is the multiplicity of the trivial representation. Its character inner product here isHence $V^G=0$. Equivalently, Frobenius reciprocity identifies this multiplicity with $⟨{\chi }_W,1_H{⟩}_H=(2-2)/8=0$.