= Solution
In the order of <conjugacy classes> used in part (c), the elements of $H$ have the following types:
$$
\begin{array}{c|c}
\text{type in }S_4&\text{elements in }H\\\hline
1&1\\
(12)&rs=(13),\ r^3s=(24)\\
(12)(34)&r^2=(13)(24),\ s=(12)(34),\ r^2s=(14)(23)\\
(123)&\text{none}\\
(1234)&r,\ r^3
\end{array}.
$$
By part (d), only $1$ and $r^2$ have nonzero <character> values. Formula (b), or equivalently the element-sum version proved there, therefore gives
$$
\chi_V(1)=[G:H]\dim W=3\cdot2=6,
\qquad
\chi_V((12)(34))=\frac{|C_G((12)(34))|}{|H|}(-2)=\frac88(-2)=-2,
$$
with zero on the other three <conjugacy classes>. Here the <centralizer> order is $24/3=8$. Thus $\chi_V=(6,0,-2,0,0)$.
Let $\varepsilon$ denote the <sign representation> and take the explicit two-dimensional $G$-<module>
$$
U=\mathbf1\oplus\varepsilon.
$$
Its <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 $\operatorname{tr}(M\otimes N)=\operatorname{tr}(M)\operatorname{tr}(N)$. Consequently
$$
\chi_{C\otimes U}
=(3,-1,-1,0,1)(2,0,2,2,0)
=(6,0,-2,0,0)=\chi_V.
$$
By <Maschke's theorem> both <complex representations> are <semisimple representations>, and <characters distinguish finite-dimensional complex semisimple representations>. Therefore
$$
\boxed{V\cong C\otimes(\mathbf1\oplus\varepsilon)
\cong C\oplus(C\otimes\varepsilon).}
$$
This 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 is
$$
\dim V^G=\langle\chi_V,1\rangle_G
=\frac{6+3(-2)}{24}=0.
$$
Hence \b[$V^G=0$]. Equivalently, <Frobenius reciprocity> identifies this multiplicity with $\langle\chi_W,1_H\rangle_H=(2-2)/8=0$.
Back to article page
