= Solution
The two <one-dimensional representations> are the <trivial representation>, $\rho_+(g)=1$, and the <sign representation>, $\rho_-(g)=\operatorname{sgn}(g)$. The first respects multiplication immediately. For the second, the <sign of a permutation> is the <determinant> of its <permutation matrix>, and multiplicativity of the <determinant> proves $\rho_-(gh)=\rho_-(g)\rho_-(h)$. They are inequivalent since a <transposition> has images $1$ and $-1$, and conjugation by a nonzero scalar cannot change a one-dimensional matrix.
Their <group homomorphism kernels> are
$$
\boxed{\ker\rho_+=S_4,\qquad \ker\rho_-=A_4}.
$$
Both are <subgroups>, indeed <normal subgroups>, because they are kernels. In the second case $A_4$ is the <alternating group> of even <permutations>, with twelve elements.
The <characters> are their scalar values, since the <trace> of a one-dimensional matrix is its only entry. In the class order from part (b),
$$
\chi_+=(1,1,1,1,1),\qquad \chi_-=(1,-1,1,1,-1).
$$
The <character inner product> is the group average of $\chi_+\overline{\chi_-}$, so
$$
\boxed{\langle\chi_+,\chi_-\rangle
=\frac1{24}(1-6+3+8-6)=0}.
$$
Each has squared norm $(1+6+3+8+6)/24=1$, in agreement with <character orthogonality> for distinct <irreducible representations>.
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