= Solution
Use the composition convention $(\sigma\tau)(i)=\sigma(\tau(i))$. Then $R_+(\sigma)R_+(\tau)e_i=e_{\sigma(\tau(i))}=R_+(\sigma\tau)e_i$. Since the <sign of a permutation> is multiplicative, the same calculation gives $R_-(\sigma)R_-(\tau)=R_-(\sigma\tau)$. Both send the identity to $I$ and have inverses given by $\sigma^{-1}$, so both are <group representations>.
The <trace> of a <permutation matrix> counts fixed points, and multiplying that <matrix> by its <sign of a permutation> multiplies its <trace> by that sign. Thus the two <characters> are
$$
\chi_+=(4,2,0,1,0),\qquad
\chi_-=(4,-2,0,1,0).
$$
Let $u=e_1+e_2+e_3+e_4$ and $W=\{x\in\mathbb C^4:\sum_ix_i=0\}$. Both are <invariant subspaces>, and $\mathbb C^4=\mathbb Cu\oplus W$. On $\mathbb Cu$, $R_+$ is the <trivial representation> and $R_-$ is the sign <representation>. Subtracting their <characters> yields
$$
\boxed{\chi_3=\chi_+-\chi_0=(3,1,-1,0,-1),\qquad
\chi_4=\chi_--\chi_1=(3,-1,-1,0,1).}
$$
The <character inner product> verifies that it is an <irreducible representation>:
$$
\langle\chi_3,\chi_3\rangle
=\frac{9+6+3+0+6}{24}=1.
$$
The squares of the entries of $\chi_4$ are identical, so its <character inner product> with itself is also one. Their values at a transposition differ, so these <irreducible representations> are inequivalent. Therefore
$$
\boxed{R_+=R_0\oplus R_3,\qquad R_-=R_1\oplus R_4,\qquad R_4=R_1\otimes R_3.}
$$
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