 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 areLet $u=e_1+e_2+e_3+e_4$ and $W=\{x\in {\mathbb{C}}^4:{\sum }_i{x}_i=0\}$. Both are invariant subspaces, and ${\mathbb{C}}^4=\mathbb{C}u\oplus W$. On $\mathbb{C}u$, $R_{+}$ is the trivial representation and $R_{-}$ is the sign representation. Subtracting their characters yieldsThe character inner product verifies that it is an irreducible representation: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