Clifford correspondence for A3 in S3

An operator representing $r$ satisfies $r^3=1$ and is diagonalizable, since $x^3-1$ has three distinct roots over $\mathbb{C}$. In an irreducible $N$-module an eigenline is invariant under $r$ and hence under $N$, so the module is that line. This gives precisely the three displayed characters. The relation $sr{s}^{-1}=r^{-1}$ interchanges $\lambda$ and ${\lambda }^{-1}$, while $N$ fixes every type. Thus their inertia is $N$, while the trivial type has inertia $G$.

The trivial character extends trivially to $S_3$. The quotient $S_3\mathrm{/}{A}_3$ is cyclic of order two. An irreducible representation of this quotient is an eigenline for its generator, with eigenvalue $1$ or $-1$, so its two characters inflate to the trivial and sign characters. Gallagher gives exactly these characters above the trivial normal type.

The inertia criterion makes ${\mathrm{Ind}}_{A_3}^{S_3}\lambda$ irreducible. To identify it, let $P=\{(x_1,x_2,x_3)\in {\mathbb{C}}^3:x_1+x_2+x_3=0\}$ with the left permutation action. The vectors $v=(1,{\zeta }^2,\zeta )$ and $w=(1,\zeta ,{\zeta }^2)$ form a basis of $P$; $rv=\zeta v$, $rw={\zeta }^{-1}w$, and $sv=w$, $sw=v$. Any invariant line would be one of the distinct $r$-eigenlines, but $s$ swaps them, so $P$ is irreducible. Since it lies over $\lambda$ and inertia equals $N$, Clifford correspondence identifies it with the induced module.

The displayed eigenbasis gives $P{\mathrm{\mid }}_N\cong \lambda \oplus {\lambda }^{-1}$, each once, so its ramification is one and its normal restriction is reducible and not isotypical. The two normal-type orbits exhaust all types; the orbit partition therefore shows that the trivial, sign, and standard modules are the complete list of irreducible $S_3$-modules.