Clifford Theory over Normal Subgroups — Examples

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.

For $(a,b)\in N\times Q$, conjugation sends $(n,1)$ to $(a^{-1}na,1)$ in the left-character convention. Characters are invariant under inner conjugation of $N$ by similarity of matrices, so every element fixes $\theta$ and inertia is $G$. The map $(n,q)\mapsto S(n)$ is multiplicative and restricts to the original action. The quotient map $(n,q)\mapsto q$ identifies $G\mathrm{/}N$ with $Q$.

Gallagher applies to this extension. Its tensor with an inflated quotient module has exactly the displayed action, hence is $S⊠U$. The bijection gives irreducibility, exhaustivity, and uniqueness of the parameter. Restricting to $N$ makes the second factor trivial and gives $\dim U$ copies of $S$, which is the ramification. When $Q=1$ this is just $S$; when $N=1$ it is just $U$.

For $N=1$, the trivial group has just its one-dimensional trivial irreducible: every subspace is invariant, so a nonzero irreducible space has dimension one. Every $G$-module restricts to its dimension many copies of this type. Conjugation fixes it, so $I=G$, its isotypical component is the whole module, and induction from $G$ to itself is identity. The ramification is therefore $\chi (1)$.

The trivial normal type extends to the trivial representation of $G$. Gallagher with $G\mathrm{/}1=G$ sends $\eta$ to $1\cdot \eta =\eta$, recovering all of $\mathrm{Irr}(G)$. A higher-degree irreducible thus has homogeneous but reducible restriction to $1$.

For $N=G$, characters are fixed by inner conjugation, so $I=G$. The restriction of an irreducible to the same group is itself, hence it contains $\theta$ exactly when it equals $\theta$, then with multiplicity one. Clifford correspondence is the singleton identity. The module affording $\theta$ extends to itself, and Gallagher has the trivial quotient $G\mathrm{/}G$, again giving only $\theta$. If $G=1$, these two endpoint descriptions agree and all degrees and indices are one.