= Solution
Let $V$ be an irreducible $G$-module with character $\chi\in\operatorname{Irr}(G\mid\theta)$, and let $W=V_\theta\ne0$ be its $\theta$-<isotypic component>. It is invariant under the <inertia group of a character> $T$. The sum of its $G$-translates is a nonzero $G$-submodule, so irreducibility makes it all of $V$. Translates associated with different cosets $gT$ have distinct $N$-types and hence form a <direct sum>:
$$
V=\bigoplus_{g\in\mathcal R}gW.
$$
To see that $W$ is irreducible over $T$, let $0\ne W_0\subseteq W$ be a $T$-submodule. Then $\sum_{g\in\mathcal R}gW_0$ is a nonzero $G$-submodule of $V$, hence equals $V$. Its $\theta$-isotypic part is exactly $W_0$, whereas the corresponding part of $V$ is $W$. Thus $W_0=W$.
Let $\xi$ be the <irreducible character> of $W$ as a $T$-module. It lies over $\theta$, and the natural map of <induced representations>
$$
\mathbb C[G]\otimes_{\mathbb C[T]}W\longrightarrow V,\qquad g\otimes w\longmapsto gw
$$
is an isomorphism: it identifies each summand with the corresponding $gW$, and these summands are direct and exhaust $V$. Consequently $\chi=\xi^G$, proving surjectivity of induction on the stated sets.
For injectivity, the construction in (a) shows that the $\theta$-isotypic component of $\operatorname{Ind}_T^G W$ is exactly $1\otimes W$, with its original $T$-action. Isomorphic induced $G$-modules therefore have isomorphic intrinsic $\theta$-components as $T$-modules. Since complex representations with the same character are isomorphic, $\xi_1^G=\xi_2^G$ forces $\xi_1=\xi_2$. Hence the <Clifford correspondence> is the bijection
$$
\boxed{\operatorname{Irr}(T\mid\theta)\xrightarrow{\ \xi\mapsto\xi^G\ }\operatorname{Irr}(G\mid\theta),}
$$
whose inverse takes the character of the $\theta$-isotypic component.
Back to article page
