Induced Brauer character

Question

Let H is a subgroup of G. How we can show that $(1_{H^0})^G$ is reducible Brauer character of G. Where $1_{H^0}$ is principal Brauer character of H.

Answer 1

Your question should follow from the following standard result from Mackey theory:

Let $H$ be a finite index subgroup of a finite group $G$, and let $(\sigma, W)$ be a (finite dimensional, complex) representation of $H$, and consider the induced representation $\operatorname{Ind}_H^G \sigma$ of $G$. This is the vector space of all functions $\varphi: G \rightarrow W$ such that $\varphi(hg) = \sigma(w) \varphi(g)$ for all $h \in H$ and $g \in G$, on which $G$ acts by right translation: $g \cdot \varphi(x) = \varphi(xg)$.

Let $\operatorname{Hom}_{\mathbb C}(W,W)$ be the set of linear transformations from $W$ to itself, and consider the set $\mathcal I$ of functions $$A: G \rightarrow \operatorname{Hom}_{\mathbb C}(W,W)$$

satisfying the condition

$$ A(h_1gh_2) = \sigma(h_1) \circ A(g) \circ \sigma(h_2)$$

for all $h_1, h_2 \in H, g \in G$. For each $A \in \mathcal I$, define $\Phi \in \operatorname{Hom}_G(\operatorname{Ind}_H^G \sigma, \operatorname{Ind}_H^G \sigma)$ by

$$\Phi(\varphi)(g) = \sum\limits_{x \in H \backslash G} A(gx^{-1}) \varphi(x).$$

Then $A \mapsto \Phi$ defines a bijection from $\mathcal I$ onto the set $\operatorname{Hom}_G(\operatorname{Ind}_H^G \sigma, \operatorname{Ind}_H^G \sigma)$ of intertwining operators from $\operatorname{Ind}_H^G \sigma$ to itself.

Let $V = \operatorname{Ind}_H^G 1_H$, where $1_H$ is the trivial representation of $H$ over the complex numbers. I assume by your question, you're asking equivalently why $V$ is never an irreducible representation of $G$.

To say that $V$ is irreducible is the same as saying that the only intertwining operators from $V$ to itself are scalar multiplication maps. Thus, you just need to show that there are nontrivial intertwining operators from $V$ to itself.

Answer 2

As Nate mentioned, this can be seen from Frobenius reciprocity since this module can be lifted to the integers. However, we can also directly find a nonzero, proper submodule.

Suppose $G$ is a finite group, $H$ is a subgroup of $G$ with index $n$, and $p$ is a prime. Let $\rho:G \to \operatorname{GL}_n(\mathbb{F}_p)$ be given by sending a group element $g$ to the permutation matrix corresponding to the permutation induced by $G$ on the cosets of $H$. We can choose a basis of $V=(\mathbb{F}_p)^n$ starting with an arbitrary $\vec{e}_H$ and defining $\vec{e}_{Hg} = \vec{e}_{H} \cdot g$. We have $\vec{e}_{H} \cdot h = \vec{e}_H$ for all $h \in H$ and $\vec{e}_{Hg} \cdot k = \vec{e}_{H} \cdot g \cdot k = \vec{e}_{Hgk}$.

This representation has character $(1_H)^G$.

This representation is reducible if $H$ is a proper subgroup of $G$ (otherwise $n=1$ and we get the irreducible trivial representation). We show this by giving an explicit nonzero proper submodule:

Let $W = \left\{ \alpha \cdot \left( \sum_{g \in T} \vec{e}_{Hg} \right) : \alpha \in \mathbb{F}_p \right\}$ where $T$ is a transversal of $H$ in $G$.

This is a one-dimensional trivial submodule of $V$. If $n > 1$, then it is a proper submodule and $V$ is reducible.

Answer 3

There is a version of Frobenius reciprocity that holds over any field:

If $G$ is a finite group, and $H$ is a subgroup of $G$ then $Ind_H^G$ and $Res_H^G$ are adjoint functors between $Rep_k(H)$ and $Rep_k(G)$ for any field $k$.

If you don't like this functor language, this concretely means that $$Hom_G(Ind_H^G(V), W) \cong Hom_H(V, Res_H^G(W))$$ and $$Hom_H(Res_H^G(W), V) \cong Hom_G(W, Ind_H^G(V))$$

For all representations $V$ and $W$ over $k$ of $H$ and $G$ respectively.

Applying these when both $V$ and $W$ are the trivial representation shows that $Ind_H^G(1)$ has a trivial submodule as well as a trivial quotient.