Brian Bi
$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\span}{span} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\ad}{ad} \newcommand\d{\mathrm{d}} \newcommand\pref{(\ref{#1})}$
$\DeclareMathOperator{\Ind}{Ind}$

Problem 5.8.4 Let's start by recalling that the elements of $$\Ind^G_K V$$ are functions $$G \to V$$ and the elements of $$\Ind^G_H \Ind^H_K V$$ are functions $$G \to (H \to V)$$. These functions must satisfy certain constraints, but writing out the types helps us to identify the required isomorphism. It must have the type $$\varphi : (G \to V) \to (G \to (H \to V))$$. This suggests that we might try: \begin{equation*} \varphi(f) := g, h \mapsto f(hg) \end{equation*}

Let's now check that $$\varphi$$ actually maps from $$\Ind^G_K V$$ to $$\Ind^G_H \Ind^H_K V$$. Let $$f \in \Ind^G_K V$$, so that for all $$k \in K, g \in G$$, we have $$f(kg) = kf(g)$$. For each $$h, h' \in H, g \in G$$, we have: \begin{equation*} \varphi(f)(hg)(h') = f(h'hg) \end{equation*} and \begin{align*} (h(\varphi(f)(g)))(h') &= \varphi(f)(g)(h'h) \\ &= f(h'hg) \end{align*} Therefore, for all $$g \in G, h \in H$$, we have $$\varphi(f)(hg) = h(\varphi(f)(g))$$, which is one of the requirements to have $$\varphi(f) \in \Ind^G_H \Ind^H_K V$$. The other requirement is that each of the maps $$\varphi(f)(g)$$ actually lies in $$\Ind^H_K V$$. For this, note that for each $$k \in K$$, we have $$(\varphi(f)(g))(kh) = f(khg) = kf(hg) = k(\varphi(f)(g)(h))$$, as required.

Next, we show that $$\varphi$$ is a linear map. Let $$F$$ denote the base field. If $$\alpha \in F$$, then for each $$f \in \Ind^G_K V$$, we have $$\varphi(\alpha f) = (g, h \mapsto (\alpha f)(hg)) = (g, h \mapsto \alpha f(hg)) = \alpha(g, h \mapsto f(hg)) = \alpha \varphi(f)$$, and for each $$f_1, f_2 \in \Ind^G_K V$$, we have $$\varphi(f_1 + f_2) = (g, h \mapsto (f_1 + f_2)(hg)) = (g, h \mapsto f_1(hg) + f_2(hg)) = (g, h \mapsto f_1(hg)) + (g, h \mapsto f_2(hg)) = \varphi(f_1) + \varphi(f_2)$$, as required.

Next, we show that $$\varphi$$ is injective. Suppose $$\varphi(f) = 0$$ for some $$f \in \Ind^G_K V$$. Then for all $$g \in G, h \in H$$, we have $$\varphi(f)(g)(h) = 0$$. This implies that $$f(hg) = 0$$ for all $$g, h$$. But the set of elements of the form $$hg$$ is all of $$G$$, so $$f$$ vanishes identically.

For $$V$$ finite-dimensional, we could immediately conclude that $$\varphi$$ is surjective because its domain and codomain have the same dimension over $$F$$. There doesn't seem to be anything in the text implying that $$V$$ is assumed finite-dimensional, so we will show surjectivity directly. Suppose $$j \in \Ind^G_H \Ind^H_K V$$. Then let $$f : G \to V$$ be defined by $$f(g) = j(g)(1)$$. We then see that \begin{align*} \varphi(f)(g)(h) &= f(hg) \\ &= j(hg)(1) \\ &= h(j(g))(1) \\ &= j(g)(1\cdot h) \\ &= j(g)(h) \end{align*} therefore $$\varphi(f) = j$$. Also, for each $$k \in K$$, \begin{align*} f(kg) &= j(kg)(1) \\ &= k(j(g))(1) \\ &= j(g)(1\cdot k) \\ &= j(g)(k\cdot 1) \\ &= k(j(g)(1)) \\ &= kf(g) \end{align*} therefore $$f \in \Ind^G_K V$$ as required.

Having established that $$\varphi$$ is a linear isomorphism, we have one final task, namely to show that $$\varphi$$ is an isomorphism of $$G$$-modules. So let $$f \in \Ind^G_K V$$ and let $$g, g' \in G, h \in H$$. Then \begin{align*} (\varphi(g'f))(g)(h) &= (g'f)(hg) \\ &= f(hgg') \\ &= \varphi(f)(gg')(h) \\ &= (g'\varphi(f))(g)(h) \end{align*} so $$\varphi(g'f) = g'\varphi(f)$$ as required.

Exercise 5.8.5 The required isomorphism $$\varphi : \Ind^G_K \mathbb{C}_\chi \to \mathbb{C}[G]e_\chi$$ is given by: \begin{equation*} \varphi(f) := \sum_{g \in G} f(g^{-1}) g \end{equation*} Note that: \begin{align*} \varphi(f)e_\chi &= \frac{1}{|K|}\sum_{g \in G} \sum_{k \in K} \chi(k)^{-1} f(g^{-1}) gk \\ &= \frac{1}{|K|} \sum_{g \in G} \sum_{h \in gK} \chi(g^{-1}h)^{-1} f(g^{-1})h \\ &= \frac{1}{|K|} \sum_{h \in G} \sum_{g : h \in gK} \chi(h^{-1}g) f(g^{-1})h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{g \in hK} \chi(h^{-1}g) f(g^{-1})\right] h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{k \in K} \chi(h^{-1}hk) f((hk)^{-1})\right]h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{k \in K} \chi(k) f(k^{-1}h^{-1})\right] h \\ &= \frac{1}{|K|} \sum_{h \in G} \left[\sum_{k \in K} \chi(k) k^{-1} f(h^{-1})\right] h \\ &= \frac{1}{|K|} \sum_{h \in G} \sum_{k \in K} f(h^{-1})h \\ &= \frac{1}{|K|} \sum_{h \in G} |K| f(h^{-1})h \\ &= \sum_{h \in G} f(h^{-1})h \\ &= \varphi(f) \end{align*} therefore $$\varphi(f) \in \mathbb{C}[G]e_\chi$$, as required.

It is obvious that $$\varphi$$ is a linear map. If $$\varphi(f) = 0$$, then it is easy to see that $$f = 0$$, so $$\varphi$$ is injective.

Notice that for all $$g \in G, k \in K$$, we have \begin{align*} gk e_\chi &= \frac{1}{|K|} gk \sum_{k' \in K} \chi(k')^{-1} k' \\ &= \frac{1}{|K|} g \sum_{k' \in K} \chi(k')^{-1} kk' \\ &= \frac{\chi(k)}{|K|} g \sum_{k' \in K} \chi(k)^{-1} \chi(k')^{-1} kk' \\ &= \frac{\chi(k)}{|K|} g \sum_{k' \in K} \chi(kk')^{-1} kk' \\ &= \frac{\chi(k)}{|K|} g \sum_{k'' \in K} \chi(k'')^{-1} k'' \\ &= \chi(k) g e_\chi \end{align*} therefore $$\mathbb{C}[G]e_\chi$$ is $$|G|/|K|$$-dimensional with a basis given by $$g_1 e_\chi, \ldots, g_{|G|/|K|} e_\chi$$ where the $$g_i$$'s belong to different left $$K$$-cosets of $$G$$. Since $$\mathbb{C}[G]e_\chi$$ has the same dimension (as a vector space) as $$\Ind^G_K \mathbb{C}_\chi$$, the map $$\varphi$$ is surjective.

Finally, we must show that $$\varphi: \Ind^G_K \mathbb{C}_\chi \to \mathbb{C}[G]e_\chi$$ is an isomorphism of $$G$$-modules. Let $$f \in \Ind^G_K \mathbb{C}_\chi, g \in G$$. Then \begin{align*} \varphi(gf) &= \sum_{h \in G} (gf)(h^{-1})h \\ &= \sum_{h \in G} f(h^{-1}g)h \\ &= g\sum_{h \in G} f(h^{-1}g)g^{-1}h \\ &= g\sum_{h' \in G} f(h'^{-1})h' \\ &= g\varphi(f) \end{align*} as required.