There are a couple of different approaches to induced representations that are commonly found in the literature. The one that we will give here is the one that generalizes to topological groups, such a Lie groups. The other, based on the tensor product, is less versatile.

Let $G$ be a finite group, $H$ a subgroup and $(\pi, V)$ a representation of
$H$. We will define $V^G$ to be the vector space of all functions $f : G
\longrightarrow V$ such that $f (h x) = \pi (h) f (x)$ when $h \in H$ and $x
\in G$. Define, for $g \in G$
\[ (\pi^G (g) f) (x) = f (x g) . \]
Thus $g$ acts on $V^G$ by *right translation*.

**Exercise 4.2.1:***
Check that if $f \in V^G$ and $g \in G$ then $\pi^G (g) f \in V^G$. Also
check that
\[ \pi^G (g_1 g_2) = \pi^G (g_1) \pi^G (g_2), \]
so that $(\pi^G, V^G)$ is a representation of $G$.
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**Theorem 4.2.1:***
( Frobenius reciprocity.) Let $(\pi, V)$ be a
representation of $H$ and let $(\sigma, W)$ be a representation of $G$. We
have a vector space isomorphism
\[ \operatorname{Hom}_G (W, V^G) \cong \operatorname{Hom}_H (W, V) . \]
In this isomorphism the $G$-module homomorphism $\Phi : W \longrightarrow
V^G$ corresponds to the $H$-module homomorphism $\phi : W \longrightarrow
V$, where we may express $\Phi$ in terms of $\phi$ and $\phi$ in terms of
$\Phi$ by the following formulae.
\[ \phi (w) = \Phi (w) (1), \hspace{2em} \Phi (w) (g) = \phi (\sigma (g) w)
. \]
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**Proof. **(Click to Expand/Collapse)

We first check that if $\Phi : W \longrightarrow V^G$ is a $G$-module
homomorphism, then $\phi (w) = \Phi (w) (1)$ defines an $H$-module
homomorphism. Indeed, we have, for $h \in H$
\[ \phi (\sigma (h) w) = \Phi (\sigma (h) w) (1) = (\pi^G (h) \Phi (w)) (1)
= \Phi (w) (1 \cdot h) = \Phi (w) (h \cdot 1) = \pi (h) \Phi (w) (1), \]
where we have used the definition of $\phi$, the assumption that $\Phi$ is a
$G$-module homomorphism, the definition of $\Phi^G$, the identity $1 \cdot h
= h \cdot 1$, and the assumption that $\Phi (w) \in V^G$. This equals $\pi
(h) \phi (w)$, so $\phi$ is an $H$-module homomorphism.

We leave the reader to complete the proof (Exercise 4.2.2).

**Exercise 4.2.2:***
Complete the above proof as follows.
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(a) Show that if $\phi : W \longrightarrow V$ is an $H$-module homomorphism then $\Phi (w) (g) = \phi (\sigma (g) w)$ defines an element of $V^G$, and that $\Phi : W \longrightarrow V^G$ is a $G$-module homomorphism.

*
(b) Show that the two constructions $\phi \longmapsto \Phi$ and $\Phi
\longmapsto \phi$ are inverse maps between $\operatorname{Hom}_G (W, V^G)$ and
$\operatorname{Hom}_H (W, V) .$
*

**Theorem 4.2.2:***
If $\chi$ is the character of the representation $(\pi, V)$ of $H$, then
$\chi^G$ is the character of the representation $(\pi^G, V^G)$ of $G$.
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**Proof. **(Click to Expand/Collapse)

If $\chi'$ is the character of $\pi^G$, and if $\theta$ is the character of
the representation $(\sigma, W)$ of $G$, then by
Proposition 2.4.4 and Theorem 4.2.1 we have
\[ \left\langle \chi', \theta \right\rangle_G = \dim \; \operatorname{Hom}_G (W,
V^G) = \dim \; \operatorname{Hom}_H (W, V) = \left\langle \chi, \theta
\right\rangle_H . \]
By Frobenius reciprocity for characters, we have $\left\langle \chi', \theta
\right\rangle_G = \left\langle \chi^G, \theta \right\rangle_G$ for every
character $\theta$ of $G$. In particular, if $\theta$ is irreducible, this
mean that $\theta$ has the same multiplicity in the decompositions of
$\chi'$ and $\chi^G$ into irreducibles, and so they are equal.

**Exercise 4.2.3:***
Suppose that $(\pi, V)$ is a representation of $H$. Let $x \in G$, and let
$v \in V$. Let
\[ f_{x, v} (g) = \left\{ \begin{array}{ll}
\pi (h) x & \text{if $g = h x$ for some $h \in H$;}\\
0 & \text{if $g \notin H x$.}
\end{array} \right. \]
Prove that $f_{x, v} \in V^G$.
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A *monomial matrix* is one that has a unique nonzero element in each
row and column. If that entry equals $1$, the matrix is called a
*permutation matrix*. Thus if
\[ A = \left(\begin{array}{cccc}
0 & 3 & 0 & 0\\
0 & 0 & - 1 & 0\\
0 & 0 & 0 & 1\\
4 & 0 & 0 & 0
\end{array}\right), \hspace{2em} P = \left(\begin{array}{cccc}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0
\end{array}\right) \]
both are monomial matrices and $P$ is a permutation matrix.

**Exercise 4.2.4:***
Suppose that $\dim (V) = 1$. Fix a nonzero element $v \in V$. Let $x_1,
\cdots, x_m$ be a complete set of coset representatives for $H \backslash
G$. Show that $f_{x_i, v}$ are a basis of $V^G$. Show that the endomorphism
$\pi (g)$ of $V^G$ is a permutation matrix with respect to this basis.
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