# Induction of Characters

Induction of representations will be a major theme of the next chapter. Here we just prove enough to help us compute some character tables.

We will use ad hoc methods here to prove the existence of induced characters, without really revealing the nature of the induced representation. In the next chapter we will give a better definition of the induced representation, by constructing the induced representation itself. Here we will simply define the character of the representation by giving its formula, and use the orthogonality relations to prove that it is a character.

More specifically, let $H$ be a subgroup of the finite group $G$, and let $(\sigma, V)$ be a representation of $H$. There is a representation $(\sigma^G, V^G)$ of $G$ that is "induced'' from $(\sigma, V)$. Let $\theta$ be the character of $(\sigma, V)$, and let $\theta^G$ be the character of $(\sigma^G, V^G)$. In this section we will define $\theta^G$ by a formula, then prove that it is the character of a representation, without actually constructing the representation $(\sigma^G, V^G)$.

The character $\theta$ is a function on $H$. We will extend $\theta$ to a function $\dot{\theta}$ on $G$. (Look closely to see the difference between $\theta$ and $\dot{\theta}$.) We define $\dot{\theta} (g) = \left\{ \begin{array}{ll} \theta (g) & \text{if g \in H,}\\ 0 & \text{otherwise} . \end{array} \right.$ Define $\theta^G (g) = \frac{1}{|H|} \sum_{x \in G} \dot{\theta} (x g x^{- 1}) .$

If $H$ is of small index in $G$, this formula is inefficient to compute with, so we will rewrite it more efficiently.

Lemma 2.5.1: Let $x_1, \cdots, x_d$ be a set of right coset representatives of $H$, so that $G = H x_1 \cup \cdots \cup H x_d \hspace{2em} (\operatorname{disjoint}),$ and $d = [G : H]$. Then
 $\theta^G (g) = \sum_{i = 1}^d \dot{\theta} (x_i g x_i^{- 1}) .$ (2.5.1)

Proof. (Click to Expand/Collapse)

We have
 $\text{\dot{\theta} (h g h^{- 1}) = \dot{\theta} (g) if h \in H} .$ (2.5.2)

Indeed, both sides of (2.5.2) vanish if $g \notin H$, and if $g \in H$, both sides are equal because $\theta$ is a class function on $H$. Now it follows that the value of $\dot{\theta} (x g x^{- 1})$ depends only on the coset $H x$, so instead of summing over all elements of $G$ then dividing by $|H|$, we need only choose one representative of each coset.

We will make use of the inner products on both $L^2 (G)$ and $L^2 (H)$. To distinguish them, we will subscript them, so $\left\langle f_1, f_2 \right\rangle_G = \frac{1}{|G|} \sum_{g \in G} f_1 (g) \overline{f_2 (g)}, \hspace{2em} \left\langle f_1, f_2 \right\rangle_H = \frac{1}{|H|} \sum_{g \in H} f_1 (g) \overline{f_2 (g)} .$

Theorem 2.5.1: The function $\theta^G$ is the character of a representation of $G$. Its degree is $[G : H] \theta (1)$. If $\chi$ is any character of $G$, and $\chi_H$ is its restriction to $H$, then
 $\left\langle \theta^G, \chi \right\rangle_G = \left\langle \theta, \chi_H \right\rangle_H$ (2.5.3)

The equation (2.5.3) is known as Frobenius reciprocity. It is extremely important and powerful. Even if $\theta$ is irreducible, we will see by example that $\theta$ may or may not be irreducible.

Proof. (Click to Expand/Collapse)

We first note that it is obvious that $\theta^G$ is a class function. It is clear from the formula for $\theta^G$ that degree of $\theta^G$ is $[G : H]$ times $\theta (1)$. We will first prove (2.5.3), then deduce the fact that $\theta^G$ is a character.

We have \begin{eqnarray*} \left\langle \theta^G, \chi \right\rangle_G & = & \frac{1}{| G|} \sum_{g \in G} \theta^G (g) \overline{\chi (g)}\\ & = & \frac{1}{| G|} \frac{1}{| H|} \sum_{g \in G} \sum_{x \in G} \dot{\theta} (x g x^{- 1}) \overline{\chi (g)}\\ & = & \frac{1}{| G|} \frac{1}{| H|} \sum_{x \in G} \sum_{g \in G} \dot{\theta} (x g x^{- 1}) \overline{\chi (g)} \end{eqnarray*} Now with $x$ fixed, we may replace $g$ by $x^{- 1} g x$, and noting that $\chi (x^{- 1} g x) = \chi (g)$ since $\chi$ is a class function, we get $\left\langle \theta^G, \chi \right\rangle_G = \frac{1}{| G|} \frac{1}{| H|} \sum_{x \in G} \sum_{g \in G} \dot{\theta} (g) \overline{\chi (g)} .$ Now the summand is independent of $x$, so we may drop the summation over $x$ and the division by $|G|$, to obtain $\left\langle \theta^G, \chi \right\rangle_G = \frac{1}{| H|} \sum_{g \in G} \dot{\theta} (g) \overline{\chi (g)} = \frac{1}{| H|} \sum_{g \in H} \theta (g) \overline{\chi (g)},$ where we have used the fact that $\dot{\theta}$ vanishes off $H$. Now (2.5.3) is proved.

It remains to be proved that $\theta^G$ is a character. Let $(\pi_1, V_1), \cdots, (\pi_h, V_h)$ be representatives of the isomorphism classes of the irreducible $G$-modules, and $\chi_1, \cdots, \chi_h$ be their characters. Since the $\chi_i$ are a basis of the complex vector space of class functions, and since $\theta^G$ is a class function, we can write $\theta^G = \sum_{i = 1}^h a_i \chi_i,$ where the $a_i$ are complex numbers, and if we can show that $a_i$ are nonnegative integers, it will follow that $\theta^G$ is a character, namely it will be the character of the representation on the module $a_i V_1 \oplus \cdots \oplus a_h V_h .$ By orthogonality, $a_i = \left\langle \theta^G, \chi \right\rangle_G$, and now by (2.5.3), we have $a_i = \left\langle \theta^G, \chi \right\rangle_G = \left\langle \theta, \chi_H \right\rangle_H .$ This expression shows that $a_i$ is a nonnegative integer by Proposition 2.4.4, since both $\theta$ and $\chi_H$ are characters of $H$.