3 Induced representations 3.2 Frobenius reciprocity 3.4 Example: D₄ to S₄

3.3 Characters

Let $H\subset G$ be finite groups, and let $(\rho,V)$ be a representation of $H$ with character $\chi$ . We use Frobenius reciprocity to determine the character $\operatorname{Ind}_{H}^{G}\chi$ of $\operatorname{Ind}_{H}^{G}\rho$ .

Theorem 3.12.

Suppose that we are in the above situation. Suppose that $\mathcal{C}$ is a conjugacy class of $G$ , and let $\mathcal{C}\cap H=\mathcal{D}_{1}\cup\ldots\cup\mathcal{D}_{n}$ where $\mathcal{D}_{1},\ldots,\mathcal{D}_{n}$ are conjugacy classes of $H$ . Then

(\operatorname{Ind}_{H}^{G}\chi)(\mathcal{C})=\frac{|G|}{|H|}\sum_{i=1}^{n}% \frac{|\mathcal{D}_{i}|}{|\mathcal{C}|}\chi(\mathcal{D}_{i}).

Proof.

Let $1_{\mathcal{C}}$ be the indicator function of $\mathcal{C}$ . Then for any class function $\chi$ on $G$ ,

\left\langle\chi,1_{\mathcal{C}}\right\rangle_{G}=\frac{|\mathcal{C}|}{|G|}% \chi(\mathcal{C}).

So, combining this with Frobenius reciprocity,

	$\displaystyle(\operatorname{Ind}_{H}^{G}\chi)(\mathcal{C})$	$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\operatorname{Ind}_{H}^{G}% \chi,1_{\mathcal{C}}\right\rangle_{G}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\chi,1_{\mathcal{C}}\right% \rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\left\langle\chi,1_{% \mathcal{D}_{i}}\right\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{% \|H\|}\chi(\mathcal{D}_{i})$
		$\displaystyle=\frac{\|G\|}{\|H\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{\|\mathcal{% C}\|}\chi(\mathcal{D}_{i})$

which gives the claimed formula. ∎

Remark 3.13.

We derived the character formula from Frobenius reciprocity. It is also possible to go the other way around: prove the character formula directly, then derive Frobenius reciprocity as a consequence. Personally, I find that knowing Frobenius reciprocity is the easiest way to remember the character formula.

Remark 3.14.

If $g\in G$ , we write $C_{G}(g)$ for the centralizer of $g$ :

C_{G}(g)=\{x\in G:x^{-1}gx=g\}.

By the orbit-stabiliser theorem, if $\mathcal{C}(g)$ is the conjugacy class of $g$ , then

\frac{|G|}{|\mathcal{C}(g)|}=|C_{G}(g)|

giving an interpretation for some of the factors in the above formula.

Exercise 3.15.

Let $H\subset G$ be groups and let $\chi$ be a character of $H$ . If $\dot{\chi}(g)=\chi(g)$ if $g\in H$ and $0$ otherwise.

1.

Show that the formula for the induced character may be rewritten

$(\operatorname{Ind}_{H}^{G}\chi)(g)=\frac{1}{|H|}\sum_{x\in G}\dot{\chi}(x^{-1% }gx).$
2.

If $g_{1}H,\ldots,g_{r}H$ are the left cosets of $H$ in $G$ , show that

$(\operatorname{Ind}_{H}^{G}\chi)(g)=\sum_{i=1}^{r}\dot{\chi}(g_{i}^{-1}gg_{i}).$

Example 3.16.

We continue with the example of the dihedral group. Let $H=C_{n}\subset G=D_{n}$ , and let $\psi:H\to\mathbb{C}^{\times}$ be a homomorphism with $\psi(r)=\omega$ . Then:

•

$\operatorname{Ind}_{H}^{G}\psi(e)=[G:H]=2$ .
•

$\operatorname{Ind}_{H}^{G}\psi(s)=\operatorname{Ind}_{H}^{G}\psi(rs)=0$ as the conjugacy class of $s$ or $r s$ does not intersect $H$ .
•

If $0<i<n/2$ , then the conjugacy class $\{r,r^{-1}\}$ of $G$ splits into two conjugacy classes, $\{r\}$ and $\{r^{-1}\}$ , of $H$ . We have

$\operatorname{Ind}_{H}^{G}(\psi)(r^{i})=2(\frac{1}{2}\psi(r^{i})+\frac{1}{2}% \psi(r)^{-i})=\omega^{i}+\omega^{-i}.$
•

If $i=n/2$ , then the conjugacy class $\{r^{n/2}\}$ of $G$ remains as a single conjugacy class of $H$ and

$\operatorname{Ind}_{H}^{G}(\psi)(r^{n/2})=2\psi(r^{n/2})=2\omega^{n/2}.$

Taking $\omega\neq\pm 1$ we again obtain all the irreducible two-dimensional characters of $D_{n}$ .

	$\displaystyle(\operatorname{Ind}_{H}^{G}\chi)(\mathcal{C})$	$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\operatorname{Ind}_{H}^{G}% \chi,1_{\mathcal{C}}\right\rangle_{G}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\chi,1_{\mathcal{C}}\right% \rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\left\langle\chi,1_{% \mathcal{D}_{i}}\right\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{% \|H\|}\chi(\mathcal{D}_{i})$
		$\displaystyle=\frac{\|G\|}{\|H\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{\|\mathcal{% C}\|}\chi(\mathcal{D}_{i})$