3.16. Lecture 16

3.16.1. The character formula

Let $H\leq G$ be finite groups and $(\pi,V)$ a representation of $H$ . We now use Frobenius reciprocity to give a formula for $\operatorname{Ind}_{H}^{G}\chi_{\pi}$ .

Theorem 3.16.1.

Let ${\mathcal{C}}$ be a conjugacy class of $G$ . Let $H$ be a subgroup of $G$ and write

{\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

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

Proof.

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

\langle f,\mathbbold{1}_{\mathcal{C}}\rangle_{G}=\frac{|{\mathcal{C}}|}{|G|}f(% {\mathcal{C}}).

Combining this with Frobenius reciprocity (Corollary 3.15.2), we have

	$\displaystyle(\mathrm{Ind}_{H}^{G}\chi_{\pi})({\mathcal{C}})$	$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\langle\mathrm{Ind}_{H}^{G}\chi_{\pi}% ,\mathbbold{1}_{\mathcal{C}}\rangle_{G}$
		$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\langle\chi_{\pi},\operatorname{Res}_% {H}^{G}\mathbbold{1}_{\mathcal{C}}\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\sum_{i=1}^{n}\langle\chi_{\pi},% \mathbbold{1}_{{\mathcal{D}}_{i}}\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\sum_{i=1}^{n}\frac{\|{\mathcal{D}}_{i% }\|}{\|H\|}\chi_{\pi}({\mathcal{D}}_{i})=\frac{\|G\|}{\|H\|}\sum_{i=1}^{n}\frac{\|{% \mathcal{D}}_{i}\|}{\|{\mathcal{C}}\|}\chi_{\pi}({\mathcal{D}}_{i})$

which is the claimed formula. ∎

Remark 3.16.2.

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.

Remark 3.16.3.

If $g\in G$ , we write $C_{G}(g)$ for the centraliser 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.

Example 3.16.4.

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

(i)

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

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

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

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

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

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

Taking $\omega\neq\pm 1$ we again obtain all the irreducible 2-dimensional characters of $D_{n}$ .

3.16.2. Example: D₄ to S₄

Let $H$ be the subgroup of $G=S_{4}$ isomorphic to $D_{4}$ , obtained by labeling the vertices of a square 1, …, 4 and letting $D_{4}$ act on them. In other words, $H$ is the image of the injective homomorphism $D_{4}\rightarrow S_{4}$ sending

r\mapsto(1234),s\mapsto(12)(34).

Recall that $D_{4}$ has five irreducible representations with characters as shown:

\begin{array}[]{c|rrrrr}&e&r=(1234)&r^{2}=(13)(24)&s=(12)(34)&rs=(13)\\ &1&2&1&2&2\\ \hline\cr{\mathrm{triv}}&1&1&1&1&1\\ \epsilon&1&1&1&-1&-1\\ \phi_{+}&1&-1&1&1&-1\\ \phi_{-}&1&-1&1&-1&1\\ \sigma&2&0&-2&0&0\\ \end{array}

while recall that $S_{4}$ has character table

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr{\mathrm{triv}}&1&1&1&1&1\\ \operatorname{sgn}&1&-1&1&1&-1\\ \pi&3&1&-1&0&-1\\ \operatorname{sgn}\pi&3&-1&-1&0&1\\ \rho&2&0&2&-1&0\end{array}

We want to first find out how the conjugacy classes of $S_{4}$ intersect with $D_{4}$ . The result is as follows, writing ${\mathcal{C}}_{g}$ for the conjugacy class of $g$ in $S_{4}$ , and ${\mathcal{D}}_{h}$ for the conjugacy class of $h$ in $D_{4}$ .

\begin{array}[]{l|l|l}g\in S_{4}&{\mathcal{C}}_{g}\cap D_{4}&\text{sizes}\\ \hline\cr e&{\mathcal{D}}_{e}&1\\ (12)&{\mathcal{D}}_{rs}&2\\ (12)(34)&{\mathcal{D}}_{r^{2}}\cup{\mathcal{D}}_{s}&1,2\\ (123)&\emptyset&-\\ (1234)&{\mathcal{D}}_{r}&2\\ \end{array}

We use this and the character formula to determine $\mathrm{Ind}_{H}^{G}\phi_{+}$ . The factor $|G|/|H|$ is constant, equal to 3. We obtain:

\begin{array}[]{l|r|r}&&\mathrm{Ind}_{H}^{G}\phi_{+}\\ \hline\cr e&1&3\\ (12)&6&3\cdot 2/6\cdot(-1)=-1\\ (12)(34)&3&3(1/3+2/3)=3\\ (123)&8&0\\ (1234)&6&3(-2/6)=-1\\ \end{array}

Decomposing this character, we see that

\mathrm{Ind}_{H}^{G}\phi_{+}=\operatorname{sgn}+\chi_{\rho}.

But note that we did not need to know the character table of $S_{4}$ to find the induced character.

We check that this is consistent with Frobenius reciprocity: the restriction of $\operatorname{sgn}$ to $D_{4}$ is $\phi_{+}$ , so

\langle\operatorname{Res}^{G}_{H}\operatorname{sgn},\phi_{+}\rangle_{D_{4}}=% \langle\operatorname{sgn},\mathrm{Ind}_{H}^{G}\phi_{+}\rangle_{S_{4}}=1

as required. The restriction of $\chi_{\rho}$ to $D_{4}$ is ${\mathrm{triv}}+\phi_{+}$ so

\langle\operatorname{Res}^{G}_{H}\chi_{\rho},\phi_{+}\rangle_{D_{4}}=\langle% \chi_{\rho},\mathrm{Ind}_{H}^{G}\phi_{+}\rangle_{S_{4}}=1

(we could also check that the restrictions of the other irreducible characters of $S_{4}$ to $D_{4}$ do not contain $\phi_{+}$ ).

3.16.3. Exercises

Problem 86. Let $\chi_{\sigma}$ be the irreducible degree 3 character (i.e. $\chi_{\sigma}(e)=3$ ) of $H=A_{5}$ such that

\chi_{\sigma}((12345))=\frac{1+\sqrt{5}}{2}.

Let $G=A_{6}$ , with $H$ as the subgroup of $G$ fixing 6. Compute the character $\mathrm{Ind}^{G}_{H}\chi_{\sigma}$ .

Problem 87. Let $H=\langle(12\ldots p)\rangle\subseteq S_{p}$ , for $p$ a prime, and $\psi$ a nontrivial character of $H$ .

(a)

Compute $\mathrm{Ind}_{H}^{S_{p}}\psi$ .
(b)

By considering $\langle\mathrm{Ind}_{H}^{S_{p}}\psi,\mathrm{Ind}_{H}^{S_{p}}\psi\rangle_{S_{p}}$ , show that

$(p-1)!\equiv-1\quad(\mathrm{mod}\;p).$

(Recall that if $\zeta$ is a non-trivial $p$ -th root of unity, we have $1+\zeta+\zeta^{2}+\ldots+\zeta^{p-1}=0$ .)

Problem 88. Let

G=\left\langle x,y\,|\,y^{7}=x^{3}=e,\,xyx^{-1}=y^{2}\right\rangle.

$G$ has 21 elements, and may be expressed as $\{x^{i}y^{j}\,|\,0\leq i\leq 2,\,0\leq j\leq 6\}$ .

(a)

Show that $H=\langle y\rangle$ is a normal subgroup of $G$ .
(b)

By considering representations lifted from $G/H$ and induced from $H$ , find the character table of $G$ .

	$\displaystyle(\mathrm{Ind}_{H}^{G}\chi_{\pi})({\mathcal{C}})$	$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\langle\mathrm{Ind}_{H}^{G}\chi_{\pi}% ,\mathbbold{1}_{\mathcal{C}}\rangle_{G}$
		$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\langle\chi_{\pi},\operatorname{Res}_% {H}^{G}\mathbbold{1}_{\mathcal{C}}\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\sum_{i=1}^{n}\langle\chi_{\pi},% \mathbbold{1}_{{\mathcal{D}}_{i}}\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|{\mathcal{C}}\|}\sum_{i=1}^{n}\frac{\|{\mathcal{D}}_{i% }\|}{\|H\|}\chi_{\pi}({\mathcal{D}}_{i})=\frac{\|G\|}{\|H\|}\sum_{i=1}^{n}\frac{\|{% \mathcal{D}}_{i}\|}{\|{\mathcal{C}}\|}\chi_{\pi}({\mathcal{D}}_{i})$