1 Representation theory of finite groups 1.3 Example: dihedral groups 1.5 Maschke’s theorem and complete reducibility

1.4 Schur’s Lemma

We move on to more theoretical considerations. Let $G$ be a group.

Theorem 1.32.

Let $V$ and $W$ be two irreducible finite-dimensional complex representations of $G$ . Let $T:V\to W$ be a $G$ -homomorphism. Then

1.

Either $T$ is an isomorphism or $T=0$ .
2.

If $V=W,$ then

$T=\lambda\mathrm{id}_{V}$

for some scalar $\lambda\in\mathbb{C}$ .
3.

More generally,

$\dim\operatorname{Hom}_{G}(V,W)=\begin{cases}1&\text{ if $V\cong W$}\\ 0&\text{ otherwise}.\end{cases}$

Proof.

Part (1) follows directly from Lemma 1.24: suppose that $T$ is nonzero. Then $\ker(T)\neq V$ and $\operatorname{im}(T)\neq\{0\}$ . Since $V$ and $W$ are irreducible, we must have $\ker(T)=0$ and $\operatorname{im}(T)=V$ .

For (2), we can find an eigenvalue $\lambda$ of the endomorphism $T$ — here we are using that $V$ is a finite-dimensional complex vector space. Then $T-\lambda\mathrm{id}_{V}$ is also a $G$ -homomorphism with non-zero kernel. Since $V$ is irreducible, by part (1) we must have $T-\lambda\mathrm{id}_{V}=0$ . Therefore $T=\lambda\mathrm{id}_{V}$ .

For (3), by part (1) we have $\dim\operatorname{Hom}_{G}(V,W)=0$ if they are not isomorphic. So assume that they are isomorphic, so that there is an isomorphism $S:V\rightarrow W$ . If $T:V\to W$ is a $G$ -homomorphism, then $S^{-1}\circ T:V\to V$ is also a $G$ -homomorphism. By (2), $S^{-1}\circ T=\lambda\mathrm{id}_{V}$ for some $\lambda,$ so $T=\lambda S$ . We have shown that $\operatorname{Hom}_{G}(V,W)$ is one-dimensional, spanned by $S$ . ∎

As a corollary we obtain

Theorem 1.33.

Let $G$ be an abelian group. Then every finite-dimensional irreducible complex representation of $G$ is one-dimensional.

Proof.

Let $(\rho,V)$ be an irreducible representation. For $h\in G,$ set $T_{h}=\rho(h)\in\operatorname{GL}(V)$ . Then $T_{h}$ is actually a $G$ -homomorphism $V\rightarrow V$ : this is because it commutes with $\rho(g)$ for all $g\in G$ . Indeed,

T_{h}\rho(g)=\rho(h)\rho(g)=\rho(hg)=\rho(gh)=\rho(g)\rho(h)=\rho(g)T_{h}.

Hence by Schur’s Lemma, $T_{h}=\rho(h)$ acts by a scalar $\chi(h)$ on $V$ :

\rho(h)v=\chi(h)v

for all $v\in V$ and a non-zero scalar $\chi(h)$ .

But now, any non-zero $v\in V$ spans a $G$ -invariant subspace. Since $V$ is irreducible, this implies that $V$ is one-dimensional, and $\chi=\rho$ is a homomorphism $G\rightarrow\mathbb{C}^{\times}$ . ∎

Remark 1.34.

It is possible to give an alternative proof of this using the fact from linear algebra that any commuting set of linear maps from a finite dimensional vector space to itself has a simultaneous eigenvector.

In Schur’s Lemma and Theorem 1.33, we didn’t assume $G$ was finite but we did assume the irreducible representation $V$ was finite-dimensional. If, in fact, $G$ is finite, then $V$ is automatically finite-dimensional.

Proposition 1.35.

Any irreducible representation of a finite group is finite-dimensional.

Proof.

Let $V$ be an irreducible representation of the finite group $G,$ and choose $v\in V$ nonzero. Then the subspace $V^{\prime}$ spanned by $\{g\cdot v:g\in G\}$ is finite dimensional, and is preserved by $G$ (why?). As $V$ is irreducible, we have $V^{\prime}=V,$ so $V$ is finite dimensional. ∎

A homomorphism $\chi:G\rightarrow\mathbb{C}^{\times}$ is often called a character (though this will later cause an unfortunate clash of notation). If $G$ is abelian, then the group

\hat{G}=\{\text{homomorphisms }\chi:G\rightarrow\mathbb{C}^{\times}\}

is called the character group, or dual group, of $G$ . It is a group under the operation $(\chi_{1}\chi_{2})(g)=\chi_{1}(g)\chi_{2}(g)$ .

Example 1.36.

Let $G=C_{n}$ be a cyclic group of order $n$ . Then $\hat{G}\cong C_{n}$ .

Indeed, pick a generator $g$ of $G$ and let $\omega=e^{2\pi i/n}$ be a primitive $n$ th root of unity. Then a character $\chi$ of $G$ is determined uniquely by $\chi(g),$ which must be an $n$ th root of unity $\omega^{a}$ . If we let $\chi_{a}\in\hat{G}$ be the homorphism such that $\chi_{a}(g)=\omega^{a},$ then the map

a\mapsto\chi_{a}

determines a group isomorphism $\mathbb{Z}/n\mathbb{Z}\rightarrow\hat{G}$ . This is a homomorphism because

\chi_{a+b}(g)=\omega^{a+b}=\omega^{a}\omega^{b}=\chi_{a}(g)\chi_{b}(g).

In fact, if $G$ is any finite abelian group, then $\hat{G}\cong G$ . You can prove this using the cyclic case and the fundamental theorem of finite abelian groups.

For arbitrary groups $G$ the same method of proof gives:

Proposition 1.37.

Let $(\rho,V)$ be an irreducible finite-dimensional representation of $G$ and let

Z=Z(G)=\{z\in G:zg=gz\text{ for all $g\in G$}\}

be the center of $G$ . Then $Z$ acts on $V$ as a character: there is a character $\chi:Z\rightarrow\mathbb{C}^{\times}$ such that

\rho(z)v=\chi(z)v.

for all $z\in Z$ and $v\in V$ .

We call $\chi$ the central character of $\rho$ .

Proof.

Homework; mimic the proof that irreducible representations of abelian groups are one-dimensional. ∎

Finally, we can use our classification of the irreducible representations of abelian groups to get a bound on the dimension of the irreducible representations of any finite group.

Proposition 1.38.

Let $G$ be a finite group, let $A$ be an abelian subgroup of $G$ and let $(\rho,V)$ be an irreducible representation of $G$ . Then

\dim V\leq\frac{|G|}{|A|}=[G:A].

Proof.

Restrict the representation to $A$ and find an irreducible $A$ -subrepresentation $W$ of $V$ . By Theorem 1.33, $W$ is one-dimensional, spanned by a vector $v$ . So there is a character $\chi$ of $A$ such that

\rho(h)v=\chi(h)v

for all $h\in A$ . Now $\{\rho(g)v:g\in G\}$ spans a subrepresentation of $V,$ hence is equal to $V$ by irreducibility. Write $g_{1}A,g_{2}A,\ldots,g_{r}A$ for the left cosets of $A,$ where $r=[G:A]$ . Then for $h\in A,$ we have

\rho(g_{i}h)v=\rho(g_{i})\rho(h)v=\rho(g_{i})\chi(h)v=\chi(h)(\rho(g_{i})v)

But this implies that $V=\left\langle\{\rho(g)v:g\in G\}\right\rangle$ is already spanned by

\{\rho(g_{i})v:i=1,\ldots,r\},

so has dimension at most $r=[G:A]$ . ∎

Example 1.39.

The group $D_{n}$ has an abelian subgroup $C_{n}$ of index two, and so every irreducible representation of $D_{n}$ has dimension at most $2$ .