Lecture 6

Solution: For $n$ odd Proposition 3.1 gives us a complete list of irreducibles and their dimensions. Thus we have

as required. For $n$ even we consult the table in the solution of Problem 26. Thus we have

as required.

Solution:

1. (a)
   ​

   $\Phi(\alpha T+\beta S)=(\alpha T+\beta S)(e)=\alpha T(e)+\beta S(e)=\alpha\Phi(T)+\beta\Phi(S)$.

2. (b)

   	$\displaystyle\Psi(\alpha v+\beta u)\left(\sum_{g\in G}a_{g}g\right)=\sum_{g\in G}a_{g}\pi(g)(\alpha v+\beta u)$
   	$\displaystyle=\alpha\sum_{g\in G}a_{g}\pi(g)v+\beta\sum_{g\in G}a_{g}\pi(g)u=\alpha\Psi(v)\left(\sum_{g\in G}a_{g}g\right)+\beta\Psi(u)\left(\sum_{g\in G}a_{g}g\right),$

   ​

   hence $\Psi(\alpha v+\beta u)=\alpha\Psi(v)+\beta\Psi(u)$.

3. (c)
   $$\Phi(\Psi(v))=\Psi(v)\big{(}e\big{)}=1\pi(e)(v)=v.$$

Solution: We know that $D_{3}$ has 2 one-dimensional irreducible representations and one two-dimensional irreducible, hence by Theorem 6.6, each one-dimensional representation occurs a single time in the decomposition, and the two-dimensional representation appears twice. Since each $\lambda(g)$ acts a permutation on the basis of ${\mathbb{C}}(G)$ given by $e,r,r^{2},s,sr,sr^{2}$, we quickly see that the trivial representation is given by $\big{(}\lambda,{\mathbb{C}}(e+r+r^{2}+s+sr+sr^{2})\big{)}$. Similarly, $\lambda(s)$ swaps $r^{j}$ and $sr^{j}$, while $\lambda(r)$ moves $r^{j}$ to $r^{j+1}$ and $sr^{j}$ to $sr^{j+2}$ In the other one-dimensional representation, $\lambda(r)$ acts as $1$, and $\lambda(s)$ acts as $-1$. Since $\lambda(r)$ permutes $e,r,r^{2}$, these must all have the same coefficients in the linear combination that gives a spanning vector for the other one-dimensional representation, for example $e+r+r^{2}$ applying $\lambda(s)$ to this gives $s+sr+sr^{2}$, so for $\lambda(s)$ to act as $-1$, the vector must be $e+r+r^{2}-s-sr-sr^{2}$. The other one-dimensional representation is therefore $\big{(}\lambda,{\mathbb{C}}(e+r+r^{2}-s-sr-sr^{2})\big{)}$. Now we search for eigenvectors for $\lambda(r)$ coming from the two copies of the two-dimensional representation. From the classification of the irreducible representations, we know that the eigenvalues are $\omega,\omega^{2}$, where $\omega=e^{2\pi i/3}$. Just studying the $r$-action on the cyclic group gives two obvious eigenvectors: $e+\omega^{2}r+\omega r^{2}$ is an eigenvector with eigenvalue $\omega$, and $e+\omega r+\omega^{2}r^{2}$ is an eigenvector with eigenvalue $\omega^{2}$. Applying $\lambda(s)$ to these gives the vectors $s+\omega^{2}sr+\omega sr^{2}$ and $s+\omega sr+\omega^{2}sr^{2}$. These are also eigenvectors for $\lambda(r)$:

and similarly

We’ve thus identified two two-dimensional irreducible subrepresentations:

It remains to show that the direct sum of all these subspaces is ${\mathbb{C}}(D_{3})$. This follows directly from the observation that they are all orthogonal with respect to the standard inner product on the natural basis (cf. Problem 27).

Theorem 6.7. If every irreducible representation of a finite group $G$ is one-dimensional, then $G$ is Abelian.

Hint: Consider the representation $(\lambda,{\mathbb{C}}G)$. Show that $\ker\lambda=\{e\}$, and thus $G$ is isomorphic to $\lambda(G)<\operatorname{GL}({\mathbb{C}}G)$. Use Theorem 6.6 to show that there is a basis of ${\mathbb{C}}G$ for which the matrices of all $\lambda(g)$ are diagonal, hence $\lambda(G)$ is Abelian.

Remark: Combining this with Proposition 3.1 and Theorem 4.6 gives the following nice result:

Theorem 6.8 A finite group $G$ is Abelian if and only if all its irreducible representations are one-dimensional.

Solution: We show that for $(\lambda,{\mathbb{C}}G)$, we have $\ker\lambda=\{e\}$. Suppose that $g\in\ker\lambda$, i.e. $\lambda(g)={\,\mathrm{Id}}$. Then $\lambda(g)v=v$ for all $v\in{\mathbb{C}}G$. In particular, for each basis vector $h\in G$, we have $\lambda(g)h=h$, i.e. $gh=h$ for all $h\in G$, hence $g=e$. Thus by the isomorphism theorem, $G\cong\lambda(G)/\ker(\lambda)=\lambda(G)$. Since each irreducible representation of $G$ is one-dimensional, by Theorem 6.6 we have the decomposition

and for each $i=1,\ldots,|G|$, we have

where $\psi_{i}:G\rightarrow{\mathbb{C}}^{\times}$ is a homomorphism. Writing the matrix of $\lambda(g)$ with respect to the basis $v_{1},\ldots,v_{|G|}$ gives

The map $g\mapsto[\lambda(g)]$ is thus an isomorphism from $G$ to $[\lambda(G)]$, which is a subgroup of the diagonal matrices in $\operatorname{GL}_{|G|}({\mathbb{C}})$. The invertible diagonal matrices are an Abelian subgroup of $\operatorname{GL}_{|G|}({\mathbb{C}})$, so $G$ is isomorphic to a subgroup of an Abelian group, and is hence Abelian.

Solution:

1. (a)
   ​

   The proof is very similar to that of Lemma 4.1. As $V$ is a finite-dimensional vector space over ${\mathbb{C}}$, $T$ has an eigenvalue $\lambda\in{\mathbb{C}}$. Consider the morphism $T-\lambda\,{\,\mathrm{Id}}$. Then for any $a\in A$,

   $$\pi(a)(T-\lambda{\,\mathrm{Id}})=\pi(a)T-\lambda\pi(a)=T\pi(a)-\lambda\pi(a)=(T-\lambda{\,\mathrm{Id}})\pi(a);$$

   $T-\lambda{\,\mathrm{Id}}$ is therefore in $\operatorname{Hom}_{A}(V)$. Let $v\in\ker(T-\lambda{\,\mathrm{Id}})$, then

   $$(T-\lambda{\,\mathrm{Id}})\pi(a)v=\pi(a)(T-\lambda{\,\mathrm{Id}})v=\pi(a)0=0,$$

   for all $a\in A$, so $\big{(}\pi,\ker(T-\lambda{\,\mathrm{Id}})\big{)}$ is a submodule of $(\pi,V)$. The subspace $\ker(T-\lambda{\,\mathrm{Id}})\subseteq V$ is non-zero, since any $\lambda$-eigenvector of $T$ is in $\ker(T-\lambda{\,\mathrm{Id}})$, and so $(\pi,V)$ being irreducible then gives $V=\ker(T-\lambda{\,\mathrm{Id}})$, hence $T=\lambda{\,\mathrm{Id}}\in{\mathbb{C}}{\,\mathrm{Id}}$.

2. (b)
   ​

   Let $(\pi,W)$ be a nontrivial subrepresentation of $(\pi,V)$. Then $\pi(g)w=w$ for all $w\in W$. Thus for an arbitrary element of ${\mathbb{C}}G$ we have

   $$\pi\left(\sum_{g\in G}z_{g}g\right)w=\sum_{g\in G}z_{g}\pi(g)w$$

   and, as each $\pi(g)w$ is in $W$, $W$ is closed under the action of ${\mathbb{C}}G$. Thus $(\pi,W)$ is a proper submodule of $(\pi,V)$.

   Now let $(\pi,V)$ be an irreducible representation of $G$. Then there exists a $v\in V$ such that $\mathrm{span}\{\pi(g)v|g\in G\}=V$. Then by considering each $g\in G$ as an element of ${\mathbb{C}}G$, in other words set $z_{g}=1$ and all other coefficients to zero, we have the span of $\pi(g)v$s is all of $V$ hence $(\pi,V)$ is an irreducible module.

Solution:

1. (a)
   ​

   We first note that $Z(A)$ is nonempty as $0$ is always in the centre. Additionally

   $$({\bf a}+\lambda{\bf b}){\bf c}={\bf a}{\bf c}+\lambda{\bf b}{\bf c}={\bf c}{\bf a}+\lambda{\bf c}{\bf b}={\bf c}({\bf a}+\lambda{\bf b})$$

   for all ${\bf a},{\bf b}\in Z(A)$, $\lambda\in k$ and ${\bf c}\in A$, so $\mathbb{Z}(A)$ is a vector subspace. Furthermore

   $$({\bf a}{\bf b}){\bf c}={\bf a}({\bf b}{\bf c})={\bf a}({\bf c}{\bf b})=({\bf a}{\bf c}){\bf b}=({\bf c}{\bf a}){\bf b}={\bf c}({\bf a}{\bf b})$$

   for all ${\bf a},{\bf b}\in Z(A)$ and ${\bf c}\in A$, so $\mathbb{Z}(A)$ is closed under the product in $A$ and thus is a subalgebra.

2. (b)
   ​

   Let $a(1)+b(12)+c(13)+d(23)+e(123)+f(132)$ be in $Z({\mathbb{C}}S_{3})$, then multiplying by $(12)$ on both sides,

   $$a(12)+b(1)+c(132)+d(123)+e(23)+f(13)=a(12)+b(1)+c(123)+d(132)+e(13)+f(23),$$

   and comparing coefficients e have that $c=d$ and $e=f$. Repeating this with other elements of $S_{3}$ gives us that $b=c=e$ and thus

   $$Z({\mathbb{C}}S_{3})=\left\{a(1)+b\sum_{\sigma\in S_{3}}\sigma\;\Bigg{|}\;a,b\in{\mathbb{C}}\right\}$$
3. (c)
   ​

   Let $z\in Z(A)$, then $\pi(z)\in\operatorname{Hom}(V)$ and $\pi(a)\pi(z)=\pi(z)\pi(a)$ for all $a\in A$. Thus we can apply Schur’s lemma for algebra modules, Problem 41(a), and we have $\pi(z)\in{\mathbb{C}}{\,\mathrm{Id}}$. Let $\psi:Z(A)\rightarrow{\mathbb{C}}$, where $\psi(z)$ is the coefficient of ${\,\mathrm{Id}}$ in $\pi(z)$. As $\pi(z+\lambda w)=\pi(z)+\lambda\pi(w)$ and $\pi(wz)=\pi(zw)=\pi(z)\pi(w)=\pi(w)\pi(z)$ for all $z,w\in Z(A)$ and all $\lambda\in{\mathbb{C}}$, $\psi$ is a homomorphism of commutative algebras.