The Group Algebra and Representations of Finite Groups: Examples and False Statements

By [L1], the field $\mathbb{C}$ satisfies the scalar-endomorphism condition of [L2], so it is a splitting field for the finite group $G$. Since $G$ is abelian, [L2] makes every irreducible complex representation of $G$ one-dimensional.

By [L3], every irreducible representation of $G$ is equivalent to a normalized degree-one representation on $\mathbb{C}$, hence to a homomorphism $\chi :G\to {\mathbb{C}}^{\times }$. Because $G=⟨g⟩$, such a homomorphism is determined by the value $\lambda =\chi (g)$, and the relation $g^n=e$ forces ${\lambda }^n=1$. Conversely, if ${\lambda }^n=1$, then ${\chi }_{\lambda }(g^m):={\lambda }^m$ is well defined and multiplicative.

By [L4], the polynomial $t^n-1$ splits over $\mathbb{C}$ and has exactly $n$ distinct roots there, namely the elements of ${\mu }_n(\mathbb{C})$. So step 2.1 produces exactly $n$ equivalence classes of irreducible complex representations, represented by the normalized characters ${\chi }_{\lambda }$ with $\lambda \in {\mu }_n(\mathbb{C})$.

Put $u:=[e]+[s]$ and $v:=[e]-[s]$. By [L1], $s\cdot u=[s]+[e]=u$ and $s\cdot v=[s]-[e]=-v$.

Because $\mathrm{char}k\mathrm{\ne }2$, the two vectors $u$ and $v$ are linearly independent and span $k[C_2]$: one has $[e]=\frac{1}{2}(u+v)$ and $[s]=\frac{1}{2}(u-v)$. Therefore $ku$ is the trivial line of [L2], $kv$ is the sign line of [L2], and the regular representation is their direct sum.

By [L1], the action of $S_3$ on $\{e_1,e_2,e_3\}$ extends to ${\mathbb{C}}^3$ by permuting coordinates. The vector $(1,1,1)$ is fixed by every permutation, so $L$ is invariant. The coordinate sum $x+y+z$ is also permutation-invariant, so $U$ is invariant and ${\mathbb{C}}^3=L\oplus U$.

Suppose $W\subseteq U$ is a nonzero invariant line, and choose $0\mathrm{\ne }v=(a,b,c)\in W$. Because $W$ is invariant under the transposition $\tau =(12)$, one has $\tau (v)=\lambda v$ for some scalar $\lambda$, and ${\tau }^2=1$ forces $\lambda =\pm 1$. If $\lambda =1$, then $a=b$ and the relation $a+b+c=0$ gives $v=a(1,1,-2)$. If $\lambda =-1$, then $a=-b$ and $c=0$, so $v=a(1,-1,0)$.

The $3$-cycle $\sigma =(123)$ sends $(1,1,-2)$ to $(1,-2,1)$ and $(1,-1,0)$ to $(0,1,-1)$, and neither image is a scalar multiple of the original vector. Thus neither of the two possibilities from step 2.1 can span an invariant line. So $U$ has no proper nonzero invariant line, and by [L2] it is irreducible.

For $g,x\in G$, left multiplication sends the coset $xH$ to $(gx)H$, so $G$ acts on the finite set $G\mathrm{/}H$ by $g\cdot (xH):=(gx)H$.

By [L2], the induced representation on the basis vectors $e_{xH}$ of $k^{(G\mathrm{/}H)}$ is $g\cdot e_{xH}=e_{(gx)H}$. So the matrix of $g$ in this basis is a permutation matrix recording the induced permutation of the cosets.

The subgroup $⟨g⟩$ of [L3] is finite because it is a subset of the finite group $G$, so some least integer $n\ge 1$ satisfies $g^n=e$. Because $g\mathrm{\ne }e$, this least $n$ is at least $2$. The vectors $[e]-[g]$ and $[e]+[g]+\cdots +[g^{n-1}]$ are both nonzero, since they are sums of distinct basis vectors from [L1].

Using [L1] and [L2], $([e]-[g])([e]+[g]+\cdots +[g^{n-1}])=([e]+[g]+\cdots +[g^{n-1}])-([g]+[g^2]+\cdots +[g^n])=[e]-[g^n]=0.$ So $k[G]$ has nonzero zero divisors.

For each $q\in Q_8$, define $L_q:\mathbb{H}\to \mathbb{H}$ by $L_q(x)=qx$. By [L2], quaternion multiplication is distributive and real scalars commute with every quaternion, so $L_q$ is $\mathbb{R}$-linear. By [L3], the underlying vector space is $4$-dimensional.

Because each $q\in Q_8$ is nonzero by [L1], [L2] gives an inverse $q^{-1}$ in $\mathbb{H}$, and $L_{q^{-1}}$ is the inverse of $L_q$. So every $L_q$ is an invertible linear map.

The action laws hold: $1\cdot x=x$, and $(pq)\cdot x=(pq)x=p(qx)=p\cdot (q\cdot x)$ by associativity from [L2]. Therefore left multiplication is a finite-dimensional real representation of $Q_8$.

A direct multiplication gives $A^2+A+I_2=0$, hence $A^3=I_2$. Therefore $g\mapsto A$ defines a real representation of $C_3=⟨g⟩$ on $V={\mathbb{R}}^2$.

Let $T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$. The condition $TA=AT$ from [L2] is equivalent to $c=-b$ and $d=a$, so the commuting endomorphisms are exactly the matrices $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a,b\in \mathbb{R}$.

Let $W\subseteq V$ be a nonzero invariant line, and choose $0\mathrm{\ne }v\in W$. Then $Av=\lambda v$ for some $\lambda \in \mathbb{R}$. Applying the polynomial identity from step 1.1 gives $({\lambda }^2+\lambda +1)v=0$, so ${\lambda }^2+\lambda +1=0$. But $4({\lambda }^2+\lambda +1)=(2\lambda +1{)}^2+3>0$ by [L3], impossible. Hence no nonzero proper invariant line exists, so the representation is irreducible by [L1].

The map $\mathrm{\Phi }:\mathbb{C}\to {\mathrm{End}}_{C_3}(V)$ given by $\mathrm{\Phi }(a+bi)=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ is bijective by the uniqueness in [L4], and the multiplication formula in [L4] matches matrix multiplication of these $2\times 2$ matrices. Therefore ${\mathrm{End}}_{C_3}(V)$ is a copy of $\mathbb{C}$.

By [L1], every element of the nontrivial group $G$ acts as the identity in the trivial representation.

Since some element of $G$ is not the identity, [L2] shows that this representation is not faithful. Therefore the stated claim is false.

By [L1], choose nonzero elements $x,y\in k[G]$ with $xy=0$.

If $k[G]$ were a field, then [L2] would make $x$ invertible, and multiplying $xy=0$ by $x^{-1}$ would give $y=0$, contradicting step 1.1. So the claim is false.

Applying [L1] with $n=2$ gives a degree-one representation of $C_2$ with $g\mapsto -1$.

This representation is not trivial because the generator acts by $-1\mathrm{\ne }1$. Therefore the stated claim is false.

By [L1], over the field $\mathbb{R}$ there exists an irreducible representation whose endomorphism ring is $\mathbb{C}$.

Since $\mathbb{C}\mathrm{\ne }\mathbb{R}$, that representation contradicts the stated claim. Therefore the claim is false.