Maschke's Theorem, Complete Reducibility and the Structure of $k[G]$ — Examples

The group $G$ is abelian and has three elements, so each element forms its own conjugacy class. Hence [L1] gives exactly three irreducible complex representations, with degrees $d_1,d_2,d_3$, and [L2] gives $d_1^2+d_2^2+d_3^2=3.$

Each $d_i$ is a positive integer, so the only way three positive squares can sum to $3$ is $d_1=d_2=d_3=1$. Applying [L3], all three Wedderburn factors are $1\times 1$ matrix algebras, so $\mathbb{C}[G]\cong M_1(\mathbb{C})\times M_1(\mathbb{C})\times M_1(\mathbb{C})={\mathbb{C}}^3\mathrm{.}$

The conjugacy classes of $S_3$ are $\{e\},\{(12),(13),(23)\},\{(123),(132)\}\mathrm{.}$ Hence [L1] gives exactly three irreducible complex representations, with degrees $d_1,d_2,d_3$, and [L2] gives $d_1^2+d_2^2+d_3^2=6.$

Each $d_i$ is a positive integer. Since $1^2+1^2+2^2=6$ and any larger square would already exceed $6$, the only possible degree multiset is $1,1,2$. Applying [L3], the Wedderburn factors are therefore $M_1(\mathbb{C})$, $M_1(\mathbb{C})$, and $M_2(\mathbb{C})$, so $\mathbb{C}[S_3]\cong \mathbb{C}\times \mathbb{C}\times M_2(\mathbb{C})\mathrm{.}$

In $Q_8$, the elements $1$ and $-1$ are central by [L2]. Also $ji{j}^{-1}=ji(-j)=(-k)(-j)=kj=-i$, so $i$ is conjugate to $-i$; similarly $ki{k}^{-1}=-i$, and conjugation by $\pm 1$ or $\pm i$ fixes $i$. Thus the conjugacy class of $i$ is $\{i,-i\}$. The same calculation with cyclic permutations of $i,j,k$ gives the further classes $\{j,-j\}$ and $\{k,-k\}$. Hence the conjugacy classes of $Q_8$ are $\{1\},\{-1\},\{i,-i\},\{j,-j\},\{k,-k\}\mathrm{.}$

In $D$, the relations of [L1] show that $r^2$ is central, while $sr{s}^{-1}=r^{-1}=r^3,rs{r}^{-1}=r^{2}s,r(rs)r^{-1}=r^{3}s\mathrm{.}$ Therefore the conjugacy classes of $D$ are $\{1\},\{r^2\},\{r,r^3\},\{s,r^{2}s\},\{rs,r^{3}s\}\mathrm{.}$ So both $Q_8$ and $D$ have five conjugacy classes.

By [L4], each group has five irreducible complex representations. By [L5], their degrees $d_1,\dots ,d_5$ satisfy $d_1^2+\cdots +d_5^2=8.$ Since every $d_i\ge 1$, the only possible multiset is $1,1,1,1,2$.

Applying [L6] to each group gives one $2\times 2$ factor and four $1\times 1$ factors, so $\mathbb{C}[Q_8]\cong {\mathbb{C}}^4\times M_2(\mathbb{C})\text{and}\mathbb{C}[\mathrm{Dih}(C_4)]\cong {\mathbb{C}}^4\times M_2(\mathbb{C})\mathrm{.}$

The group $G$ is abelian with four elements, so it has four conjugacy classes. Hence [L1] gives four irreducible complex representations, with degrees $d_1,d_2,d_3,d_4$, and [L2] gives $d_1^2+d_2^2+d_3^2+d_4^2=4.$

Every $d_i$ is positive, so the only way four positive squares can sum to $4$ is $d_1=d_2=d_3=d_4=1$. Then [L3] says each irreducible occurs in the regular representation with multiplicity equal to $1$, so the regular representation splits as the direct sum of those four one-dimensional summands.

Every line in $V$ is invariant because the action is trivial. Any nonzero line is irreducible by [L2], since a one-dimensional vector space has only the subspaces $0$ and itself. Therefore both displayed decompositions are decompositions into irreducible subrepresentations, and they are different because $k{e}_1\mathrm{\ne }k(e_1+e_2)$.

On every nonzero line the action is trivial, so each line is equivalent to the one-dimensional trivial representation of [L1]. Hence every irreducible summand of $V$ has the same type, and the sum of all irreducible summands of that type is all of $V$. By [L3], this is the unique isotypic component.

Let $N=J-I$, so $N^2=0$. A short induction gives $J^m=(I+N{)}^m=I+mN$ for every $m\ge 0$, and in $k=\mathbb{Z}\mathrm{/}p$ this yields $J^p=I$ because $p=0$. Therefore sending the generator $g$ to $J$ defines a two-dimensional representation of $C_p$ over $k$ in the sense of [L2]. The line $L=k{e}_1$ is invariant because $J{e}_1=e_1$.

Every one-dimensional complement to $L$ has the form $M=k(a{e}_1+e_2)$ for some $a\in k$. But $J(a{e}_1+e_2)=(a+1)e_1+e_2\mathrm{.}$ If this vector lay in $M$, then $(a+1)e_1+e_2=\lambda (a{e}_1+e_2)$ for some $\lambda \in k$. Comparing the $e_2$-coefficients gives $\lambda =1$, and then comparing the $e_1$-coefficients gives $a+1=a$, impossible in a field. So $L$ has no invariant complement. This refutes the claim.

By [L1], the coefficient ring $\mathbb{Z}\mathrm{/}p$ is a field, so $A$ is a finite-dimensional algebra over a field. By [L2], ${\dim }_{\mathbb{Z}\mathrm{/}p}A=\mathrm{\mid }{C}_p\mathrm{\mid }=p\mathrm{.}$

The field $\mathbb{Z}\mathrm{/}p$ has characteristic $p$, and $\mathrm{\mid }{C}_p\mathrm{\mid }=p$, so [L3] applies and shows that $A$ is not semisimple. Therefore a finite-dimensional algebra over a field need not be semisimple.

The example [L1] exhibits one completely reducible representation with two different decompositions into irreducible subrepresentations.

By [L2], this does not contradict isotypic uniqueness: the isotypic block is canonical, but the individual irreducible splitting is not. Therefore the displayed example refutes the claim.

From [L1], the irreducible complex representations of $S_3$ have degrees $1$, $1$, and $2$. Their sum is $1+1+2=4$.

But $\mathrm{\mid }{S}_3\mathrm{\mid }=6$, so the unsquared sum is not the group order. Indeed [L2] gives the correct identity $1^2+1^2+2^2=6.$ Therefore the claim is false.

By [L1], the relevant value is $r=5$, while the irreducible degrees of $Q_8$ are only $1,1,1,1,2$.

None of those degrees equals $5$, so the claim fails even for $Q_8$. Therefore a group with $r$ conjugacy classes need not have an irreducible representation of degree $r$.