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

Choose the projection $P$ from [A2] and define $P_G:=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_{g\in G}\rho (g)P\rho (g{)}^{-1}\mathrm{.}$ Each summand is $k$-linear, so $P_G$ is a $k$-linear endomorphism of $V$.

For every $h\in G$, $\rho (h)P_G\rho (h{)}^{-1}=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_{g\in G}\rho (hg)P\rho (hg{)}^{-1}=P_G,$ because left multiplication by $h$ permutes the finite set $G$. Thus $P_G$ is $G$-equivariant.

Each summand $\rho (g)P\rho (g{)}^{-1}$ maps $V$ into $\rho (g)(W)=W$, so $P_G(V)\subseteq W$ by [L1]. If $w\in W$, then $\rho (g{)}^{-1}w\in W$ by [L1], hence $P(\rho (g{)}^{-1}w)=\rho (g{)}^{-1}w$ and therefore $\rho (g)P\rho (g{)}^{-1}(w)=w$ for every $g$. Summing gives $P_G(w)=w$. So $P_G$ has image exactly $W$.

Put $U:=\ker P_G$. Since $P_G$ is $G$-equivariant, $U$ is a subrepresentation. For every $v\in V$ one has $v=P_G(v)+(v-P_G(v)),$ with $P_G(v)\in W$ and $v-P_G(v)\in U$. If $x\in W\cap U$, then step 3.1 gives $x=P_G(x)=0$, so the sum is direct. Hence $V=W\oplus U$.

If $V=0$, then [L1] makes $V$ completely reducible as the empty direct sum.

Assume now that ${\dim }_{k}V>0$, and as induction hypothesis suppose every representation of smaller dimension is completely reducible. Among the nonzero subrepresentations of $V$, choose one with least positive dimension, using [A1]. It is irreducible by [L2], since any proper nonzero subrepresentation would have smaller positive dimension. Call this irreducible subrepresentation $W$. By [L3], there is a subrepresentation $U$ with $V=W\oplus U$.

The subrepresentation $W$ is nonzero, so ${\dim }_{k}U<{\dim }_{k}V$. The induction hypothesis therefore makes $U$ completely reducible. Using [L1] to expand that decomposition and adjoining the irreducible summand $W$, one gets a direct-sum decomposition of $V$ into irreducible subrepresentations. Hence $V$ is completely reducible.

By [L1] and [L2], the regular representation of $G$ on $k[G]$ is finite-dimensional. Therefore [L3] makes it completely reducible.

Expanding that term with [L4], the left regular representation is an internal direct sum of irreducible subrepresentations. By [L5], those are exactly simple left $k[G]$-submodules. So the left regular module ${}_{k[G]}k[G]$ is an internal direct sum of simple submodules.

By [L6], that is exactly the definition that $k[G]$ is a semisimple ring.

Assume, for contradiction, that $k[G]=I_G\oplus J$ for some left $k[G]$-submodule $J$. Since $\epsilon ([e])=1$ by [L1], one has $[e]\mathrm{\notin }{I}_G$, so $J\mathrm{\ne }0$. The restriction $\epsilon {\mathrm{\mid }}_J$ is injective because $\ker (\epsilon {\mathrm{\mid }}_J)=J\cap I_G=0$. It is also nonzero, for otherwise $J\subseteq I_G$. Choose $x\in J$ with $\epsilon (x)\mathrm{\ne }0$, and replace $x$ by $\epsilon (x{)}^{-1}x$ so that $\epsilon (x)=1$.

For each $g\in G$, the element $gx=[g]x$ lies in $J$ because $J$ is a $k[G]$-submodule by [L2]. Also $\epsilon (gx)=\epsilon ([g])\epsilon (x)=1\cdot 1=1$ by [L1], so $\epsilon (gx-x)=0$ and therefore $gx-x\in I_G$. Since both $gx$ and $x$ lie in $J$, one also has $gx-x\in J$. Thus $gx-x\in I_G\cap J=0$, so $gx=x$ for every $g\in G$.

Write $x={\sum }_{h\in G}{a}_h[h]\mathrm{.}$ For any $g\in G$, step 2.1 gives ${\sum }_{h\in G}{a}_h[h]=x=gx={\sum }_{h\in G}{a}_h[gh]={\sum }_{h\in G}{a}_{g^{-1}h}[h]\mathrm{.}$ Comparing coefficients in the basis of $k[G]$ shows $a_h=a_{g^{-1}h}$ for all $g,h$, so all coefficients are equal to one scalar $a\in k$. Hence $1=\epsilon (x)={\sum }_{h\in G}a=\mathrm{\mid }G\mathrm{\mid }a=0$ by [L1] and [A1], a contradiction. Therefore no such complement $J$ exists.

Assume, for contradiction, that $k[G]$ is semisimple. Then [L2] makes the left regular module ${}_{k[G]}k[G]$ semisimple. It is generated by $[e]$, so [L3] makes it a finite direct sum of simple submodules. In particular it is a finite-length module.

Applying [L4] to that finite-length semisimple module shows that every submodule of ${}_{k[G]}k[G]$ has a complementary submodule. In particular the augmentation ideal $I_G$ has a complement.

Step 2.1 contradicts [L1]. Therefore $k[G]$ is not semisimple.

By [L3], choose irreducible subrepresentations $U_1,\dots ,U_n$ with $V=U_1\oplus \cdots \oplus U_n\mathrm{.}$ Group these summands by equivalence class: for each class represented by $S_i$, let $W_i$ be the direct sum of those $U_j$ equivalent to $S_i$. Then $V=W_1\oplus \cdots \oplus W_r,$ and each $W_i$ is contained in $V_{(S_i)}$ by [L1].

Fix $i$ and let $U\le V$ be any irreducible subrepresentation equivalent to $S_i$. Write ${\pi }_j:V\to U_j$ for the projection attached to step 1.1. If $U_j$ is not equivalent to $S_i$, then ${\pi }_j{\mathrm{\mid }}_U:U\to U_j$ is an intertwiner between non-equivalent irreducibles, so [L2] makes it zero. Hence the projection of $U$ onto ${⨁}_{j:U_j\cong ̸S_i}{U}_j$ is zero, and therefore $U\le W_i$. Since this holds for every such $U$, the defining sum [L1] satisfies $V_{(S_i)}\le W_i$. Together with step 1.1, this gives $V_{(S_i)}=W_i$.

Step 2.1 shows that each grouped block $W_i$ is exactly the isotypic component $V_{(S_i)}$, so it depends only on the equivalence class of $S_i$, not on the chosen irreducible splitting. Since the $W_i$ already form a direct sum in step 1.1, the displayed isotypic decomposition is unique.

By [L1], the ring $k[G]$ is semisimple. It is also nonzero because the basis element $[e]$ is its identity, so [L2] gives $k[G]\cong {\prod }_{i=1}^r{M}_{n_i}(D_i)$ for positive integers $r,n_i$ and division rings $D_i$.

For each factor, let $S_i$ be the simple module supplied by [L3]; under the ring isomorphism of step 1.1 it becomes a simple left $k[G]$-module, and then [L4] turns it into an irreducible representation of $G$ over $k$. By [L5] and [L6], every $k[G]$-module endomorphism of $S_i$ is scalar multiplication by an element of $k$.

Identify $S_i$ with the column module $D_i^{n_i}$. If $f:S_i\to S_i$ is $M_{n_i}(D_i)$-linear, write $e_1,\dots ,e_{n_i}$ for the standard column basis and let $d\in D_i$ be the first entry of $f(e_1)$. Since the matrix units $E_{aa}$ kill every coordinate except the $a$-th and $E_{a1}{e}_1=e_a$, linearity forces $f(e_a)=e_{a}d$ for every $a$. Hence $f(x)=xd$ for every column $x$, and composition corresponds to multiplication in the opposite order. So ${\mathrm{End}}_{k[G]}(S_i)\cong D_i^{\mathrm{o}\mathrm{p}}$. Step 2.1 therefore gives $D_i^{\mathrm{o}\mathrm{p}}\cong k$, hence $D_i\cong k$ because a field is canonically isomorphic to its opposite ring. Replacing each $D_i$ in step 1.1 by $k$ yields the claimed decomposition.

By [L1], the left regular $k[G]$-module is isomorphic to the left regular module of ${\prod }_{i=1}^r{M}_{n_i}(k)$. As a module over that product, the regular module splits as the direct sum of the factor regular modules, and [L3] decomposes factor $i$ into $n_i$ copies of the column module $k^{n_i}$. Thus the regular representation [L5] is a direct sum of finitely many simple modules, with the factor-$i$ simple occurring exactly $n_i$ times.

By [L2], those factor column modules give all simple module isomorphism classes, one for each factor. Translating with [L4], there are finitely many irreducible representations of $G$, one for each factor, and the factor-$i$ representation has degree $n_i$ because its underlying vector space is $k^{n_i}$. Therefore the regular representation contains each irreducible representation with multiplicity equal to its degree.

By [L1], the regular representation is a direct sum of ${\dim }_k{V}_i$ copies of each $V_i$, and [L2] says its total dimension is $\mathrm{\mid }G\mathrm{\mid }$.

Taking dimensions in step 1.1 gives $\mathrm{\mid }G\mathrm{\mid }={\dim }_{k}k[G]={\sum }_{i=1}^r{\dim }_k(V_i^{\oplus {\dim }_k{V}_i})={\sum }_{i=1}^r({\dim }_k{V}_i{)}^2\mathrm{.}$ This is the required identity.

Let $X=(x_{ab})\in Z(M_n(k))$. For each $i$, let $E_{ii}$ be the diagonal matrix with a single $1$ in position $(i,i)$. Because $E_{ii}X=X{E}_{ii}$ and [L1] computes matrix products entrywise, the $(a,b)$-entry comparison shows $x_{ab}=0$ whenever $a\mathrm{\ne }b$. So $X$ is diagonal.

Write $X=\mathrm{diag}(d_1,\dots ,d_n)$. For $i\mathrm{\ne }j$, let $E_{ij}$ be the matrix unit with a single $1$ in position $(i,j)$. Then [L1] gives $E_{ij}X=d_j{E}_{ij},X{E}_{ij}=d_i{E}_{ij}\mathrm{.}$ Since $X$ is central, these are equal, so $d_i=d_j$ for all $i,j$. Thus $X=\lambda I_n$ for one scalar $\lambda \in k$. Conversely every scalar matrix commutes with every matrix because scalar multiplication is entrywise and $I_n$ is the identity from [L1]. Hence the center consists exactly of the scalar matrices.

Let $C$ be a conjugacy class and $h\in G$. Using [L3], $[h]\widehat{C}[h{]}^{-1}={\sum }_{g\in C}[hg{h}^{-1}]\mathrm{.}$ Because conjugation by $h$ permutes the elements of $C$, this sum is again $\widehat{C}$. Hence $[h]\widehat{C}=\widehat{C}[h]$ for every basis element $[h]$, so $\widehat{C}\in Z(k[G])$ by [L1] and [L3].

Now let $x={\sum }_{g\in G}{a}_g[g]$ be any central element. For every $h\in G$, centrality gives $[h]x=x[h]$, so multiplying on the right by $[h{]}^{-1}$ and using [L3] yields $[h]x[h{]}^{-1}=x$. Comparing coefficients in the basis $\{[g]\}$ shows $a_{hg{h}^{-1}}=a_g$ for all $g,h\in G$. Thus the coefficient function $g\mapsto a_g$ is constant on conjugacy classes, and $x$ is a $k$-linear combination of the class sums.

Distinct conjugacy classes are disjoint subsets of $G$, so their class sums have disjoint supports in the basis $\{[g]\}$. Therefore a linear relation among class sums forces every coefficient to vanish. Combined with step 2.1, this shows that the class sums form a basis of $Z(k[G])$.

By [L1], there is one basis vector of $Z(k[G])$ for each conjugacy class of $G$.

The dimension of a finite-dimensional vector space is the number of vectors in any basis, so step 1.1 identifies ${\dim }_{k}Z(k[G])$ with the number of conjugacy classes.

By [L5], the dimension of $Z(k[G])$ is the number of conjugacy classes of $G$.

Write the product decomposition of [L1]. An element of a direct product is central exactly when each coordinate is central in its own factor, so [L4] gives $Z(k[G])\cong Z\text{\hspace{0.17em}⁣}({\prod }_{i=1}^m{M}_{n_i}(k))\cong {\prod }_{i=1}^{m}Z(M_{n_i}(k))\cong k^m\mathrm{.}$ Hence ${\dim }_{k}Z(k[G])=m$. By [L2], the product ring has exactly $m$ simple left module isomorphism classes, and [L3] translates these exactly into the irreducible representations of $G$. So the number of irreducible representations is also $m$.

Steps 1.1 and 1.2 are the two computations of the same dimension, so the number of irreducible representations equals the number of conjugacy classes.

Let $m=\mathrm{ord}(g)$ and let $H=⟨g⟩$, so $H$ is a finite cyclic group of order $m$. Restrict $\rho$ to $H$. Since $\mathrm{char}k=0$, the integer $m$ is nonzero in $k$, so [L1] makes the restricted representation completely reducible. Also [L3] and [L2] show that $k$ is a splitting field for the finite group $H$, and $H$ is abelian. Therefore [L4] makes every irreducible $H$-summand one-dimensional.

Choose a basis of each one-dimensional $H$-summand and concatenate these bases. Each summand is invariant under $\rho (g)$ and one-dimensional, so $\rho (g)$ acts on it by a scalar. Hence the matrix of $\rho (g)$ in the concatenated basis is diagonal. Therefore $\rho (g)$ is diagonalisable over $k$.