Group representations give a concrete reason to study module categories and bimodules. A group action by linear maps extends uniquely to a module action over a ring built from the group. A familiar operation on representations, induction, then becomes tensoring with a bimodule.

Conventions: fields, groups, and dimension

Fix a field \(k\), with \(0\ne1\), and a group \(G\), with identity \(e\). Neither \(G\) nor the dimension of a representation is assumed finite. A \(k\)-vector space means a left \(k\)-module. The group \(\mathrm{GL}_k(V)\) consists of invertible \(k\)-linear maps \(V\to V\), with composition as multiplication. A representation is a homomorphism \(\rho:G\to\mathrm{GL}_k(V)\).

7.1. Building the group algebra

Definition 7.1. The group algebra

The vector space \(k[G]=k^{(G)}\) consists of formal sums \(\sum_{g\in G}a_g[g]\) with coefficients in \(k\) and finite support. The symbols \([g]\) are its standard basis. Addition and scalar multiplication act on coefficients. Define multiplication by

\[\left(\sum_g a_g[g]\right)\left(\sum_h b_h[h]\right)=\sum_{g,h}a_gb_h[gh].\]

All sums here are finite; repeated basis symbols are combined by adding their coefficients. The proposed identity is \([e]\).

Alphabeta Math: group rings

Worked example for Definition 7.1

Theorem 7.2. All ring laws

This multiplication makes \(k[G]\) an associative unital ring. The map \(k\to k[G]\), \(a\mapsto a[e]\), is an injective unital ring map whose image is central. Thus \(k[G]\) is a \(k\)-algebra, meaning a ring equipped with such a central scalar map. Each \([g]\) is a unit, with inverse \([g^{-1}]\). Moreover, \(k[G]\) is commutative exactly when \(G\) is abelian.

Alphabeta Math: the algebra laws · commutativity

Worked example for Theorem 7.2

Proof.

The additive group is abelian by the direct-sum construction. If the factors have supports \(A,B\subseteq G\), their product has support inside \(AB=\{gh:g\in A,h\in B\}\), which is finite. For each \(u\in G\), its coefficient is the finite sum \(\sum_{gh=u}a_gb_h\), restricted to those supports. This defines a unique finite-support function, so multiplication is well defined.

For three factors \(x=\sum_g a_g[g]\), \(y=\sum_h b_h[h]\), \(z=\sum_\ell c_\ell[\ell]\), expansion gives

\[(xy)z=\sum_{g,h,\ell}(a_gb_h)c_\ell[(gh)\ell],\qquad x(yz)=\sum_{g,h,\ell}a_g(b_hc_\ell)[g(h\ell)].\]

The group and field associativity laws identify corresponding summands; finiteness permits regrouping, proving associativity. Expanding \(x(y+z)\) and \((x+y)z\) and using the two field distributive laws proves both ring distributive laws. Multiplication by \([e]\) on either side leaves every basis symbol fixed, hence every sum fixed; it is the identity. Also \([g][g^{-1}]=[g^{-1}][g]=[e]\).

The scalar map preserves addition, multiplication, and identities, and is injective because the coefficient at \(e\) of \(a[e]\) is \(a\). For every basis symbol, \((a[e])[g]=a[g]=[g](a[e])\), and bilinearity proves centrality for every sum. If \(G\) is abelian, the product formula is symmetric by commutativity of \(k\). Conversely, if \(k[G]\) is commutative, \([gh]=[g][h]=[h][g]=[hg]\). Distinct basis symbols are distinct functions, so \(gh=hg\) for every \(g,h\).

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If \(G=\{e\}\), the scalar map is an isomorphism \(k\cong k[G]\). The algebra of a nontrivial finite group can have zero divisors even though \(k\) is a field; an explicit instance appears below.

7.2. Representations are exactly modules

Definition 7.3. The representation category

An object of \(\mathrm{Rep}_k(G)\) is a pair \((V,\rho)\), with \(V\) a \(k\)-vector space and \(\rho:G\to\mathrm{GL}_k(V)\) a representation. A morphism \(f:(V,\rho)\to(W,\sigma)\), called an intertwiner, is a \(k\)-linear map satisfying \(f\rho(g)=\sigma(g)f\) for all \(g\in G\). Identity maps and composites satisfy the same equation.

Alphabeta Math: group actions on modules · intertwiners

Worked example for Definition 7.3

Counterexample: An involution need not split into its eigenspaces in characteristic two

Theorem 7.4. The representation–module correspondence

There are inverse functors between \(\mathrm{Rep}_k(G)\) and \(k[G]\text{-}\mathrm{Mod}\). The correspondence preserves the underlying abelian group and identifies intertwiners with module morphisms. It holds for arbitrary groups and arbitrary-dimensional representations.

Alphabeta Math

Worked example for Theorem 7.4

Counterexample: An involution need not split into its eigenspaces in characteristic two

Proof.

From a representation define the action

\[\left(\sum_g a_g[g]\right)v=\sum_g a_g\rho(g)v.\]

This is a finite sum. It is additive in the algebra argument and in \(v\) by linearity of each \(\rho(g)\). For \(x=\sum_g a_g[g]\) and \(y=\sum_h b_h[h]\),

\[x(yv)=\sum_{g,h}a_gb_h\rho(g)\rho(h)v=\sum_{g,h}a_gb_h\rho(gh)v=(xy)v.\]

The equality uses \(k\)-linearity and the homomorphism law. Also \([e]v=\rho(e)v=v\), so this is a unital left module action. Its restriction to \(a[e]\) is the original scalar multiplication. Any extending module action must satisfy this formula by additivity and the action of \(a[e][g]\), proving uniqueness.

Conversely, a \(k[G]\)-module \(M\) becomes a \(k\)-vector space by \(a\cdot m=(a[e])m\). Define \(\rho(g)m=[g]m\). This map is additive, and centrality of \(a[e]\) gives \([g]((a[e])m)=(a[e])([g]m)\), so it is \(k\)-linear. Its inverse is the action of \([g^{-1}]\), and \(\rho(g)\rho(h)=\rho(gh)\), giving a representation. The two constructions restore the original actions: they agree on the scalar map and on each \([g]\), which together determine every finite sum.

An intertwiner commutes with the action of each sum by linearity and the intertwining equation, so is a \(k[G]\)-module map. Conversely, a module map commutes with every \(a[e]\), so is \(k\)-linear, and with every \([g]\), so intertwines. Both functors keep the same underlying functions on morphisms; hence they preserve identities and composites and are inverse.

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The finite support in \(k[G]\), rather than finite dimension of \(V\), makes the formula meaningful. By Theorem 5.6 this also makes \(\mathrm{Rep}_k(G)\) an abelian category: kernels and quotients of intertwining maps carry the induced group actions, as their corresponding submodules and quotient modules do.

7.3. The two-element group

Example 7.5. An action is an involution

Let \(C_2=\{e,t\}\), with \(t^2=e\), and write \(1=[e]\), \(u=[t]\). Every element of \(k[C_2]\) is uniquely \(a+bu\), with \(u^2=1\), and

\[(a+bu)(c+du)=(ac+bd)+(ad+bc)u.\]

A representation is exactly a vector space equipped with a linear map \(T\) satisfying \(T^2=1\): this condition makes \(T\) invertible and supplies all the group relations. The corresponding action is \((a+bu)v=av+bTv\).

The field has characteristic two when \(1+1=0\). Otherwise \(2=1+1\) is nonzero and has an inverse, denoted \(1/2\). In this latter case, every \(v\in V\) has the decomposition

\[v=\frac{v+Tv}{2}+\frac{v-Tv}{2},\qquad T\left(\frac{v+Tv}{2}\right)=\frac{v+Tv}{2},\quad T\left(\frac{v-Tv}{2}\right)=-\frac{v-Tv}{2}.\]

The eigenspaces \(V_+=\{v:Tv=v\}\) and \(V_-=\{v:Tv=-v\}\) are subspaces by linearity of \(T\). They have zero intersection, since \(w=-w\) implies \(2w=0\), so this is a direct-sum decomposition in every dimension. It comes directly from the module action of \((1+u)/2\) and \((1-u)/2\).

In characteristic \(2\), division by \(2\) is impossible. On \(k^2\), let \(N(x,y)=(y,0)\) and \(T=1+N\). Since \(N^2=0\) and \(2N=0\), \(T^2=1\), giving a representation even though \(T\ne1\). In the algebra, \(u-1\ne0\) by independence of the two basis symbols, but \((u-1)^2=u^2-2u+1=0\). This is a concrete nonzero element whose square is zero.

7.4. Induction is tensoring with a bimodule

Proposition 7.6. The induction bimodule

For a subgroup \(H\subseteq G\), extending finite-support functions by zero embeds \(k[H]\) into \(k[G]\) as a unital subring. Thus \(k[G]\) is a \(k[G]\)-\(k[H]\) bimodule by left and right multiplication. Define

\[\operatorname{Ind}_H^G(V)=k[G]\otimes_{k[H]}V.\]

This is a \(G\)-representation by Theorem 7.4. If \(W\) is a \(G\)-representation, let \(\operatorname{Res}_H^G(W)\) denote the same vector space with its action restricted to \(H\). There is a natural bijection

\[\operatorname{Hom}_{k[G]}(\operatorname{Ind}_H^G V,W)\cong\operatorname{Hom}_{k[H]}(V,\operatorname{Res}_H^G W).\]

Alphabeta Math: induction by tensoring · extension and restriction of scalars

Worked example for Proposition 7.6

Proof.

Products of basis symbols from \(H\) stay in \(H\), so the embedding preserves multiplication; it preserves addition and the identity, and is injective by its coefficients. Associativity in \(k[G]\) verifies the bimodule compatibility. The tensor construction now supplies the left \(k[G]\)-action, with \([g](x\otimes v)=([g]x)\otimes v\), and therefore the group action.

Given \(F:k[G]\otimes_{k[H]}V\to W\), define \(f(v)=F(1\otimes v)\). It is additive, and for \(a\in k[H]\), balancing and \(k[G]\)-linearity give \(f(av)=F(a\otimes v)=af(v)\), so it is \(k[H]\)-linear. Conversely, given such \(f\), the map \((x,v)\mapsto xf(v)\) is biadditive and \(k[H]\)-balanced because \((xa)f(v)=x f(av)\). The tensor property gives \(F(x\otimes v)=xf(v)\), which is \(k[G]\)-linear by associativity. The formulas are inverse: one recovers \(f\) at \(1\otimes v\), and every \(k[G]\)-linear \(F\) satisfies \(F(x\otimes v)=xF(1\otimes v)\). They commute with maps of \(V,W\) by these same formulas, proving naturality.

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For \(H=\{e\}\) and \(G=C_2\), induction takes a vector space \(V\) to \(V\oplus V\). Explicitly, \((a+bu)\otimes v\mapsto(av,bv)\) is balanced over \(k\); its inverse sends \((v,w)\) to \(1\otimes v+u\otimes w\). These formulas are inverse by additivity. The action of \(t\) exchanges the two coordinates, since \(u(1\otimes v+u\otimes w)=u\otimes v+1\otimes w\). For \(V=k\), this is the two-dimensional regular representation.

This example realizes the series' functor–bimodule correspondence before its general proof. Chapter 8 now identifies precisely which abelian categories are module categories, and Chapter 9 identifies precisely which functors between them are tensor functors.

Sources and further reading

Worked examples

Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.

Worked example 1. Multiplying in the algebra of a two-element group

Applies to: Definition 7.1, Theorem 7.2.

Let \(k=\mathbb R,G=\{e,s\}\) with \(s^2=e\). Then \((a[e]+b[s])(c[e]+d[s])=(ac+bd)[e]+(ad+bc)[s]\). For instance \(([e]+2[s])(3[e]-[s])=[e]+5[s]\). This is the ring \(\mathbb R[t]/(t^2-1)\) under \([s]\mapsto[t]\): every residue has a unique linear representative, so the map is bijective and preserves multiplication. Hence associativity, distributivity, and the unit follow directly from the polynomial quotient. Scalars \(a[e]\) are central and injectively embedded, and \([s]^{-1}=[s]\). This algebra is commutative, agreeing with the abelian group hypothesis. In contrast in \(\mathbb R[S_3]\), \([(12)][(23)]\ne[(23)][(12)]\), since the two products are different basis elements.

Worked example 2. An involution and its intertwiners

Applies to: Definition 7.3, Theorem 7.4.

For the same \(G\), let \(V=\mathbb R^2\) and \(\rho(s)=\operatorname{diag}(1,-1)\). Since \(\rho(s)^2=I\), this is a representation. The associated module action is \((a[e]+b[s])(x,y)=((a+b)x,(a-b)y)\); multiplying two such actions gives exactly the product computed above. Conversely the action of \([s]\) recovers \(\rho(s)\). If \(f=\begin{pmatrix}p&q\\r&t\end{pmatrix}\), then \(f\rho(s)=\rho(s)f\) forces \(q=r=0\), by comparing off-diagonal entries over \(\mathbb R\). Thus the intertwiners are precisely diagonal maps. These are also exactly the maps preserving the displayed module action, checking the correspondence on morphisms as well as objects.

Worked example 3. Inducing from the trivial subgroup

Applies to: Proposition 7.6.

Take \(H=\{e\}\subset C_2\), \(V=\mathbb R\). The subgroup algebra is the scalar subring \(\mathbb R[e]\), so \(\mathbb R[C_2]\otimes_{\mathbb R}\mathbb R\cong\mathbb R[C_2]\), with basis \([e]\otimes1,[s]\otimes1\); \(s\) exchanges them. For any representation \(W\), a linear map \(V\to W\) is determined by \(w\), the image of 1. Its induced module map is \([g]\otimes a\mapsto a\rho_W(g)w\). Conversely a module map is determined by the image of \([e]\otimes1\), since \([s]\otimes1=s([e]\otimes1)\). These inverse formulas prove the claimed natural Hom bijection in this concrete induction example.

Counterexamples and limits of the hypotheses

Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.

Counterexample 1. An involution need not split into its eigenspaces in characteristic two

Related statements: Definition 7.3, Theorem 7.4.

Claim refuted. Every representation of a finite group is a direct sum of invariant lines, or every involution is diagonalizable over its field.

Proof.

Let \(k=\mathbb F_2\), \(G=C_2=\{e,s\}\), and \(V=k^2\). Set \(\rho(s)=\begin{pmatrix}1&1\\0&1\end{pmatrix}=I+N\), with \(N\ne0,N^2=0\). In characteristic two, \((I+N)^2=I+2N+N^2=I\), so this is a genuine representation, and the chapter’s correspondence makes it a \(k[G]\)-module. Its fixed vectors solve \(N(x,y)=(y,0)=0\), giving only the line \(ke_1\). Every invariant line would have scalar action \(\lambda\) satisfying \(\lambda^2=1\), whose only root in \(\mathbb F_2\) is 1. Thus every invariant line lies in \(ke_1\), and no eigenbasis exists.

More concretely a complement to \(ke_1\) would be spanned by \(e_2+ae_1\). Its image is \(e_2+(a+1)e_1\), a different vector and not a scalar multiple over \(\mathbb F_2\). Hence the exact sequence with submodule \(ke_1\) and quotient \(V/ke_1\) does not split as representations. The representation–module correspondence remains valid; semisimplicity is an extra assertion requiring extra hypotheses.