GROUP RINGS

E. L. Lady

Let G be a not-necessarily-abelian group and M a module over a commutative ring R. We say that G acts on M if there is a function G × M → M (which we write as multiplication) such that 1m = m (where 1 here denotes the identity element of G), and for r ∈ R, g ∈ G, g(rm) = r(gm), and g_{1}(g_{2}m) = (g_{1}g_{2})m. (note: The use of the symbol 1 to denote both the identity in R and the identity element of G is an inconsistency which turns out not to create problems.)

We define the group ring R[G] as follows: As an R-module, R[G] is the free R-module on the basis {g | g ∈ G}. Thus every element of R[G] is uniquely represented in the form ∑_{g∈G} r_{g} g with, as usual, almost all r_{g} trivial. The group operation in G then gives a product defined on all pairs of basis elements. It is easy to see that this extends uniquely to an associative and distributive multiplication on R[G] that makes R[G] into an R-algebra. Note that the identity element in G is also the identity of R[G]. In fact, we can identify R as a subring of R[G] in such a way that all possible identity elements coincide.

Group rings have been the object of extensive study. The classical text is Donald Passman, Infinite Group Rings (1971). (See also, D. Passman, What is a group ring? Amer. Math. Monthly 83 (1976), 173–84 and Passman, The Algebraic Structure of Group Rings (1977). )

One of the main points to notice is that an R-module M together with a group action of G on M is both notationally and conceptually essentially the same thing as an R[G]-module.

There is no requirement here that G be finite. However for the case of finite groups, there is a smoothing operation on morphisms which is extremely useful.

Lemma. If G is a finite group and |G| is invertible in R, then for R[G]-modules M and N there is an operator that takes R-linear maps φ ∈ Hom_{R}(M, N) to R[G]-linear maps φ̃ ∈ Hom_{R[G]}(M, N) such that the following are true:

(1) If φ is R[G]-linear then φ̃ = φ.

(2) If φ ∈ Hom_{R}(M, N) and ψ ∈ Hom_{R[G]}(N, P) then ψ̃φ = ψφ̃. proof: For φ ∈ Hom_{R}(M, N) and m ∈ M, define φ̃(m) = \frac{1}{|G|} ∑_{g∈G} g^{−1}φ(gm). To see that φ̃ is R[G]-linear it suffices to note that for g_{1} ∈ G,

φ̃(g_{1}m) = \frac{1}{|G|} ∑_{G} g^{−1}φ(gg_{1}m)

= g_{1} \frac{1}{|G|} ∑(gg_{1})^{−1}φ(gg_{1}m)

= g_{1}φ̃(m),

since multiplication by g_{1} permutes the elements of G, so that the family of elements {gg_{1}}_{g∈G} is simply G.

Now if φ is R[G]-linear, then

φ̃(m) = \frac{1}{|G|} ∑_{G} g^{−1}φ(gm)

= \frac{1}{|G|} ∑_{G} g^{−1}gφ(m)

= \frac{1}{|G|} ∑_{G} φ(m) = φ(m).

Also note that if φ is R-linear and ψ is R[G]-linear then

ψ̃φ(m) = \frac{1}{|G|} ∑_{G} g^{−1}ψφ(gm)

= ψ (\frac{1}{|G|} ∑_{G} g^{−1}φ(gm))

= ψφ̃(m).  ✓

Proposition. If |G| is invertible in R and P is an R[G]-module which is projective as an R-module, then P is a projective R[G]-module. proof: We will show that if M is an R[G]-module and ψ: M → P is a R[G]-linear epimorphism then ψ splits. Since ψ is R-linear and P is a projective R-module, there exists φ ∈ Hom_{R}(P, M) such that ψφ = 1_{P} . Then by the previous Lemma there exists an R[G]-linear map φ̃: P → M such that ψφ̃ = ψ̃φ = ˜1_{P} = 1_{P} . Therefore ψ is a split epimorphism of R[G]-modules. ✓ Maschke’s Theorem. If K is a field whose characteristic is zero or relatively prime to |G|, then K[G] is a semi-simple ring. proof: It suffices to prove that every K[G]-module P is projective. But P is a projective K-module, since K is a field. Thus the result follows from the preceding proposition. ✓

Remark. If G is a finite group and K is a field whose (non-zero) characteristic divides |G|, then K[G] is never semi-simple. proof: Let z = ∑_{G} g. Note that for all g′ ∈ G, g′z = z = zg′ (why?). It follows that z is in the center of K[G]. It also follows that z^{2} = |G| z = 0 since |G| is a multiple of char K. Thus for all r ∈ R, (1 + rz)(1 − rz) = 1 − r^{2}z^{2} = 1, so 1 − rz is left invertible. It follows that z is in the Jacobson radical of K[G]. Since z ≠ 0, thus K[G] is not semi-simple. ✓

For g ∈ G, the conjugacy class of g is {h^{−1}gh | h ∈ G}. If G is infinite, then a conjugacy class may be either finite or infinite. If C is a finite conjugacy class, then it is easily seen that ∑_{C} g is an element of the center of R[G].

HW Proposition. The center of R[G] is free as an R-module with a basis consisting of those elements ∑_{C} g, where C ranges over the finite conjugacy classes of G.

For any group algebra R[G] there exists a unique R-algebra morphism ε: R[G] → R such that ε(g) = 1 for all g ∈ G. This is called the augmentation map. Ker ε is called the augmentation ideal of R[G].

HW Proposition. (1) The augmentation ideal is the ideal of R[G] generated by all

elements g − 1 for g ∈ G.

(2) Let I be the augmentation ideal of R[G]. For g_{1}, g_{2} ∈ G, g_{1}g_{2} − 1 ≡ (g_{1} − 1) + (g_{2} − 1) (mod I^{2}).

(3) Let [G, G] be the commutator subgroup of G, i. e. the subgroup generated by all elements g_{1}g_{2}g_{1}^{−1}g_{2}^{−1} and let I be the augmentation ideal of ℤ[G]. Then G/[G, G] ≈ I/I^{2} .

(4) If G and G′ are (not necessarily finite) abelian groups and ℤ[G] ≈ ℤ[G′] then G ≈ G′.