A group ring is an algebraic structure that combines concepts from group theory and ring theory. It is constructed by "attaching" a ring to a group, allowing you to study the group’s structure using ring-like operations.
Definition8.39.
Let \(G = (G, \cdot)\) be a group and let \(R\) be a commutative ring with \(1\neq 0\) (often \(R\) is taken to be a field). Let \(R[G]\) be the collection of formal expressions of the form indicated below, where the \(+\) operation is assumed to be commutative:
Equivalently, a typical element of \(R[G]\) can be written as \(\sum_{g \in G} r_g g\text{,}\) where \(r_g \in R\) for all \(g\) and \(r_g = 0\) for all but a finite number of \(g\)’s.
With these definitions, \(R[G]\) is a ring, called the group ring of \(G\) with coefficients in \(R\text{.}\) It is a unital ring with identity \(1_Re_G\text{.}\) If \(R=F\) is a field then \(F[G]\) is an \(F\)-vector space.
Remark8.40.
Some authors write \(R[G]\) as \(RG\text{,}\) but the notation \(R[G]\) is more standard.
Proposition8.41.
Let \(R\) be any commutative ring and \(G\) a group. Then \(R[G]\) is commutative if and only if \(G\) is abelian.
Example8.42.
Let \(Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}\) denote the group of quaterinons and \({\mathbb{R}}\) the field of real numbers, and let us consider the group ring \({\mathbb{R}}[Q_8]\text{.}\) Actually, the notation here is not so good since \((-1) k\) is easily confused with \(1 (-k)\text{,}\) and, even worse, things like \(1 \cdot 1\text{,}\)\(1 \cdot (-1)\) are highly confusing. So, let us rename the elements of \(Q\text{,}\) so that
so that \(e\) is what we were writing as \(1\text{,}\)\(e'\) is what we were writing as \(-1\text{,}\)\(i'\) is what we were writing as \(-i\text{,}\) etc. So, for example, we now have \(i^2 = e'\) in this group.
\({\mathbb{R}}[Q_8]\) is a non-commutative ring, and you might guess that it is the same as the quaterinons \(\mathbb{H}\) defined above, but it can’t be: \({\mathbb{R}}[Q_8]\) is \(8\)-dimensional as a \({\mathbb{R}}\)-vector space whereas \(\mathbb{H}\) is \(4\) dimensional. In fact \({\mathbb{R}}[Q_8]\) is not a division ring, since it has zero divisors: \((e - i)(e + i + i^2 + i^3) =0\) and so neither of the two factors can be units.
The problem is that \((-1) e \in R[Q_8]\) is not the same thing as \(1 e'\in R[Q]\text{,}\) but we want them to be the same in \(\mathbb{H}\text{.}\) Once we learn about quotient rings, we will be able to show that \(\mathbb{H}\) is the quotient of \(R[Q_8]\) by the ideal generated by \(e' + e\text{.}\) Roughly this means we mod out by the relation \(e' \sim -e\) and all consequences of this relation. For example, once one imposes this equivalence relation, the element
\begin{equation*}
e + i + i^2 + i^3 = e + i + e' + e' i = (e + e') + i(e + e')
\end{equation*}
becomes the zero element.
Proposition8.43.
For any commutative ring \(R\text{,}\) the inclusion \(i:G \hookrightarrow R[G]\) given by \(i(g)= 1_Rg\) lands in \(R[G]^\times\) and induces a homomorphism of abelian groups \(G \to R[G]^{\times}\text{.}\)
Note that \((1_R g)(1_R h) = 1_R (gh)\) by definition of multiplication in \(R[G]\text{.}\)
Lemma 2.18. If \(F\) is a field, \(F[G]\) is an \(F\)-vector space and \(G\) is a basis, so that \(\operatorname{dim}_F(F[G]) = \#G\text{.}\)
Example 2.5. Take \(G = S_3\) and \(R = {\mathbb{R}}\text{.}\) Then \({\mathbb{R}}[S_3]\) is a six dimensional real vector space with basis \(\{e, (1 \, 2), (1 \, 3), (2 \, 3), (1 \, 2 \, 3), (1 \, 3 \, 2) \}\text{.}\) An element is any expression of the form
where \(r_1, \dots, r_6\) are real numbers. This ring has some zero divisors — for example
\begin{equation*}
(e - (1 \, 2))(e + (1 \, 2)) = e - (1 \, 2) + (1 \, 2) - (1 \, 2)^2 = e - e = 0.
\end{equation*}
Here we are abusing notation a bit — for example, \(- (1 \, 2)\) is really \((-1_R) (1 \, 2)\text{.}\) In general, \(1_R g\) is just written as \(g\) in \(R[G]\) and \((-r)g\) is just written as \(- rg\text{,}\) since \((-r)g\) is the additive inverse of \(rg\text{.}\)
Exercise 2.19.
We identify \(G\) as a subset of \(R[G]\) in the obvious way (by identifying \(1_R g\) with \(g\)).
thus \(i(g)\) is a unit in \(R[G]\) with inverse \(i(g^{-1})\text{.}\) This shows that \(\operatorname{Im}(i)\subseteq R[G]^\times\text{.}\)
The formula \((1_R g)(1_R h) = 1_R (gh)\) also gives that the map \(g\mapsto 1_Rg\) is group homomorphism. ◻
Question8.44.Motivation: The What, Where, and Why of Groups.
Why are group rings important? Where are they found? What are they used for?
Answer.
(Co)homology.
To compute group cohomology, we often work with projective resolutions of \(G\)-modules, which are constructed from the group ring \(Z[G]\text{.}\)
Representation Theory.
The group ring \(R[G]\) acts on vector spaces, allowing one to study how groups can be represented by linear transformations. This is fundamental for studying symmetries and representations of finite and infinite groups. A representation of \(G\) on a vector space \(V\) can be viewed as a module over the group ring \(R[G]\text{,}\) which means that \(V\) becomes a left \(R[G]\)-module, and the action of \(g∈G\) on a vector in \(V\) corresponds to multiplication by \(g\) as an element of \(R[G]\text{.}\)
Module Theory.
Studying modules over group rings helps in understanding structures such as projective modules, simple modules, and injective modules.
Algebraic Number Theory.
Group rings are used in the study of class field theory and Galois groups. They are also employed in the study of cyclotomic fields and class numbers, where group rings like \(Z[G]\) arise naturally.
K-Theory.
In algebraic \(K\)-theory, the classification of projective modules over a ring \(R\) is central. For a group ring \(R[G]\text{,}\) projective modules over \(R[G]\) correspond to certain representations of \(G\text{.}\) The \(K\)-theory groups \(K_n(R[G])\) (for \(n≥0n\)) provide information about the structure of these modules. In particular, \(K_0(R[G])\) classifies finitely generated projective modules up to isomorphism, which has connections to representations of \(G\text{.}\)
SubsectionWhat is a Polynomial Ring?
Example8.45.
Group rings give lots of cool examples of rings, but we will now just focus on the boring case when \(G\) is a free abelian group (written with multiplicative notation) generated by \(x_1, x_2, \dots, x_n\text{.}\) In this case an element of \(G\) may be written uniquely as \(x_1^{e_1}, \dots, x_n^{e_n}\) for \(e_1, \dots, e_n \in\mathbb{Z}\text{.}\) For any commutative ring \(R\) a typical element of \(R[G]\) is thus
This is a Laurent polynomial in the variables \(x_1, \dots, x_n\) with \(R\)-coefficients. Say \(n = 1\) and let \(x = x_1\text{,}\) so that \(G = \langle x \rangle\text{.}\) Then \(a x^{-3} + b x^{1} + c + dx^5\) with \(a,b,c,d \in R\) is a representative example of an element of \(R[G]\text{.}\) Addition is by combining like powers of \(x\text{.}\) Multiplication is uniquely determined by the fact that it must satisfy the distributive law and \(x^i x^j = x^{i +j}\) for \(i, j \in \mathbb{Z}\text{.}\)
It is clear that from the rules for \(+\) and \(\cdot\) that if we consider those elements with \(e_i \geq 0\) for all \(i\) in a Laurent polynomial ring, we obtain a subring:
Definition8.46.
Let \(G\) be a free abelian groups with generators \(x_1, \dots, x_n\text{.}\) For any commutative ring \(R\text{,}\) the polynomial ring in \(x_1, \dots, x_n\), written \(R[x_1, \dots, x_n]\text{,}\) is the subring of \(R[G]\) consisting of (finite) sums of the form
Example 2.8. If \(n = 1\text{,}\) letting \(x = x_1\text{,}\) then \(R[x]\) consists of all expressions of the form \(\sum_{i=0}^\infty r_i x^i\) with \(r_i = 0\) for all but a finite number of \(i\text{.}\)
Definition 2.22. Let \(R\) be a commutative ring with \(1 \neq 0\) and let \(M\) be a monoid (set endowed with a binary operation that is associative and has an identity). The monoid ring\(R[M]\) is the set of formal expressions
\begin{equation*}
R[m]=\left\{\sum_{m \in M} r_mm \mid r_m \in R, m \in ,M r_m = 0 \text{ for all but a finite number of }m\right\},
\end{equation*}
Definition 2.23. The polynomial ring on \(n\) variables \(x_1,\ldots,x_n\) with coefficients in \(R\) is the monoid ring \(R[x_1,...,x_n] = R(\mathbb{N}^n)\) on the free abelian monoid \(\mathbb{N}^n\) where each \(x_i\) is identified with \((0,\ldots, 0,1,0,\ldots, 0)\) with the 1 in the \(i\)-th position.
Remark 2.24. When \(R = {\mathbb{R}}\text{,}\) one is tempted to think of \({\mathbb{R}}[x]\) as being a subring of the ring \(\mathcal C({\mathbb{R}})\) of continuous, real-valued functions that are defined on all of \({\mathbb{R}}\text{.}\) This is not technically true: elements of \({\mathbb{R}}[x]\) are just formal expressions, not functions. But there is an injective ring homomorphism (see below)
