WHAT IS A GROUP RING? D. S. PASSMAN 1. Introduction.Let K be a field. Supposewe are given some three element set {a, (, y} and we are asked to form a K-vector space V with this set as a basis. Then certainlywe merely take V to be the collection of all formal sums a ca+ b ,(+ c y with a, b, c E K. In the same way if we were originallygiven four, five or six element sets we would again have no difficultyin performingthis construction.After awhile, however, as the sets get larger,the plus sign becomes tedious and at that point we would introducethe E notation. In general,if we are given a finite set S, then the K-vector space V with basis S consistsof all formalsums ),2s aa *a with coefficientsa,a E K. Finally,there is no real difficultyin letting S become infinite. We merely restrictthe sums Y2a ca to be finite, by which we mean that only finitely many nonzero coefficientsaa, can occur. Of course, addition in V is given by (E a, ))+ (nba a) = E(aa+ba)b a and scalar multiplicationis just b( , a, a) (ba,,) cta. Moreover,by identifying 3 E S with the element ,B'= ab, a E V,where b, = 1 and b",= 0 for a , we see that V does indeed have this copy of S as a basis and our original problem is solved. Now, how do we multiplyelements of V? Certainlythe coefficient-by-coefficientmultiplication a, .a) 0 & . a)=(a,) a is exceedinglyuninteresting and other than that no naturalchoice seems to exist. So we are stuck. But suppose finallythat we are told that S is not just any set, but ratherthat S is in fact a multiplicative group. We would then have a natural multiplicationfor the basis elements and by way of the distributivelaw this could then be extended to all of V. Let us now start again.Let K be a field and let G be a multiplicativegroup, not necessarilyfinite. Then the group ring K[G] is a K-vector space with basis G and with multiplicationdefined distributivelyusing the given multiplicationof G. In other words, for the latter we have a. x E (axby) (xy)- Ecz z, xExeG (Ebyy EG y) x,yEG zeG where CZ= axby xbx-lz. xy=z xEG Certainlythe associativelaw in G guaranteesthe associativityof multiplicationin K[G] so K[G] is a ring and in fact a K-algebra. It is clear that these easily defined group rings offer rather attractive objects of study. Furthermore,as the name implies, this study is a meeting place for two essentially differentdisciplines and indeed the resultsare frequentlya rathernice blendingof group theory and ring theory. Suppose for a moment that G is finite so that K[G] is a finite dimensionalK-algebra. Since the study of finite dimensionalK-algebras (especially semisimple ones over algebraicallyclosed fields) is in far better shape than the study of finitegroups, the groupring K[G] has historicallybeen used as a tool of grouptheory. This is of coursewhat the ordinaryand modularcharacter theory is all about (see [21for example).On the other hand,if G is infinitethen neitherthe grouptheory nor the ringtheory is particularlyadvanced and what becomes interestinghere is the interplaybetween the two. Our main concernin this articlewill be with infinitegroups and in the followingwe shall discussand prove a few 173 174 D. S. PASSMAN [March selected results. These results, however, are by no means representativeof the entire scope of the subject,but ratherwere chosen becausetheir proofs are both elegant and elementary.Few references will be given here, but the interested reader is invited to consult the book [5] or the more recent surveys, [4] and [6]. 2. Zero divisors.Before we go any further,two observationsare in order.First, once we identify the elements of G with a basis of K[G], then the formalsums and productsin , ax x are actually ordinarysums and products.In particular,we shall drop the dot in this notation. Second, if H is a subgroupof G then, since H is a subset of the basis G, its K-linear span is clearlyjust K[H]. Thus K[H] is embedded naturallyin K[G]. Now let x be a nonidentityelement of G so that G contains the cyclic group (x ). Then K[G] contains K[(x)] and we briefly consider the latter group ring. Suppose first that x has finite order n > 1. Then 1, x,_ , xn-' are distinct powers of x and the equation (1 -X) (1 + x + * + Xn1) = 1-Xn = O shows that K[(x)], and hence K[G], has a properdivisor of zero. On the other hand, if x has infinite order then all powersof x are distinctand K[(x)] consistsof all finite sums of the form E aix. Thus this group ring looks somethinglike the polynomialring K[x] and indeed every element of K[(x)] is just a polynomialin x divided by some sufficientlyhigh power of x. Thus K[(x)] is containedin the rational function field K(x) and is therefore an integral domain. Now what we have shown above is the following.If G has a nonidentityelement of finite order, a torsion element, then K[G] has a nontrivialdivisor of zero, but if G has no nonidentityelement of finite order, then there are at least no obvious zero divisors.Because of this, and with franklyvery little additionalsupporting evidence, it was conjecturedthat G is torsionfree if and only if K[G] has no zero divisors. Remarkablythis conjecture has held up for over twenty-fiveyears. We still know very little about this problem.In fact all we know is that the conjectureis true for some rathersimple classes of groupsas, for example,free groups,or abeliangroups. In the lattercase the proof is even quite easy. Thus, suppose that G is torsionfree abelian,and let a, P be elements of K[G] with a/3 = 0. Then, clearly, a and ,B belong to K[H] for some finitely generated subgroup H C G and, by the fundamentaltheorem of abeliangroups, H is just the directproduct of the infinite cyclic groups (x,), (x2), - and (x,). It is then quite easy to see that K[H] is contained naturally between the polynomialring K[xX,x **, x,] and the rationalfunction field K(x1, x2, ** , x,). Indeed, K[H] is just the set of all elements in K(x1, X2,* * , x,) which can be written as a polynomial in xI, x2, , x, divided by some sufficientlyhigh power of (xI x2... x,). Thus K[H] is certainly an integral domain and hence, clearly, either a = 0 or / = 0. Actually the best result to date on this conjectureconcerns supersolvablegroups. Here for the first time nontrivialring theory comes into play but the proof is unfortunatelytoo complicatedto give here. On a more positive note, there is a variant of the zero divisor problem which we can handle effectively.A ring R is said to be prime if aR/ = 0 for a, : E R impliesthat a = 0 or : = 0. Clearly this agrees with the usual definitionin the commutativecase. In addition,this concept makes more sense arddis more importantfrom a ring-theoreticpoint of view than the very much more stringent condition of no zero divisors.For example, the matrixring Knis alwaysprime, even though it does have zero divisors for n -' 2. The main result of interest here is as follows. THEOREM I. The group ring K[G] is prime if and only if G has no nonidentityfinite normal subgroup. The proof in one directionis quite easy. Suppose H is a nonidentityfinite normalsubgroup of G and set a = Elh EH h, the sum of the finitelymany elements of H. If h E H then hH = H so ha = a and thus a 2 ha=IHIa. hEH 1976] WHAT IS A GROUP RING? 175 In particular,if : = IH I1 - a then afp = 0. Furthermore,since H is normal in G, we have for all x E G, Hx = xH, and hence ax = xa. Thereforea commuteswith a basis of K[G], so it is centraland we have aK[G]f = K[G]af = 0. Finally, since a and,, are clearly not zero we see that K[G] is not prime. The converse directionis much more difficultand as a starterwe must ask ourselveshow we are going to find a finite normalsubgroup in G. Observe that if H is such a subgroupand if h E H, then certainly h has finite order, and furthermore,for all x E G we have hX = x-1hx E x-1Hx = H. Thus all G-conjugatesof h are containedin H and so certainlythere are only finitelymany distinct ones. This, therefore, leads us to define two interestingsubsets of G, namely, A(G) = {x E 0G x has only finitely many conjugates in G} and Al(G) = {x E G x has only finitely many conjugates in G and x has finite order}. At this point a certainamount of grouptheory obviously comes into play. While the proofs are by no means difficultit does seem inappropriateto offer them here. Therefore we just tabulate the necessary facts below. See [5, Lemma 19.3] for details. LEMMA 1. Let G be a groupand let A(G) and Al(G) be definedas above. Then (i) A(G) and Al(G) are normalsubgroups of G. (ii) A(G) : AI(G) and the quotientgroup A(G)IA+(G) is torsionfree abelian. (iii) I+(0) # (1) if and only if G has a nonidentityfinite normalsubgroup. Part(iii) above certainlyseems to answerour firstquestion. Now observethat if x is any element of G, then the G-conjugatesof x are in a naturalone-to-one correspondencewith the right cosets of CG(x), the centralizer of x.