SEHGAL, SUDARSHANK.1969

CAN. J. MATH.Vol. 21, pp. 410-413

ON THE ISOMORPHISM OF INTEGRAL GROUP RINGS. I

SUDARSHAN K. SEHGAL

Reprinted fromThe Canadian Journal of Mathematics

ON THE ISOMORPHISM OF INTEGRALGROUP RINGS. I

SUDARSHAN K. SEHGAL

1. Introduction. In this note we study the question of automorphismsof the integral group ring Z(G) of a finite group G. We prove that if G isnilpotent of class two, any automorphism of Z(G) is composed of an auto-morphism of G and an inner automorphism by a suitable unit of Q(G), thegroup algebra of G with rational coefficients. In § 3, we prove that if twofinitely generated abelian groups have isomorphic integral group rings, thenthe groups are isomorphic. This is an extension of the classical resul t ofHigman (2) for the case of finite abelian groups. In the last section we give anew proof of the fact that an isomorphism of integral group rings of finitegroups preserves the lattice of normal subgroups. Other proofs are given in(1; 4).

2. Automorphisms of group rings. We shall use the following theoremof Glauberman (a proof can be found in (4 or 6)). All groups in this sectionare finite.

THEOREMI (Glauberman). Let ():Z(G) --+Z(H) be an isomorPhism. Let Kxbe a class sum in G, i.e., sum of distinct conjugates of an element x of G. Then8(Kx) = ::f:: KII for some y E H.

PROPOSITION 1. Let () be an automorPhism of Z(G). Let Ci, I ~ i ~ r, bethe conjugacy classes and K i the corresponding class sums of G. Suppose that8(Ki) = Ki" 1 < i, if < r, and that there exists an automorPhism (J" of G suchthat (T(Ct) = Ct, for all I < i ~ r. Then we can find a unit 'Y E Q(G) such that

8(g) = 'Yg"'Y-l for all g E G.

Proof. Extend (J" and 8 to Q(G) in the natural way. The centre of Q(G)which is generated by the class sums Ki' I ~ i < r, is kept fixed elementwiseby (J"-I().Since Q(G) is semi-simple, we can write

Q(G) = SI EB S2 EB . . . EB SI!

as a direct sum of simple rings St. Let I = el + e2 + . . . + ell where ei E Stare central idempotents. Then

St = eiQ(G) and «(T-l())(Si) = Si' I < i ~ t.

Received November 28, 1967. The author wishes to thank the Canadian National ResearchCouncil for partial support during the time this work was done.

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INTEGRAL GROUP RINGS. I 411

As U-l() keeps the cen tre of S i fixed elemen twise, it acts on S i as an innerautomorphism by some ai E Si; see (5). It follows that u-l() is an innerautomorphism of Q(G) by a = al + a2 + . . . + at. Thus, we have forg E G, that

(U-l() (g) = aga-l, ()(g) = 'Yg'T'Y-l, where'Y = u(a).

Remark. Let H be a normal subgroup of G. Let /1(H) be the kernel of thenatural map Z(G) - Z(GjH). Then /1(H) is the two-sided ideal generatedby the elements (1 - h), where h runs over H.

PROPOSITION2. Let J.Lbe an automorphism of Z (G), where G is a nilPotentgroup of class two. Let Ki' 1:::; i ~ r, be the class sums of G. Suppose thatp.(K i) = K i', 1 ~ i ~ r. Then there exists an automorphism u of G which,when extended to Z(G), satisfies u(Ki) = Ki, for all 1 ~ i ~ r.

Proof. We first observe that a class sum of a nilpotent group of class twois of the form gO, where H = LhEH hand H is a subgroup of the derivedgroup G'. Let J.L(g)= 'Y. Then J.L(Kg) = J.L(gH) = 'YHI, where HI is anothersubgroup. Now 'YHI = Kgl = glH2. We claim that HI = H2. Observe thatjHIIgIH2 = IHII'YHI = 'YHI'HI = glH2.HI and IHlIH2 = H2.HI. There-fore, HI is contained in H2, and by symmetry, HI = H2. Thus, we have thatJ.L(Kg) = 'YHI = glHl. Next, we claim that there exists a g2 E G such that'Y = g2 mod /1(HI)/1(G). Since 'YHI = glHI, we have that

'Y = gl + L (1 - h)t(h)hEHl

= gl + L (1 - h)nh mod /1(Hl)/1(G)h

(where nh E Z)

=gl + 1 - n hnh mod /1(HI)/1(G)h

= gin h-rthmod /1(HI)/1(G).h

Thus, 'Y = g2 mod /1(HI)/1(G), and hence 'Y= g2 mod /1(G')/1(G). It has beenproved independently by Jackson (3) and Whitcomb (6) that given -y asabove, there exists a unique g-yE G such that 'Y= g-ymod /1(G')/1(G) and'Y- g-yis an isomorphism;\: J.L(G)- G. Because of uniqueness, it follows thatg-y = g2. Since g-y= 'Ymod /1(HI)/1(G), we have that g-yHI = 'YHI' Letu(g) = ;\J.L(g). Then u is an automorphism of G and

u(Kg) = AJ.L(Kg) = A('YHl) = ;\('Y)HI = g-yHl = 'YHI = J.L(Kg).

This completes the proof.

N QWwe have the following theorem.

THEOREM 2. Let ()be an automorPhism of Z (G), where G is a nilPotent groupof class two. Then there exists an automorphism A of G and a unit 'Y of Q(G)such that ()(g) = ::I::-ygX'Y-lfor all g E G.

412 SUDARSHAN K. SEHGAL

Proof. Let us write C(a) = Lu au if a = Lu aug is an element of Z(G).Clearly, C(O(g)) = ::l::1 for any element g of G. Normalize 0 by definingJI.(g) = C(O(g)) .O(g). By linear extension, JI. becomes an automorphism ofZ(G) which maps class sums to class sums. The theorem now follows fromthe last two propositions.

3. Group rings of finitely generated abelian groups. Higman proved(in 1940) that the only units of finite order in the group ring Z(G) of a finiteabelian group are ::l::g,g E G. From this, it follows that only isomorphicfinite abelian groups have isomorphic integral group rings. We extend thisresult to finitely generated abelian groups. We first prove the following lemma.

LEMMA1. The only units of finite order in the group ring Z (G) of an arbitraryabelian group G are of the form ::l::t,where t is a torsion element of G.

Proof. We can assume without loss of generality that G is finitely generated.Write G = T X F, where T is torsion and F is free. Further, F =(Xl) X . . . X (XI)' A typical element g of G can be written uniquely asg = t'Xlalx2a2. . . Xlal, where ai E Z and t E T. Define di(g) = jail. Supposethat 'Y = mlgl + m2g2 + . . . + msgs is such that 'Yn = 1. We have to provethat 'Y = ::l::t.Let

We shall use the group

H = (X12nlH) X (X22n2H) X . . . X (XI2nIH)

ni = max dt(gj).l:;;;j:;;;s

which is of finite index in G. Clearly, (-y)n = 1, where 'Yis the image of 'Yinthe projection Z(G) ~ Z(G/H), and by Higman's result, 'Y= ::l::xH,X E G.Since gl, . . . , gs belong to different cosets of H, it must be that 'Y= ::l::gi,where gi E T.

THEOREM 3. Suppose that G and H are finitely generated abelian groups. ThenZ (G) ~ Z (H) imPlies G "" H.

Proof. Write G = T X (Xl) X . . . X (Xl) and H = Tl X (Yl) X . . . X (Ym),as in the lemma. Let 0: Z(G) ~Z(H) be the given isomorphism. For t E T,set JI.(t) = tl E Tl if Bet) = ::l::tl and set JI.(Xi) = O(Xi)' By linear extensionwe obtain an isomorphism JI.:Z(G) ~ Z(H) such that JI.(T) = Tl' We con-clude that Z(G/T) ro...JZ(H/Tl). Since Z(G/T) is free of zero divisors andthe degree of transcendency over Q of its field of quotients is l, it followstha t 1 = m and that G ~ H.

4. Central idempotents and normal subgroups. In this section, allgroups are finite. Let N be a normal subgroup of G; then N = LnEN n is acentral element with the property that (JV")2= IN!. N. We characterize allN, where N is a normal subgroup of G as certain elements of Z(G) with thisproperty. More precisely, we prove the following proposition.

INTEGRAL GROUP RINGS. I 413

PROPOSITION3. Let l' = LUEG'Yugbe a central element of Z(G) such thatl' e = 1, L 'Yu ~ 0 and 1'2 = m'Y,where m is a natural number. Then l' = L hEH h,where H is a normal subgroup of G.

Proof. For a typical element a = LUEG aug of Z(G), write a* = LUEG aug-l.I t is clear that (a + (3)* = a* + {3*and (a{3)*= {3*a*,and thus we havethat (-y*)2 = m'Y* and (1'1'*)2 = m2'Y'Y*. We also remark that Lu 'Yu = m.By the eigenvalues (trace) of an element a of Z(G) we mean the eigenvalues(trace) of R(a), where a --* R(a) is the regular representation of Z(G).Clearly, l' is diagonaIizable. The only possible eigenvalues of l' are 0 and m.Since 'Ye = 1, we have that trace('Y) = (G:l), and therefore exactly(G: 1) - (G: 1)/m eigenvalues of l' are zero. Thus, at least (G: 1) - (G: 1)/meigenvalues of 1'1'* are zero, and therefore

(" 2) 2 (G:l)(G:l) .L.: 'Yu = trace('Y'Y*) ~ m .--:;;:;-.We conclude that (Lu 'Yu2) ~ m = (Lu 'Yu), and therefore 'Yu = 0 or 1. Since1'2 = m'Y, it follows that l' = LhEH h, where H is a subgroup. Since l' iscentral, H is normal and the proof is complete.

THEOREM 4. Let (J:Z(G) --* Z(H) be a normalized isomorPhism (i.e.,C«(J(g» = 1 for all g E G). Then there exists a one-lo-one correspondencebetween the normal subgroups of G and H which preserves order, union, andintersection.

Proof. Let N be a normal subgroup of G. Then 9 = LnEN n satisfies(9)2 = INI9. Let (J(9) = l' = Lu 'Yug; then 1'2 = INhand C('Y) ~ O. Also,'Ye = 1, due to the fact that for a unit of finite order {3= L {3ug, {3e~ 0implies {3= ::l::e (see (1) for the proof); therefore, {3e= 0 whenever {3= (J(x)for some x ~ e in G. Thus, by the last proposition, we have that l' = if,where M is a normal subgroup of Hand IMI = INI. The remainder of theresult follows from the fact that for two normal subgroups N1 and N2 of G,91.92 = mR2 if and only if N1 is a subgroup of N2.

REFERENCES

1. J. A. Cohn and D. Livingstone, On the structure of group algebras. J, Can. J. Math. 17(1965), 583-593.

2. G. Higman, The units of group rings, Proc. London Math. Soc. 46 (1940), 231-248.3. D. A. Jackson, Ph.D. Thesis, Oxford University, Oxford, 1967.4. D. S. Passman, IsomorPhic groups and group rings, Pacific J. Math. 15 (1965), 561-583.5. B. L. van der Waerden, Modern algebra (Ungar, New York, 1950).6. A. Whitcomb, Ph.D. Thesis, University of Chicago, Chicago, Illinois, 1967.

University of Alberta,Edmonton, Alberta;Ohio State University,Columbus, Ohio