Az Eszterházy Károly Tanárképző Főiskola Tudományos Közleményei. 1997. Sectio Mathematicae. (Acta Academiae Paedagogicae Agriensis : Nova series ; Tom. 24)

SZAKÁCS, A., Unitary subgroup of the Sylow 2-subgroup of the group of normalized units in an infinite commutative group ring

Unitary subgroup of the Sylow 2-subgroup of the group of. 91 Therefore x

2 = a

2 and the statement is proved for the cyclic group G of order 4. Let G be a non-cyclic group of exponent 4 and order greater than 4. Then G can be presented as a direct product of a suitable group H and the cyclic group (b

) which order divides 4. Suppose that b is an element of second order. Then every x £ V(KG) can be written in the form x = £o + x\b, where XQ,X\ £ KH. If x is a unitary element then XX* = XQXQ* + X\X\* + (xo*Xi + X

0 Xi*)b = 1 and the equations xqXq* + xixi* = 1, x

Q*xi -f Zo^i* — 0 hold. Hence (reo + £i)(zo* + = 1 and y = x

0 + xi £ V+(KH). By the induction hypothesis, y

2 = h

2 for some h £ H

. Obviously x

2 = h

2

. Let 6 be an element of order 4. The element x = x

0 + x

2b

2 + (xi + x

3b

2)b (xi £ KH, i = 0,1,2,3) of the group V(KG) is unitary if and only if , , f (x

0 + x

2b

2)(x

0* + x

2*b

2) + (xi + x

3b

2)(x^ + x

3*b

2) = 1, 1 ) \(x

0+-x

2b

2){

X

l* + x

3*b

2) = 0. Let x{

xo + x

2b

2) = 7 denote the sum of coefficients of the element XQ -f x

2b

2

. Then \(

xi + x

3b

2) = 1 + 7 and from the second equation of (1) we have that 7

(1 + 7) = 0. Since K without zero divisors, it follows that 7 = 0 or 7 = 1 i.e. one of the elements XQ + x

2b

2 or x\ + x

3b

2 is invertible. Hence for the unitary element x either XQ = x

2b

2 or X\ = x

3b

2

. If Xo = x

2 b

2 then, by (1), the element y = x\ + x

3b

2 is unitary in the group ring of the group H = H x (b

2). Then, by the induction hypothesis, y

2 = h

2 for some h £ H and obviously x

2 = y

2b

2 — h

2b

2 £ G

2

. If x\ = x

3b

2 then y = +x

2b

2 £ V*(KH) and x

2 = y

2 £ G

2

. So (V*(KG))

2 = G

2 for a finite group G of exponent 4. Suppose that G is a group of exponent 2

n+

1 (n > 1) and the statement is proved for the groups of exponent less than 2

n+

1

. It is easy to see that (V+(KG))

2 C V*(KG

2). From this, useing the induction hypothesis (V^KG

2))

27

1'

1 = (G

2f

n~\ we have that V^KGf

n C G

2"

. The reverse inclusion is obvious and the lemma is proved for a finite group G. Let G be an infinite abelian group of exponent 2

n+

1 (n > 0) and x £ V*(KG). Then the subgroup H = (supp x) of the support of x is finite and, by the statement proved in above, x

2" £ ii

2"

. This completes the proof of the lemma.