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

94 Attila Szakács The equation x

a = x^z (^

GVV

/

Í

') is impossible since from it follows that z = x

ax

u = 1 + (o; -f + g) — x

a+i

y, and, by proved in above, x

a

+„ £ V

2

. Obviously \M\ = \K\ and therefore fo(W) = \K\. Let us contruct the set M in case n > 0. Since, by Lemma 1, f

n(W

2) = f

n(G) when G

n+i = 1, it follows that we can assume that G

n+\ / 1. Let \G

n\ / 4. Then the set G

n \ G

n+ \ has neither element a, which order is not divisible by 2, or element b of order 2

r > 4, or has a subgroup (c : c

4 = 1) X (d

: d

2 — 1). Obviously in the first case a

2 $ (g). If in the other cases we put a = b,g = b

2 or a = c, g = d respectively then the condition a

2 (£ (g) holds and we have the above considered case Ai

). Let G

n = (a : a

4 = 1) and y

a = 1 + o(a + 1). Obviously the element Xa = VaVa* = 1 + (a + Q

2)(a + a

3) is unitary. Let L denote a subset of K

n that has a unique representative in every subset of the form {a, 1 -f a} C K

n. Then the elements of the set M = {x

a = y

ay

a* = 1 + (a + a

2)(a -fa

3): O/aGi} belong to the different cosets of the group W

2" [KG)[2] by the subgroup W

2 (KG)[

2]. Really, if x

a coincides with x

v (a,z/ £ L), then a + a

2 = v v

2 . Hence the equation (a + u)(l + a + i/) = 0 holds, but in the ring without zero divisors this is possible for the different a and v only in the case v — 1 + a, what contradicts to the choice of the elements of the set L. Obviously \M\ = \L\ = |A'

n|. By Lemma 1, + 1 = (a

2). If x

aW

2 1 = x

uW

2 +

1 / x

u) we get the contradictinally equation 1 + (a + a

2 )(a -f a

3) = a

2 + (v + + a

3). Therefore / x„W

2n+

1 for x

a ^ x„ the case A

2) is considered. Let A

3) holds, i.e. G

n = (g). Then G

n+

i = 1. If n = 0 then W(KG) = V

2{KG) and fo(W) = fo(V

2) = \K\. If n > 0 then, according to Lemma 1, fn(W) = f

n(G). Therefore the case A) is fully considered. Suppose now that B) holds, i.e. \G

n\ > \K

n\ and the Sylow 2-subgroup S

n of the group G

n does not coincide with the Sylow 2-subgroup of the group G

n+Then the set S

n \ S

n+1 has an element g of order q — 2

r

. Let, further on, II = H(G

n/{g)) denote the full set of representatives of the cosets of the group G

n by the subgroup (g). Let us consider two disjunct subsets III = {a e II

: a

2 £ (g)} and ü

2 = {a G II

: a

2 G {g)}