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
96 Attila Szakács Since the elements a and c belong to the distinct cosets of the group G n by the subgroup (g,v), it follows that av E G n+1 and avg 6 G n+\, but this contradicts to the choice of the element g E G n \ G n.f-i . So the case B ) is fully considered. Let C) holds, that is \G n\ > \K n\ and the Sylow 2-subgroup S n of the group G n is 2-divisible. Let us fix an element g £ 5 n[2] and choose such v E G n \ G n+i that 2 does not divide the order of element v. Since \S n\ = [S n : {</)] > |{u)| and v ^ S n, it follows that the cardinality of the full set of representatives of the cosets II = H(G n/(g, v)) of the group G n by the subgroup (g,v) coincides with \G n|. Obviously the set II decomposes to the two disjunct subsets IIx = {a E II: a 2 £ (v,g )} and n 2 = {a E II: a 2 E (v,g)}. Let I G n | = | Tlx I, E be the set which has a unique representative in every subset of the form {a, a1 } C Iii and y a = 1 + a(l + v + v1(l + g)Then the set M can be choosen in the following way: M = {x = y ay a* = 1 + (a + a~ l)(l + v + v~ l)(l + g) : a E E) . Indeed, from the equation x a = x cz (^z E V 2 + 1, a ^ c^j follows that z = I + {a + a' 1 + c + c1)^ + v + v~ l )(1 + g) E T/ r + 1 . Hence, according to the construction of the set E, the elements a and av belong to the subgroup G n+1 , but this contradicts to the condition v (£ G n+1 • Suppose now that \G n\ = |n 2|. Then v 2 ^ 1. If a 2 = v 2 for some a E n 2, then from the condition v ^ C n+i it follows that i is an even number. Let us choose in the role of the representative of the coset a(g,v } the element ai = av~ 2. Hence we can assume that the set n 2 of the representatives of the group G n by the subgroup (g,v) consists of the elements of the group Sn = Sn+ 1 • The set M = {x a = I + a(v + v _ 1)(l +g): a E n 2 } has the need property. Indeed, if x a = x cz for the distinct a, c E n 2 and for some z E V p , then 2 = x ax c = 1 + (a + c)(v + v~ l )(1 -f g) and av E G n +\ . Hence v E G n±i because - by the choice - n 2 C S n+\ , and so we get the contradiction. Therefore the case C) is fully considered and the statement is proved for a finite ordinal A = n.
