Vol. 5 No 10 - Pi Mu Epsilon

J f for each x E G and each positive integer n, there exists y E G wchthat x = ny.ble subgroups.A reduced group is me which contains no (nonzero) divisi-(Clearly the element 0 is divisible by every positiveinteger n, since 0 = no.) Now let G be a p-group for the prime p.each ordinal a, we define inductively a subgroup paG as follows:pG = {px 1 x E GI; pY+G = p(pYG); if y is a limit ordinal pYG = n pG.KYWe thus have a descending chain of subgroupsG =pG 2 zPaG =p% 2 .Note in particular if n is a positive integer and x F pnG\pnlG then xis divisible by pn in G but x 1s not divisible by pn+l in G.If x isdivisible by pn for every positive integer n, then x E puff where u isthe first infinite ordinal and x is said to have infinite height in G.In general, the element x is said to have p-height a in G, and we seth (x) = a, if x E pa~\spa+l~. If x E paG for each ordinal a then h (x) = =GGwhere - > a for each ordinal a. Since 0 e paG for each ordinal a,hG(d =The following fundamental properties are easily established.G be a p-grcup with x,y E G.If h(x) # hG(y) then hG(x + y) is thesmaller of the two heights. If h (x) = hG(y) then hG(x + 9) S hG(x).GA homomorphism cannot decrease height; that is, if f is a homomorphisma+1from G into G', then hG(x) 5 hG,(f(x)). If paG = p G, then pBG = paGfor 6 > a and paG is a divisible group.there must be an ordinal y such that pYG = { 01.If G is a reduced p-group,y is called the length of G and is denoted by \(G) = y.ForLetThe least such ordinalA bounded groupis a torsion group for which there exists a finite upper bound to theset of orders of all elements. That is, there exists a positive integern such that nx = 0 for every x in the group.Alternatively, G is abounded group if there exists a positive integer n such that nG = 0.A subgroup A of the group G Is a niee subgroup of G if and only iffor all a. We note that the subgroup A of G is nice if and only if foreach x E G there exists some a E A such that hGIA(x + A) = hG(x + a).The following characterization, given by H i l l [s], will be utilized:Theohan A:The subgroup A of a p-group G is a nice subgroup of Gif and only if each coset x + A contains an element x + a that hasp-height in G which is maximal among the elements of the coset x + A.An immediate consequence of the above theorem is the fact that anyfinite subgroup is nice.Theorem 1:Let G be a reduced abelian p-group and n a positive in-teger. Then G[pn] is nice if and only if \(GI Â u.Proof:Suppose n is an arbitrary but fixed positive integer, G[yn]is a nice subgroup of G, and that A(G) > ID. Consider the descendingchain of subgroupsG 3 pG 2.. .3 pmff 3.. .3 pWG 3 pu+'$ 3.. . .u+1Since A(i7) > u, puff # { 01 and there exists x e p^G - p G, x # 3. Notethe height of x is u. Thus for the positive integer n there exists annelement y in G such that x = p y. Moreover, for each positive integer mx E pn+m~ = pn(pmG)and hence there exists xnE pmG Such that X = p xm. NOWmnp ( x - y) = pnxm - pny = x - x = 0so x - y E G[pn] and thusmx + ~[p"] = y + G[pn] .mSince G[pn] is nice, the coset y + G[pn] contains an element y + a ofnmaximal height. Hence hG(y + a) S hG(y + g) for each g E G[p 1. Sincex E y + G[pn], h (y + a) > h (x ) 2 m for each positive integer m. Thusm G G mhG(y + a) S u. Now observe that since a e G[pn], pa = 0 and thereforepny = pnY + pna. T~USn= h (x) = h (pny) = h (pny + pna) = h(p (y +a)) 2 ID + n,G G Gwhich is absurd. (Indeed, h (px) 2 h (x) + n holds in general.) Thus ifGGG[pn] is nice then \(G) < u for each positive integer n. If \(GI < u,then the set consisting of all ordinals which serve as heights of ele-ments of G is finite and consequently any set of elements of G must con-tain an element of maximal height.non-negative integer n.In particular, G[pn] i s nice for eachNow suppose \