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Abelian Groups - László Fuchs [Springer]

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1 The Ulm-Zippin Theory 345<br />

Lemma 1.4. Assume A; C are p-groups, G is a subgroup of A, and H is a subgroup<br />

of C such that f .A; G/ D f .C; H/ for some . There exists an a 2 AŒp of height <br />

and proper with respect to G if and only if there is a c 2 CŒp of height and proper<br />

with respect to H.<br />

Proof. Observe that x 2 G./ means that x 2 AŒp is of height , and not proper<br />

with respect to G, since—we repeat—x 2 p C1 A C G is another way of saying that<br />

the coset x C G has a representative in p C1 A.<br />

ut<br />

Let A and C be reduced p-groups, and G A; H C subgroups. An<br />

isomorphism W G ! H is called height-preserving if<br />

h C .g/ D h A .g/ for every g 2 G:<br />

It is important to keep in mind that heights are always computed in the large groups.<br />

Manifestly, the restriction of an isomorphism of groups A ! C to subgroups<br />

G ! H is always height-preserving.<br />

Kaplansky–Mackey Lemma The next lemma is crucial in extending isomorphisms<br />

between subgroups to larger subgroups. It is the most essential ingredient<br />

in the proof of Ulm’s theorem below. The groups A; C in Lemma 1.5 need not be<br />

p-groups, but the Hill invariants are computed for the chosen prime p.<br />

Lemma 1.5 (Kaplansky–Mackey [1]). Let A and C be reduced groups, G a<br />

subgroup of A, and H a subgroup of C such that A=G and C=H are p-groups.<br />

Suppose W G ! H is a p-height-preserving isomorphism; furthermore, for all<br />

, f .A; G/ f .C; H/ and<br />

˛ W p AŒp=G./ ! p CŒp=H./ (11.1)<br />

are monomorphisms. If a 2 A is proper with respect to G and pa 2 G, then can<br />

be extended to a height-preserving isomorphism<br />

for a suitable c 2 C such that<br />

WhG; ai !hH; ci<br />

˛.hG; ai./=G.// DhH; ci./=H./ for all :<br />

Proof. Assume that a with h.a/ D has been chosen in its coset mod G, in addition<br />

to being proper with respect to G, to satisfy also h.pa/ >C 1 whenever possible.<br />

We distinguish two cases according as such a choice is possible or not.<br />

Case I: h.pa/ >C 1. In this case pa D pb for some b 2 A of height C 1.<br />

Then h.a b/ D , anda b is proper with respect to G, for otherwise there<br />

would exist a g 2 G with h.a b C g/ > , leading to the contradiction<br />

h.a C g/ >. By an earlier remark, a b … G./, thus there is a u 2 p CŒp

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