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Abelian Groups - László Fuchs [Springer]

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3 Completely Decomposable <strong>Groups</strong> 425<br />

is a balanced-projective resolution of G relative to A if this is a balanced-exact<br />

sequence with completely decomposable C such that A is the inclusion map<br />

˛ W A ! G.<br />

The existence of such a sequence is easily established: just take a balancedprojective<br />

resolution of G,say,0 ! H ! C !G ! 0, and choose Dr. ˚˛/W<br />

C ˚ A ! G and K D Ker . The corresponding exact sequence must be balanced,<br />

because every rank 1 torsion-free group will have the projective property relative to<br />

it (Lemma 2.3). In order to get a better understanding of relative resolutions, let us<br />

complement this observation by adding<br />

Lemma 3.3. Using the same notations, let the map W K ! C be the injection<br />

K ! A ˚ C followed by the projection into C. Then is monic and C=K Š G=A.<br />

Proof. Since jA is monic, so is . Noting that A ˚ K D A ˚ K, we evidently<br />

have<br />

C=K Š .A ˚ C/=.A ˚ K/ Š ..A C C/=K/=..A C K/=K/ Š G=A;<br />

establishing the claim.<br />

As far as the uniqueness of such relative resolutions is concerned, we can assert:<br />

ut<br />

Proposition 3.4. If (12.4) and 0 ! K 0 ! A ˚ C 0 !G ! 0 are two<br />

balanced-projective resolutions of G relative to the subgroup A .C 0 is completely<br />

decomposable/,thenK˚ C 0 Š K 0 ˚ C:<br />

Proof. Making use of the pull-back diagram<br />

0 −−−−→ K ′ −−−−→ M −−−−→ A ⊕ C −−−−→ 0<br />

⏐<br />

⏐<br />

∥<br />

↓<br />

↓φ<br />

0 −−−−→ K ′ −−−−→ A ⊕ C ′ ψ<br />

−−−−→ G −−−−→ 0,<br />

the proof is as for Theorem 3.2(ii).<br />

ut<br />

It is an immediate corollary that the group K is completely decomposable if and<br />

only if K 0 is completely decomposable.<br />

Completely Decomposable <strong>Groups</strong> The following theorem provides a satisfactory<br />

structure theorem for completely decomposable groups.<br />

Theorem 3.5 (Baer [6]). Any two decompositions of a completely decomposable<br />

group into direct sums of rank one groups are isomorphic.<br />

Proof. Let A D˚i2I A i where the A i are rational groups. It is readily seen that for<br />

any type t, the subgroups A.t/ and A .t/ are the direct sums of those A i that are of<br />

types t and > t, respectively. Thus A t D A.t/=A .t/ is isomorphic to the direct<br />

sum of those A i which are exactly of type t, andtherankofA t is precisely the

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