Abelian Groups - László Fuchs [Springer]

104 3 Direct Sums of Cyclic Groups
mod H disjoint from B 00 we can choose a representative x i 2hB 00 i. Thus B 00 [X (with
X Dfx i j i 2 Ig for some index set I) is a complete set of representatives mod H.
Next we show that jXj DjB 0 j. On one hand, rk H DjF=Hj DjYj jB 0 jrk F
implies jB 0 jDrk F DjF=Hj. On the other hand, let b be the first element of B 00 in
the chosen well-ordering of B. No two of the elements of the form b˛ b .b˛ 2 B 00 /
belong to the same coset mod H, and none of these is congruent mod H to a bˇ 2 B 00
(again, otherwise b˛ b bˇ 2 H would be associated with either b˛ or bˇ, etc.).
Thus there are at least jB 00 j many cosets of H which do not intersect B 00 ; hence,
jB 00 jjXj follows. This together with jB 00 jCjXj Drk F yields jXj Drk F: Hence
jB 0 jDjXj, so there is a bijection between the set of elements fb i g of B 0 and the set
of cosets fx i C Hg (where we have the corresponding elements carrying the same
index i). If in the basis B, b i 2 B 0 will be replaced by b i C x i , then we obtain a new
basis B ? of F which is at the same time a complete set of representatives mod H. ut
We are now able to verify the main result mentioned earlier.
Theorem 6.7 (Erdős [1]). Two presentations, F=H and F 0 =H 0 , of a torsion-free
group are equivalent if and only if rk H D rk H 0 :
Proof. To verify sufficiency, suppose rk H D rk H 0 ; as noted above, this implies
rk F D rk F 0 . We prove more than stated, viz. we show that every isomorphism
W F=H ! F 0 =H 0 is induced by an isomorphism W F ! F 0 carrying H onto H 0 .
Since A is torsion-free, both H and H 0 are pure. Ignoring the trivial case, we may
suppose that rk H is infinite. We distinguish three cases.
Case I: rk H D jAj. Then the same is true for rk H 0 . In view of Lemma 6.6,
there exist a basis B of F and a basis B 0 of F 0 which are complete sets of
representatives mod H and mod H 0 , respectively. The correspondence B ! B 0
which maps b 2 B upon b 0 2 B 0 if and only if maps the coset b C H upon
b 0 C H 0 extends uniquely to an isomorphism W F ! F 0 under which H 0 is the
image of H. Thus the two presentations are equivalent.
Case II: rk H > jAj.LetG be a free group whose rank is rk H. Replace F by F ˚ G
and F 0 by F 0 ˚ G, but keep H and H 0 . Application of Case I to A ˚ G implies
the existence of an isomorphism W F ˚ G ! F 0 ˚ G with H D H 0 inducing
. It is self-evident that F D F 0 .
Case III: rk H < jAj. There is a decomposition F D F 1 ˚ F 2 such that H F 1
and rk H D rk F 1 < rk F 2 D jAj. Thus A D F 1 =H ˚ F 2 ,and yields a
similar decomposition A 0 D F1 0 =H0 ˚ F2 0 . Case I guarantees the existence of
an isomorphism F 1 ! F1 0 mapping H upon H0 ; this along with F 2 ! F2
0
(restriction of ) yields an isomorphism W F ! F 0 .
ut
F Notes. Hill–Megibben [4] furnished another proof of Theorem 6.5 as a corollary to a more
general result which they proved on equivalent presentations of arbitrary abelian groups. F=H and
F 0 =H 0 are equivalent presentations if and only if, for each prime p,
dim.H C pF/=pF D dim.H 0 C pF 0 /=pF 0 :