Abelian Groups - László Fuchs [Springer]

308 10 Torsion Groups
Proof. We have noticed above that every subgroup of the form (10.3) is fully
invariant. Assume, conversely, that G is a fully invariant subgroup, and define
n as the minimum of the heights h.p n g/ with g running over G. The sequence
0 ; 1 ;:::; n ;::: is obviously strictly increasing (except when it reaches 1). To
verify the gap condition, suppose k C 1< kC1 for some k. Surely, there exists an
x 2 G with h.p k x/ D k , and by definition h.p kC1 x/ kC1 ; thus, this x has a gap
at k in its indicator. By Lemma 1.1, A must contain an element of order p and of
height k .
The inclusion G A. 0 ; 1 ;:::; n ;:::/is obvious. We now verify the existence
of a g 2 G such that h.p i g/ D i for i D 0;1;:::;n 1: If there is no gap
in the sequence 0 ; 1 ;:::; n 1 ,andifg 2 G satisfies h.p n 1 g/ D n 1 ,then
this g is already as desired. If there is a gap in this sequence, and if the first gap
appears between j 1 and j , then there is a g j 2 G such that h.p i g j / D i for
i D 0; 1; : : : ; j 1: If the second gap lies between k 1 and k .j < k/, then some
g 0 2 G exists with h.p i g 0 / D i for i D j;:::;k 1: By Lemma 1.1, A contains a
g k such that h.p i g k / maxfh.p i g 0 /; i C 1g for i D 0;1;:::;j 1 and h.p i g k / D
h.p i g 0 / for i j. Because of full invariance, and hence full transitivity, as well as
Corollary 1.5, g k 2 G. Thus proceeding, we construct elements g j ; g k ;:::;g` 2 G
for the gaps in 0 ; 1 ;:::; n 1 , and at the end, g D g j C g k C C g` will
satisfy h.p i g/ D i for i D 0;1;:::;n 1: Thus h.g/ h.a/ for every a 2
A. 0 ; 1 ;:::; n ;:::/ of order o.g/. Full transitivity shows that a 2 G, i.e.G
is of the form (10.3).
If . 0 ; 1 ;:::; n ;:::/ and .0 0;0 1 ;:::;0 n ;:::/ are different sequences, both
satisfying the gap condition, then let n be the first index with n ¤ n 0,say, n