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By Kennison L. S.

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Example text

Again, one has NKD 2D NMD, and thus the normalizer N , D is contained in some maximal locally finite subgroup R # M of G. NMDsD, one has arrived at a contradiction to the choice of K. Hence any two distinct maximal subgroups of G have trivial intersection. Then the subgroup ( i , j } intersects the subgroups M and N of G non-trivially, and so (i, j ) cannot be contained in any maximal subgroup of G. But the element ij has finite order (a power of 2) and it generates a normal subgroup of index two in ( i , j } .

If S is any subgroup of the group G, then every maximal p-subgroup of S is contained in at least one maximal p-subgroup of G and no two distinct elements of Max, S can lie in a single element of Max, G. Thus IMax, SI 5 JMax, GI, and the following result (of Asar [l] and B. 8. 9 Corollary. If p is a prime and G is a countable locally finite group, then the maximal p-subgroups of G are all conjugate if and only if JMax, GI < 2O '. 10 Corollary. 8, then in every factor of G the maximal p-subgroups are all conjugate.

But then H satisfies rnin by hypothesis, and thus H i s finite. Consequently, H = Hk for some k, and so if k c j < i one has Zi _c Hin ZiG Z j . Put Z = n i , k z i . Then since each of the subgroups Zi is finite and nontrivial, we have Z # 1, and the subgroup Z is contained in the centre of U i z k G i = G. This is the desired contradiction. 5 one cannot weaken the assumptions by replacing ,,ascendant” by ,,subnormal”, for in P. Hall [4]a construction is given of a countably infinite, locally finite p-group, for any prime p, in which (1) is the only abelian subnormal subgroup.

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