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Let G be a finite group. In general case, given two normal subgroups N and K of G, we need not to have N < K or K< N. The easiest example is the Klein 4-group V4 and its subgroups of order 2. So assume that G has such a property, that is the normal subgroups of G, constitute a chain with respect to inclusion. For example, simple groups, cyclic groups and symmetric groups satisfy this property. Certainly, if G has such a property, then the normal subgroups are necessarily characteristic. Also we may find out that the center of G must be cyclic. Indeed, every abelian group with this property must be a cyclic p-group (and vice versa). This also shows that G/G' is cyclic, for, the property in quotient hereditary. Please let me know, if these groups are studied before.

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What is the question? –  Emil Jeřábek Mar 14 '11 at 13:05
Your statement "Indeed, every abelian group with this property is cyclic (and vice versa)" is not correct; only cyclic groups of prime power order have this property. –  Tom De Medts Mar 14 '11 at 14:36
A minor improvement: any nilpotent group with this property is cyclic. –  ndkrempel Mar 14 '11 at 17:28
@ Tom Thank you Tom. You are right. I modifies it. –  Amin Mar 15 '11 at 13:01

1 Answer 1

up vote 4 down vote accepted

For solvable groups without Frattini chief factors, this is equivalent to each of the following (individually):

  • having a unique chief series,
  • every quotient group having a faithful primitive permutation action,
  • the upper Fitting series being a chief series
  • the lower Fitting series being a chief series

This is shown in:

Hawkes, Trevor O. "Two applications of twisted wreath products to finite soluble groups." Trans. Amer. Math. Soc. 214 (1975), 325–335. MR379657 DOI:10.2307/1997110

You might also be interested in the safari for zebra groups.

However, there are solvable groups with Frattini factors whose normal subgroups form a chain: SL(2,3) for example.

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Pazderski's ams.org/mathscinet-getitem?mr=755313 also describes these groups from a different perspective. Maybe it would also focus the search in mathoverflow.net/questions/58059/… –  Jack Schmidt Mar 14 '11 at 21:52
If a group with this property is solvable, it is also supersolvable, since all the normal subgroups are characteristic. The chief series then has factors of prime order and is unique (a group with this property has a unique minimal non-trivial normal subgroup). –  Tobias Kildetoft Mar 15 '11 at 9:24
@Jack Thank you very much for references. –  Amin Mar 15 '11 at 13:02
@ Tobias If you take G = A4, the alternating group of order 12, then it satisfies the above hypotheses . However, A4 is not supersolvable. Your assertion is true with the additional condition that Z(G) > 1. –  Amin Mar 15 '11 at 13:13

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