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This question is a generalization of a previous one: Groups such that any nontrivial normal subgroup intersects the center nontrivially (see Jack Schmidt's answer):

What finite groups $G$ have the property that $Soc(G)$ is abelian and all of the subgroups of $Soc(G)$ are normal in $G$? (the linked question talks about the case where $Soc(G)$ is central and thus is a special case).

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    @user3533: "all of its subgroups"; you mean all subgroups of $G$, or all subgroups of the socle of $g$? I'm guessing the latter, but "its" is ambiguous here. You may want to clarify by saying "all subgroups of..." instead of "all of its subgroups".2011-05-08
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    @Arturo: I mean subgroups of $Soc(G)$. Edited to clarify.2011-05-08

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The following are equivalent for an abelian subgroup N of the finite group G:

  • every subgroup of N is normal in G
  • every cyclic subgroup of N is normal in G
  • for every g in G, there is a positive integer k, such that for every x in N, xg = xk.

In this case we say that G acts as power automorphisms on N.

When N = Soc(G), then we can add another condition:

  • every minimal normal subgroup is cyclic of prime order, and those of the same order are isomorphic as modules

Suppose that not only G, but also H = G/Soc(G), and K = H/Soc(H), etc. all act as power automorphisms on their socle. In this case, not only is every minimal normal subgroup cyclic, but in fact every chief factor is cyclic, and so the group is what is called supersolvable. A finite nilpotent group is one in which every chief factor is the 1-dimensional trivial module. A finite supersolvable group is one in which every chief factor is (any) 1-dimensional module.

Power automorphisms are discussed a bit in Roland Schmidt's textbook on the Subgroup Lattices of Groups.

When all of the p-chief factors for each prime p are required to be isomorphic inside a particular normal subgroup (not the socle), then I believe this characterizes "PST" groups. In case we require it in the socle, I'm not immediately sure, but there might be something important there.

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    @Jack: Just to make sure I understood: if $G$ is a finite solvable group, then $Soc(G)$ is abelian, but there may be a subgroup of $Soc(G)$ which is not normal in $G$. But if $G$ is a finite supersolvable group then $Soc(G)$ is abelian and each subgroup of $Soc(G)$ is normal in $G$. Right?2011-05-09
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    @Jack: That's what I understood is implied by your answer. I'm making sure because it seems to me that in the solvable case the subgroups of $Soc(G)$ will be normal in $G$ too. Maybe I'm just confused, still thinking about it.2011-05-09
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    @Jack: My reasoning is that if $G$ is finite and solvable then its minimal normal subgroups are cyclic of prime order, and this is one of the "equivalent conditions" you mentioned.2011-05-09
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    @user3533: your first is exactly correct. Look at the alternating group of degree 4 (and order 12, a solvable but not supersolvable group). Its socle has order 4 and is itself a minimal normal subgroup. A sylow 3-subgroup sits on top of the socle and swirls the subgroups of order 2 around. In a solvable group every minimal **subnormal** subgroup is cyclic of prime order, and every **composition** factor is cyclic of prime order. However, minimal normal subgroups and chief factors are only of prime power order.2011-05-09
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    @Jack: I understand now. Thank you.I'm still trying to figure out why condition 4 implies the others. When condition 4 applies, the socle is a product of vector spaces and conjugation by an element of G acts on each vector space as a diagonal operator. But we need it to be a special diagonal operator, namely - multiplication by a scalar.2011-05-09
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    @user3533: You are quite correct! The 4th condition is not equivalent. 3×Sym(3) has every minimal normal subgroup cyclic of order 3, but one is central and one is inverted, and so there cannot possibly be a "k" as in the third condition. Furthermore any "diagonal" subgroup like { (x,x) : x in Z/3Z } is contained in the socle but normal. So condition 4 is weaker. I'll fix it, but then I have to ask myself if we now have a stronger condition since condition 4 really is the "supersolvable" condition, not 1=2=3.2011-05-10
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    @Jack: So the new condition 4 may be stronger than the others, but it still holds for any supersolvable group?2011-05-10
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    @Jack: I understand why in a supersolvable group, $G$, any minimal normal subgroup is cyclic of prime order, but I'm not sure why the action of $G$ on $Soc(G)$ by conjugation is an action by (universal) power automorphims.2011-05-10
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    @user3533: The new condition 4 is equivalent to the first three conditions (when N=Soc(G) is abelian). The group G=S3×S3 is supersolvable, but does not satisfy condition 4. The action on Soc(G) is not by a universal power automorphism, and the two minimal normal subgroups are not isomorphic as G/Soc(G)-modules.2011-05-11
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    @Jack: That's clear now. Thank you very much for all the help.2011-05-11