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).

  • $\begingroup$ @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". $\endgroup$ – Arturo Magidin May 8 '11 at 23:44
  • $\begingroup$ @Arturo: I mean subgroups of $Soc(G)$. Edited to clarify. $\endgroup$ – user3533 May 8 '11 at 23:53

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.

  • $\begingroup$ @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? $\endgroup$ – user3533 May 9 '11 at 16:05
  • $\begingroup$ @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. $\endgroup$ – user3533 May 9 '11 at 16:26
  • $\begingroup$ @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. $\endgroup$ – user3533 May 9 '11 at 16:49
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    $\begingroup$ @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. $\endgroup$ – Jack Schmidt May 9 '11 at 19:00
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    $\begingroup$ @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. $\endgroup$ – Jack Schmidt May 11 '11 at 4:09

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