# Finite non-abelian groups such that $Soc(G)=G$

Let $G$ be a finite group. It is possible that $Soc(G)=G$. For example: if $G=\prod_{i=1}^n \mathbb{Z}_{p_i}$ for some primes $p_1,\dotsc,p_n$ (not necessarily distinct), then $Soc(G)=G$.

Can this happen for non-abelian groups? If the answer is affirmative, can we find a condition that guarantees that $Soc(G)$ is a proper subgroup of $G$?

EDIT: It always happens if $G$ is simple, so there are non-abelian examples. The question about a general condition remains though.

EDIT: A related question is: for which groups does it hold that taking $Soc$ some finite number of times results in an abelian group. That is: $Soc(Soc(Soc(\dotsm G\dotsm)))$ is abelian for some finite number of iterations of taking the socle.

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What is the definition of $Soc(G)$? –  Asaf Karagila May 9 '11 at 12:40
@Asaf: $Soc(G)$ is the subgroup generated by all minimal nontrivial normal subgroups of $G$. –  user3533 May 9 '11 at 12:44

(All groups are finite.)

The following are equivalent:

• G is a direct product of simple groups
• G = Soc(G), that is, G is the product of its minimal normal subgroups

Such a group is called completely reducible.

From this it should be clear that Soc(Soc(G)) = Soc(G), that is, Soc is idempotent. Hence the following are equivalent:

• Soc(Soc(...(Soc(G))...)) is abelian
• Soc(G) is abelian
• Every minimal normal subgroup is (elementary) abelian

I don't know a name for such groups, but in line with the previous question, an abelian socle only requires that the minimal normal subgroups are modules at all.

If one requires an abelian socle not only of G, but also of H=G/Soc(G), and K=H/Soc(H), etc. then one gets solvable groups. So:

• nilpotent means every chief factor is a one dimensional module and central
• supersolvable means every chief factor is a one dimensional module
• solvable means every chief factor is a module

When people define iterated socles, they usually mean:

$$\newcommand{\Soc}{\operatorname{Soc}} \Soc^{n+1}(G)/\Soc^n(G) = \Soc( G/\Soc^n(G) )$$

This is sometimes called the (upper) Loewy series of G.

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The first equivalence should be a little surprising since minimal normal subgroups need not be simple (just direct products of simples). However, for a minimal normal subgroup not to be simple, there has to be something above it to swirl its factors around. If G=Soc(G), then there is nothing on top to do the mixing. –  Jack Schmidt May 9 '11 at 14:11
@Jack: "If one requires \emph{this}". Are you referring to the condition "$Soc(G)=G$" or to "$Soc(G)$ is abelian"? –  user3533 May 9 '11 at 14:18
One more caveat: it is not true that the socle is the largest normal subgroup that is a direct product of simple groups. For instance, in any non-abelian p-group of exponent p, the center is the socle, but the center is always contained as a maximal subgroup in some abelian subgroup, which is a direct product of simple groups of order p since the exponent is p. For instance, the Sylow p-subgroup of GL(3,p) is like this. –  Jack Schmidt May 9 '11 at 14:19
@user3533: this = "Soc(G) is abelian". Thanks for catching it. My phrasing was unclear. If G=Soc(G), then G/Soc(G)=1 is pretty dull. –  Jack Schmidt May 9 '11 at 14:20
@Jack: Thank you! When you say "is a module" do you mean it's an abelian group? –  user3533 May 9 '11 at 14:36
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