This is Exercise 3.3.10 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE, although the equivalence in question is mentioned here.

The Details:

Since definitions vary, on page 15, ibid., paraphrased, it states that

A subgroup $N$ of $G$ is normal in $G$ if one of the following equivalent statements is satisfied:

(i) $xN=Nx$ for all $x\in G$.

(ii) $x^{-1}Nx=N$ for all $x\in G$.

(iii) $x^{-1}nx\in N$ for all $x\in G, n\in N$.

We call a group simple if $G\neq 1$ and $1$ and $G$ are the only normal subgroups of $G$. This definition is equivalent to the one found on page 16.

Robinson defines completely reducible in terms of $\Omega$-groups on page 85; however, for $\Omega=\varnothing$, the definition is equivalent to saying

a group is completely reducible if it is a direct product of a possibly infinite family of simple groups.

The definition of the socle of $G$, which I denote as ${\rm Soc}(G)$, is on page 87:

The subgroup generated by all minimal normal subgroups of a group $G$ is called the socle: should the group fail to have any minimal normal subgroups, [. . .] the socle of $G$ is defined to be $1$.

On the same page, we find

the concept of a minimal normal subgroup of a group $G$, by which is understood a nontrivial normal subgroup that does not contain a smaller nontrivial normal subgroup of $G$.

The Question:

Let $G$ be a finite group. Show that $G$ is completely reducible if and only if it equals its socle.


I have only trivial things to add. However, it does not seem like a bunch of "iff"-statements starting with "$G$ is completely reducible" and ending with "$G={\rm Soc}(G)$" is easy to find. So I start out with,

"Suppose $G$ is completely reducible. Then $G$ is a direct product of a possibly infinite family of simple groups; it cannot be an infinite family since $G$ is finite."

Now what do I do next?

I'm guessing some form of the Krull-Remak-Schmidt Theorem (details of which can be found here) might be useful; I'm not sure why.

Once I'm done there, of course, the next task would be to show that $G={\rm Soc}(G)$ implies that $G$ is completely reducible.

I find the following comment on the answer to a similar question (linked to above) useful conceptually:

The [. . .] 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 $G={\rm Soc}(G)$, then there is nothing on top to do the mixing.

I'm hoping for a complete solution, although I would be more than content with a collection of strong hints.

This question appears to be one that I should be able to answer myself. I've given it a few days, though, and I've got little to show for it; I want to move on.

Please help :)


1 Answer 1


If $G$ is finite and completely reducible, then it’s a finite product of simple groups. Each one of these groups is a minimal normal subgroup of $G$ and together they generate $G$ so $G$ is its socle.

Assume that $G$ is its socle. Let $\mathcal{N}$ be a minimal set of minimal normal subgroups that generate $G$.

Let $S \subset \mathcal{N}$, and $N \in \mathcal{N} \backslash S$. Let $H$ be the subgroup of $G$ generated by the $N’ \in S$, then $H \triangleleft G,N \triangleleft G$, so the subgroup $K$ generated by $N$ and $H$ is exactly $HN$. Now, $H \cap N$ is a normal subgroup of $G$ contained in $N$. So if it not trivial, $N \subset H$ and thus $\mathcal{N}$ could be replaced by $\mathcal{N} \backslash \{N\}$ – contradicting minimality.

In particular, it follows that (with the same notations) $|H|=\prod_{s \in S}{|s|}$.

Consider, for each $N \in \mathcal{N}$, the subgroup $H_N$ generated by the $N’ \in \mathcal{N} \backslash \{N\}$. By the above, $H_NN=G$, $H_N \cap N$ is trivial. We have a morphism $f: G \rightarrow P=\prod_N{G/H_N}$ between groups of same cardinality. But if $N \in \mathcal{N}$, and $x \in G$, then $x \in H_NN$ and thus the element $v \in P$ which is $x$ at $N$ and the neutral element everywhere else is in the image of $f$ (in $f(N)$, precisely).

Thus $f$ is surjective and hence an isomorphism. To conclude, it remains to show that $G/H_N$ is simple, ie that there is no normal subgroup $H_N < K < G$. Indeed, if there were one, then $K=H_N (K \cap N)$ and thus $K \cap N$ would be a nontrivial normal subgroup of $G$ properly contained in $N$ – a contradiction.


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