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Let’s call a group verbally simple if it does not have any non-trivial verbal subgroup. Does there exist some sort of classification of finite verbally simple groups?

$G^n$, with $G$ being a finite simple group, is always verbally simple as it has no nontrivial characteristic subgroups and all verbal subgroups are characteristic. However those may be not the only examples...

If $G$ is verbally simple it is either abelian or perfect, as commutator subgroup is verbal.

If $G$ is abelian, then it is $C_p^n$ for some prime $p$ as for any abelian group $A$ $V_{x^q}(A)$ is a nontrivial proper subgroup for any $q$ that is a nontrivial proper divisor of $exp(A)$.

However, I do not know, how to deal with the case, when $G$ is perfect.

Chain Markov
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1 Answers1

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Actually, all finite verbally simple groups are exactly the finite characteristically simple groups, which are the groups of the form $G^n$, where $G$ is a finite simple group.

Let’s prove this statement by induction. It is trivially true for the trivial group. Now, suppose, $G$ is a non-trivial verbally simple finite group, such that any group with order less then $|G|$ is verbally simple iff it is characteristically simple. Now we only need to prove that $G$ is characteristically simple.

One can notice, that verbal subgroup of $G$ corresponding, to the set of group words $A$ can be equivalently defined, as the minimal normal subgroup $H$, such that $\frac{G}{H}$ belongs to variety, defined by $A$. That results in $G$ being verbally simple, iff $Var(\frac{G}{H}) = Var(G)$ for all proper subgroups $H$ of $G$. Now, suppose $H$ is a maximal proper normal subgroup of $G$. Then $K = \frac{G}{H}$ is simple. So, $G$ being verbally simple results in it being isomorphic to $H \times K$ (proof of this fact can be found here: Does the specific condition on a normal subgroup of a finite group imply that it is a direct factor? v2.0). Now, as $H$ is a direct factor of $G$, $H$ also has to be verbally simple. Thus, because $|H| < |G|$ it is characteristically simple by our supposition. Thus $H \cong A^{n}$ for some finite simple group $A$, and $G \cong A^n \times K$. Moreover, as non-isomorphic finite simple groups generate distinct varieties, and $Var(K) = Var(G) = Var(A)$, we can conclude, that $K \cong A$, which results in $G$ being a characteristically simple group isomorphic to $K^{n+1}$, which completes the proof of the induction step.

Chain Markov
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