Cyclic normal implies finite-pi-potentially verbal in finite

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Cyclic normal subgroup (?)) must also satisfy the second subgroup property (i.e., Finite-pi-potentially verbal subgroup (?)). In other words, every cyclic normal subgroup of finite group is a finite-pi-potentially verbal subgroup of finite group.
View all subgroup property implications in finite groups $|$ View all subgroup property non-implications in finite groups $|$ View all subgroup property implications $|$ View all subgroup property non-implications

Statement

Suppose $G$ is a finite group and $H$ is a cyclic normal subgroup of $G$. Then, there exists a finite group $K$ containing $G$ such that every prime factor of the order of $K$ and also divides the order of $G$, and $H$ is a verbal subgroup of $K$.

Related facts

Weaker facts

• Abelian hereditarily normal implies finite-pi-potentially verbal in finite
• Cyclic normal implies finite-pi-potentially characteristic in finite
• Cyclic normal implies potentially verbal in finite

Other related facts

• Homocyclic normal implies finite-pi-potentially fully invariant in finite
• Homocyclic normal implies potentially fully invariant in finite
• Central implies finite-pi-potentially verbal in finite

Facts used

1. Extending the action of quotient group on abelian normal subgroup to bigger abelian group gives rise to canonical bigger group
2. Verbality is transitive

Proof

Given: A finite group $G$, a cyclic normal subgroup $H$ of $G$.

To prove: There exists a finite group $K$ containing $G$ such that every prime factor of the order of $K$ divides the order of $G$, and $H$ is a verbal subgroup of $K$.

Proof: Let $m$ be the order of $H$ and $n$ be the order of $G$. let $h$ be the unique largest divisor of $n$ that is relatively prime to $m$. Further, let $k$ be a positive integer such that $n/h$ divides $m^{k-1}$.

Let $M$ be the cyclic group of order $m^{k}$, containing $H$ as the unique cyclic subgroup of order $m$. The action of the quotient group $G/H$ on $H$ extends to an action on $M$. Thus, by fact (1), we can construct a group $K$ containing $G$ and $M$, with $G\cap M=H$, $GM=K$, and the action of $K/M\cong G/H$ on $M$ the extension of the action we chose.

Now, consider the subgroup $V$ of $K$ generated by elements of the form $x^{hm^{k-1}}$. First, note that this subgroup is contained in $M$, because the order of $G/H$ divides the order of $G$ which in turn divides $hm^{k-1}$ by assumption. Thus, the verbal subgroup $V$ of $K$ is contained in $M$. Further, $V$ contains $H$ because the set of elements of the form $x^{hm^{k-1}}:x\in M$ itself equals $H$. Thus, $H\leq V\leq M$, with $V$ a verbal subgroup of $K$.

Since $V$ is a subgroup of $M$, it is cyclic. Every subgroup of $V$ is verbal in $V$, so $H$ is verbal in $V$. By fact (2), $H$ is verbal in $K$, completing the proof.