# 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.
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## 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 $K$.

## 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, $H \le V \le M$, 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.