Abelian direct factor implies 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., Abelian direct factor (?)) must also satisfy the second subgroup property (i.e., Potentially verbal subgroup (?)). In other words, every Abelian direct factor of finite group is a potentially verbal subgroup of finite group.
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Statement

Suppose $G$ is a finite group and $H$ is a direct factor of $G$ such that $H$ is abelian as a group. Then, there exists a group $K$ containing $G$ such that $H$ is a verbal subgroup.

Related facts

Stronger facts

• Central implies potentially verbal in finite: The proof idea is the same,but it uses a central product in place of a direct product.

Facts used

1. structure theorem for finitely generated abelian groups

Proof

Given: A finite group $G$ with direct factor $H$. Let $L$ be a normal complement to $H$ in $G$, so $G\cong H\times L$.

To prove: There exists a group $K$ containing $G$ such that $H$ is a verbal subgroup of $K$.

Proof: If $H$ is trivial, we can set $K=G$ and be done. So, we can assume $H$ is nontrivial.

1. By fact (1), we can write $H$ as a direct product of cyclic groups, say $H=C_{1}\times C_{2}\times \dots \times C_{r}$, where each $C_{i}$ is cyclic of prime power order, say $p_{i}^{k_{i}}$.
2. For each prime $p$, let $c(p)$ be a positive integer greater than the largest power of $p$ dividing the order of $L$. (Note that the same $p$ may repeat among multiple $p_{i}$s).
3. Now consider the group $M=D_{1}\times D_{2}\times \dots \times D_{r}$ where each $D_{i}$ is cyclic of order $p_{i}^{k_{i}+c(p_{i})}$. Treat $H$ as the subgroup of $M$ obtained by embedding each $C_{i}$ naturally into the corresponding $D_{i}$.
4. Define $K=M\times L$, with $G$ embedded in $K$ using the embedding of $H$ in $M$ provided above.
5. Let $n=\prod p^{c(p)}$ where the product is over the primes $p$ dividing the order of $H$. Then, it is clear that the set of $n^{th}$ powers in $K$ is precisely $H$. In particular, $H$ is a verbal subgroup of $K$.