Abelian direct factor implies potentially verbal in finite

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.

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) uses::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 $$K$$.