Verbality is not direct power-closed

Statement

It is possible to have the following:

A group $G$
A verbal subgroup $H$ of $G$
An infinite cardinal $\alpha$ (in fact, we can choose $\alpha$ to be the countable cardinal)

such that inside the direct power $G^{\alpha }$ (i.e., the $\alpha$ -fold external direct product of $G$ with itself), the group $H^{\alpha }$ is not verbal.

Related facts

Verbality is finite direct power-closed

Proof

Let $G$ be the free group of countable rank with a freely generating set $g_{1},g_{2},\dots ,g_{n},\dots$ .
Let $H$ be the verbal subgroup of $G$ given as the subset of $G$ comprising products of squares.
Let $\alpha =\omega$ , the countable cardinal.

In the group $G^{\alpha }$ , $H^{\alpha }$ is the subgroup comprising those elements such that every coordinate is a product of squares. However, the number of squares that we use for each coordinate is not bounded. Consider the element of $H^{\alpha }$ given by:

$(g_{1}^{2},g_{1}^{2}g_{2}^{2},g_{1}^{2}g_{2}^{2}g_{3}^{2},\dots )$

This element is not in the verbal subgroup comprising the products of squares, because there is no bound on the number of squares used. We would like to establish a stronger claim, namely that there is no set of words for which $H^{\alpha }$ is the verbal subgroup of $G^{\alpha }$ . PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]