Verbality is quotient-transitive

Statement
Suppose $$G$$ is a group, with $$H \le K \le G$$ subgroups. Suppose $$H$$ is a verbal subgroup of $$G$$. Note that since verbal implies normal, $$H$$ is a normal subgroup of $$G$$, so that we can talk of the quotient group $$G/H$$. Suppose we are further given that $$K/H$$ is a verbal subgroup of $$G/H$$.

Then, $$K$$ must be a verbal subgroup of $$G$$. In fact, the set of words that we can use to generate $$K$$ is obtained by taking products of the set of words used for $$H$$ in $$G$$ and the set of words used for $$K/H$$ in $$G/H$$.

Related facts

 * Verbality is transitive