Equal verbal subgroups for a subvariety not implies equal marginal subgroups

Statement
It is possible to have a group $$G$$ and two subvarieties $$\mathcal{V}_1$$ and $$\mathcal{V}_2$$ of the variety of groups such that the following hold:


 * $$V_1(G) = V_2(G)$$, where $$V_1(G)$$ and $$V_2(G)$$ denote respectively the $$\mathcal{V}_1$$-verbal subgroup and $$\mathcal{V}_2$$-verbal subgroup of $$G$$.
 * $$V_1^*(G) \ne V_2^*(G)$$, where $$V_1^*(G)$$ and $$V_2^*(G)$$ denote respectively the $$\mathcal{V}_1$$-marginal subgroup and $$\mathcal{V}_2$$-marginal subgroup of $$G$$.

Related facts

 * Equal marginal subgroups for a subvariety not implies equal verbal subgroups

Perfect and centerless
Consider an example where:


 * $$G$$ is a perfect group that is not a centerless group (see perfect not implies centerless), for instance, $$G$$ is isomorphic to special linear group:SL(2,5).
 * $$\mathcal{V}_1$$ is the variety of abelian groups.
 * $$\mathcal{V}_2$$ is the variety comprising the trivial group.

Then:


 * $$V_1(G) = [G,G]$$ is the derived subgroup of $$G$$. Since $$G$$ is perfect, $$V_1(G) = G$$.
 * $$V_2(G) = G$$.
 * Thus, $$V_1(G) = V_2(G)$$.

However:


 * $$V_1^*(G)$$ is the center of $$G$$, which is nontrivial since $$G$$ is not centerless.
 * $$V_2^*(G)$$ is the trivial subgroup of $$G$$.
 * Thus, $$V_1^*(G) \ne V_2^*(G)$$.

Example with a finite nilpotent group
Consider the example where:


 * $$G$$ is the central product of D8 and Z4.
 * $$\mathcal{V}_1$$ is the variety of abelian groups.
 * $$\mathcal{V}_2$$ is the variety of groups of exponent two (along with the trivial group).

Then:


 * $$V_1(G)$$ is the derived subgroup of $$G$$, and is isomorphic to cyclic group:Z2.
 * $$V_2(G)$$ is the Frattini subgroup of $$G$$, and it coincides with the derived subgroup of $$G$$.
 * $$V_1^*(G)$$ is the center of $$G$$, and it is isomorphic to cyclic group:Z4.
 * $$V_2^*(G) = \Omega_1(Z(G))$$ is the socle of $$G$$ (see socle equals Omega-1 of center in nilpotent p-group), and it in fact coincides with $$V_1(G) = V_2(G)$$. In particular, it is not the same as $$V_1^*(G)$$.