# Equal marginal subgroups for a subvariety not implies equal verbal 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:

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

## Proof

### Perfect and centerless

Consider an example where:

• $G$ is a centerless group that is not a perfect group (see centerless not implies perfect), for instance, $G$ is isomorphic to symmetric group:S3.
• $\mathcal{V}_1$ 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 not perfect, $V_1(G) \ne G$.
• $V_2(G) = G$.
• Thus, $V_1(G) \ne V_2(G)$.

However:

• $V_1^*(G)$ is the center of $G$, which is trivial since $G$ is centerless.
• $V_2^*(G)$ is the trivial subgroup of $G$.
• Thus, $V_1^*(G) = V_2^*(G)$.

### Example with a finite nilpotent group

Consider the example where:

Then: