Can we say anything about $\gamma_n(G) \cap \zeta_{n-1}(G)$, where $\gamma_n(G)$ is lower central series and $\zeta_{n-1}(G)$ is upper central series?

Question

Can we say anything about $\gamma_n(G) \cap \zeta_{n-1}(G)$, where $\gamma_n(G)$ is lower central series and $\zeta_{n-1}(G)$ is upper central series?

For example one can say $Z(G)\cap [G,G]$ is a subgroup of Frattini subgroup of G.

Answer 1

Proposition Let $G$ be a group, Then for all $n \geq 0$, $$\gamma_{n+1}(G) \cap \zeta_n(G) \leq \Phi(G)$$.

Proof Let $M$ be a maximal subgroup of $G$. Then either $\zeta_n(G) \leq M$ or $G=M\zeta_n(G)$. In the latter case, $\gamma_{n+1}(G)=\gamma_{n+1}(M)$. In either case, $\gamma_{n+1}(G) \cap \zeta_n(G) \leq M$, so $\gamma_{n+1}(G) \cap \zeta_n(G) \leq \Phi(G)$ $\square$.

The theorem can be generalized to varieties of groups: if $\frak{V}$ is a variety of groups then $V(G) \cap V^*(G) \subseteq \Phi(G)$, where $V(G)$ is the verbal and $V^*(G)$ is the marginal subgroup of $G$, respectively. If one takes $\frak{V}=\frak{N}_n$, the variety of all nilpotent groups of class at most $n$, then one recovers the proposition above.