When Verbal Subgroups are propers

Question

Let $w$ be a group-word, and let $G$ be a group. The verbal subgroup $w(G)$ of $G$ determined by $w$ is the subgroup generated by the set consisting of values $w(g_1, \ldots, g_n)$, where $g_1, \ldots, g_n$ are elements of $G$.

I'm tryng to prove a lema and I need that given a subgroup $H$ proper in $G$ ( $H<G$) then a verbal subgroup $w(H)$ is also proper in $w(G)$. This is true? How proof this statement?

Answer 1

Take $G = \mathbb{Z}_p$ with $p$ prime, $H = 0$, and $w(x) = x^p$.

Answer 2

Two non-trivial examples:

1. Take $G=A_5\times C_2$, $H=A_5$ and take the verbal subgroup to be the commutator subgroup, $w(x, y)=[x, y]$. Then $[A_5, A_5]=A_5$ while $[G, G]=A_5$.
2. Take $S=S_5$, $H=A_5$ and take the verbal subgroup to be the subgroup generated by all squares of elements, $w(x)=x^2$. Then as $A_5$ has index $2$ in $S_5$, $w(G)\leq A_5$. On the other hand, verbal subgroups are normal (because they are characteristic), so we must have that $w(G)=A_5$. Again using normality, $w(H)=1$ or $w(H)=H=A_5$, but $(123)\in w(H)$ (as $(132)(132)=(123)$) so $w(H)$ is non-trivial. Hence, $w(H)=A_5=w(G)$, as required.