Fully invariant implies verbal in reduced free group

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a reduced free group. That is, it states that in a Reduced free group (?), every subgroup satisfying the first subgroup property (i.e., Fully invariant subgroup (?)) must also satisfy the second subgroup property (i.e., Verbal subgroup (?)). In other words, every fully invariant subgroup of reduced free group is a verbal subgroup of reduced free group.
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Statement

Verbal statement

In a reduced free group, every fully characteristic subgroup is verbal.

Statement with symbols

Suppose $G$ is a reduced free group, i.e., $G$ is the quotient of a free group by a verbal subgroup. Then, in $G$, any fully characteristic subgroup is verbal.

Definitions used

Fully characteristic subgroup

Further information: Fully characteristic subgroup

Verbal subgroup

Further information: Verbal subgroup

Reduced free group

Further information: Reduced free group

Converse

The converse statement is true in any group: a verbal subgroup is always a fully invariant subgroup. For full proof, refer: Verbal implies fully invariant

Facts used

In any variety of algebras, if $S$ is a freely generating set for an algebra $F$, and $A$ is any other algebra in that variety, any set-theoretic map from $S$ to $A$ extends uniquely to a homomorphism from $F$ to $A$.

In particular, if $G$ is reduced free, then any set-theoretic map from a (reduced)-freely generating set of $G$ to $G$, extends uniquely to an endomorphism of $G$.

Proof

Given: A reduced free group $G$, a fully characteristic subgroup $H$ of $G$

To prove: There exists a set $W$ of words, such that $H$ is precisely the set of elements of $G$ expressible using words from $W$, by substituting elements from $G$

Proof: Since $G$ is reduced free, we can take a set of generators $x_{i},i\in I$ for $G$ coming from a freely generating set. Define $W$ as the set of all words $w$ for which there exists $h\in H$ where $h=w(x_{j_{1}},x_{j_{2}},\dots ,x_{j_{r}})$ for $j_{l}\in I$. We claim that $W$ is as required. For this, we observe two things:

One direction: trivial

Clearly, every element of $H$ is expressible using a word in $W$, substituting elements of $G$

Other direction

Consider $w(g_{1},g_{2},\dots ,g_{n})$ for some $w\in W$ and $g_{i}\in G$. We want to show that $w(g_{1},g_{2},\dots ,g_{n})\in H$.

We know that since $w\in W$, there exist $x_{j_{1}},x_{j_{2}},\dots ,x_{j_{n}}$ such that $w(x_{j_{1}},x_{j_{2}},\dots ,x_{j_{n}})\in H$. Consider the map $x_{j_{i}}\mapsto g_{i}$ and extend this to an arbitrary map from the set $\{x_{i}\}$ to the group $G$. By the universal property, this set-theoretic map extends to an endomorphism $\varphi$ of $G$, such that:

$\varphi (x_{j_{i}})=g_{i}$

Since endomorphisms preserve words, we get:

$\varphi (w(x_{j_{1}},x_{j_{2}},\dots ,x_{j_{n}}))=w(g_{1},g_{2},\dots ,g_{n})$

Now, $w(x_{j_{1}},x_{j_{2}},\dots ,x_{j_{n}})\in H$ and $H$ is a fully characteristic subgroup, so $\varphi (w(x_{j_{1}},x_{j_{2}},\dots ,x_{j_{n}}))\in H$, hence $w(g_{1},g_{2},\dots ,g_{n})\in H$, and we are done.