Verbal implies fully invariant

Statement
If $$H$$ is a verbal subgroup of $$G$$, then $$H$$ is a fully invariant subgroup (also called a fully characteristic subgroup) in $$G$$.

Intermediate properties

 * Existence-bound-word subgroup

For reduced free groups

 * Fully invariant implies verbal in reduced free group: For a reduced free group, a verbal subgroup is the same thing as a fully characteristic subgroup. The set of words corresponding to a given fully characteristic subgroup is simply the set of words in the corresponding free generators.

Facts used

 * 1) uses::Homomorphism commutes with word maps
 * 2) uses::Image of invariant set for a map under a commuting map is also invariant for the map

Hands-on proof
Given: A group $$G$$ and subgroup $$H$$. $$C$$ is a collection of words such that an element is in $$H$$ if and only if it can be expressed, as the value of a word in $$C$$ with letters taken as elements of $$G$$. In other words, $$H$$ is the union of the images of the word maps corresponding to the elements of $$C$$. (Note that the existence of such a $$C$$ is what makes $$H$$ a verbal subgroup). An endomorphism $$\varphi$$ of $$G$$ and element $$h \in H$$.

To prove: $$\varphi(h) \in H$$.

Slick proof
Note that any endomorphism is a homomorphism, so every endomomorphism of $$G$$ commutes with every word map of $$G$$ by Fact (1). Fact (2) now tells us that the image of every word map is invariant under every endomorphism of $$G$$. Since a union of invariant subsets under a bunch of functions is still invariant, we obtain that any union of images of word maps is invariant under every endomorphism. In particular, any verbal subgroup is invariant under every endomorphism.