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 with symbols
Suppose is a reduced free group, i.e., is the quotient of a free group by a verbal subgroup. Then, in , any fully invariant subgroup is verbal.
Fully invariant subgroup
Further information: Fully invariant subgroup
Further information: Verbal subgroup
Reduced free group
Further information: Reduced free group
In any variety of algebras, if is a freely generating set for an algebra , and is any other algebra in that variety, any set-theoretic map from to extends uniquely to a homomorphism from to .
In particular, if is reduced free, then any set-theoretic map from a (reduced)-freely generating set of to , extends uniquely to an endomorphism of .
Given: A reduced free group , a fully invariant subgroup of
To prove: There exists a set of words, such that is precisely the set of elements of expressible using words from , by substituting elements from
Proof: Since is reduced free, we can take a set of generators for coming from a freely generating set. Define as the set of all words for which there exists where for . We claim that is as required. For this, we observe two things:
One direction: trivial
Clearly, every element of is expressible using a word in , substituting elements of
Consider for some and . We want to show that .
We know that since , there exist such that . Consider the map and extend this to an arbitrary map from the set to the group . By the universal property, this set-theoretic map extends to an endomorphism of , such that:
Since endomorphisms preserve words, we get:
Now, and is a fully invariant subgroup, so , hence , and we are done.