I am stuck in a possibly very simple claim. Here are the terms:
X = set of generators
F(X) = free group generated by X
S = Some subset of reduced words (in F(X))
$ S^G $ = set of all conjugates of members of S (by elements of F(X) )
N = $ < S^G > $ = intersection of all subgroups of F(X) that contain all of $ S^G $
I want to show N is a normal subgroup of F(X).
It is trivial that $ x S^G x^{-1} \subseteq S^G $. From this it should follow that $ x < S^G > x^{-1} \subseteq <S^G> $. But I cannot wrap my head around this consequence.
- How does $ x S^G x^{-1} \subseteq S^G $ imply $ x < S^G > x^{-1} \subseteq <S^G> $ ?
- Is there an explicit description of the elements of N? Are they just reduced words in $ S^G $ ?
This follows from the statement: If $A \subseteq B$ then $\langle A \rangle \subseteq \langle B \rangle$. This is true because $A \subseteq B \subseteq \langle B \rangle$ shows that $\langle B \rangle$ is a subgroup containing $A$, hence it contains $\langle A \rangle$.
It is not just reduced words in $S^G$, because $S^G$ is not closed under multiplication. Imagine just the simple case where $X = \{a,b\}$ and $S = \{a\}$. Then $b a b^{-1} \in S^G$ and also $b^2 a b^{-2} \in S^G$, so their product must be in $\langle S^G \rangle$. The product is $b a b a b^{-2}$ which is not an element of $S^G$. This is a similar situation to when you study the commutator subgroup of a group: it doesn't consist of just commutators, but the subgroup generated by commutators.