Prove that every subgroup of a free group is normal

abstract-algebrafree-groupsgroup-theorynormal-subgroups

Let $F_n$ be the free group of rank $n$ and $G$ be a subgroup. (By the Nielsen-Schreier Theorem we know that $G \cong F_m$ for some $m$, but that's not really relevant I think.) I want to show $G \unlhd F_n$.

Since $G$ is a subgroup of $F_n$, it has the group presentation $G = \langle f_1, \cdots, f_n\ |\ R_1, R_2, \cdots, R_k \}$, where $f_i$ are the generators of $F_n$ and $R_j$ are relations. Then intuitively I should be able to quotient out $R_1, \cdots, R_k$, thus giving a homomorphism from $F_n$ to $F_n / G$ with kernel $G$, so $G$ is normal. But this isn't quite rigorous…how do I formalize this step?

Best Answer

I don't think this is true.

Take the cyclic subgroup $G$ generated by $ab$ in $F_2=\langle a,b\mid\ \rangle$.

Then $a(ab)a^{-1}\not\in G$, since it is a word with two consecutive $a$'s and elements of $G$ are of the form $(ab)^n, n\in\Bbb Z$ (and the writing of a word is unique, precisely because there are no relations in $F_2$).