Free Generators (Basis) for a Normal Subgroup of a Free Group

Question

Let $F=\langle a,b\rangle$ be a free group of rank $2$. Its commutator subgroup has a nice free-basis: $$[a^m,b^n], \,\,\,\,m,n\in\mathbb{Z}.$$

Instead of $[F,F]$, we consider another simplest normal subgroup. Let $N$ be the smallest normal subgroup of $F$ containing $a$. Of course, $N$ will not only contain powers of $a$ but elements $bab^{-1}$ also, and in general $waw^{-1}$ for any word $w$ in $F$.

Question: What is a free-basis of $N$?

I was thinking analogously that $\{b^iab^{-i} \colon i\in\mathbb{Z}\}$ would be a free basis. But, this is not, since in the product of these elements, the end points will be $b$ or $b^{-1}$; in particular, $(ab)a(ab)^{-1}$ can not be obtained from this set.

[As noticed by Guerin, in last paragraph, after "But, this ...." is incorrect!]

Answer 1

For the free basis property. One way to state this is that any word $w$ written as a reduced non-trivial word $w_0$ of $e_i:=b^iab^{-i}$'s is non trivial. Write your word $w_0(e_1,....)=e_{f(1)}^{\epsilon(1)}...e_{f(r)}^{\epsilon(r)}$.

I claim that if the length of $w_0(e_1,...)$ is (as a word in $a$'s and $b$'s) $l_0$ then provided that $f=b^ka^{\epsilon}b^{-k}$ is different from $e_{f(r)}^{-\epsilon(r)}$ then the length of $w_0(e_1,...)f$ is strictly greater than the length $l_0$.

Use this to show that if $length(w_0(e_1,...))=0$ (as a word in $a$'s and $b$'s) then $w_0(e_1,...)$ is a trivial word (up to reduction) in $e_i$'s.

What you have shown then is that the surjective (because of the generating set thing we discussed) morphism from $\mathbb{F}_{\mathbb{Z}}$ to $N$ sending $e_i$ to $b^iab^{-i}$ is one to one. Whence $\{b^iab^{-i}\}$ is a free basis.