Let $\Bbb Z*\Bbb Z=\langle a,b\rangle$ and $N=\{waba^{-1}b^{-1}w^{-1}:w\in\Bbb Z*\Bbb Z\}.$ Prove $\langle a,b\rangle/N$ is abelian

Question

Let $\mathbb{Z} * \mathbb{Z} = \langle a,b \rangle$ and $$N = \left\{w a b a^{-1} b^{-1}w^{-1}: w\in \mathbb{Z} * \mathbb{Z} \right\}$$ the smallest normal subgroup that contains $\left\{ a b a^{-1} b^{-1}\right\}$. I want to prove that $\langle a,b \rangle / N$ is a abelian group.

Let $w,\tilde{w} \in \mathbb{Z} * \mathbb{Z}$, then i have to show that $w\tilde{w} w^{-1}\tilde{w}^{-1} \in N$. The words on $\langle a,b \rangle$ have two possible forms $$ a^{\epsilon_1} b^{\epsilon_2}\cdots a^{\epsilon_{n-1}} b^{\epsilon_{n}}\quad \text{or } \quad b^{\epsilon_1} a^{\epsilon_2}\cdots b^{\epsilon_{n-1}} a^{\epsilon_{n}} $$

where $\epsilon_{i}\in \mathbb{Z}$. I try to make an induction on length of words, but some cases are very complicated for analyzing, I really feel that there is another way to prove this. ( I know this group is isomorphic to an abelian group, but I don't want to use that because I want to learn to understand the structure of free groups.)

Answer 1

The set $N$ is not a subgroup at all, let alone a normal subgroup. The smallest normal subgroup of $\mathbb{Z}*\mathbb{Z}$ that contains $aba^{-1}b^{-1}$ is the subgroup generated by $N$; more generally, the least normal subgroup containing the set $S$ is the collection of all finite products of conjugates of elements of $S$ and their inverses.

To see that your set $N$ is not a subgroup, note that you cannot express the identity in the form $w(aba^{-1}b^{-1})w^{-1}$. Because if you could, then you would have $aba^{-1}b^{-1} = w^{-1}(e)w = e$, which is false. So $N$ does not contain the identity.

(It is also not closed under products, as $(aba^{-1}b^{-1})^2$ is no a conjugate of $aba^{-1}b^{-1}$, nor is it closed under inverses, since $(aba^{-1}b^{-1})^{-1} = bab^{-1}a^{-1}$ is also not a conjugate of $aba^{-1}b^{-1}$, though these are a bit harder to establish.)

So let $M$ be the smallest normal subgroup that contains $aba^{-1}b^{-1}$, that is, $M=\langle N\rangle$. Then $G/M$ is generated by $Ma$ and $Mb$. To show that a group generated by a set $S$ is abelian it suffices to show that the elements of $S$ commute with each other. Here, $Ma$ clearly commutes with itself, as does $Mb$. To show that $Ma$ commutes with $Mb$, note that $(ab)(ba)^{-1} = aba^{-1}b^{-1}\in M$, so $$ (Ma)(Mb) = M(ab) = M(ba) = (Mb)(Ma).$$ Therefore, $G/M$ is indeed abelian.

Answer 2

By definition of a presentation,${}^\dagger$ we have

$$\begin{align} (\Bbb Z\ast\Bbb Z)/N&\cong\langle a,b\mid aba^{-1}b^{-1}\rangle\\ &\cong\langle a,b\mid ab=ba\rangle, \end{align}$$

which defines an abelian group because the generators commute.

In fact, the quotient defines the group $\Bbb Z\times \Bbb Z$, because $\Bbb Z\cong\langle x\mid\rangle$; see the following:

Group Presentation of the Direct Product.

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$\dagger$: As @ArturoMagidin points out in his answer, the set you describe in your question is not a subgroup; the smallest normal subgroup containing $\{aba^{-1}b^{-1}\}$ is, however, and it's what I assume you mean by $N$ here.