Group isomorphism concerning free group generated by $3$ elements.

Since my solution seems to be unnoticed, I edit it in order to make it more formal and complete.

Let $G = \langle x_1, x_2, x_3\mid x_2x_1 = x_3x_1x_2, x_3x_1 = x_1x_3, x_3x_2 = x_2x_3\rangle\ $ and $G' = (\mathbb{Z}^3, \star)$ where $\star$ is the following operation:

$$(h,l,m)\star(h',l',m') = (h+h'+lm', l + l', m + m')$$

What I have to prove is that $G\cong G'$.

Let $K$ be the normal closure of $\{x_2x_1x_2^{-1}x_1^{-1}x_3^{-1}, x_3x_1x_3^{-1}x_1^{-1}, x_2x_1x_2^{-1}x_1^{-1}\}$ in $FG^{(3)}$ then $G\cong FG^{(3)}/K$ by the definition of presentation (at least the one I use).

Now, let $a_1$, $a_2$ and $a_3$ denote the elements $(0,0,1)$, $(0,1,0)$ and $(1,0,0)$ of $G'$. They generate $G'$ since

$$\begin{align} a_3^h\star a_1^m\star a_2^l &= (1,0,0)^h\star(0,0,1)^m\star(0,1,0)^l \\ &= (h,0,0)\star(0,0,m)\star(0,l,0) \\ &= (h,0,m)\star(0,l,0) \\ &= (h,l,m)\end{align}$$

Now, we have a set of generators of cardinality $3$ so $G' \cong FG^{(3)}/K'$ for some normal subgroup $K'$ of $FG^{(3)}$.

Let $\nu\colon FG^{(3)}\to G'$ denotes the homomorphism such that $\ker(\nu) = K'$ and $\pi\colon FG^{(3)}\to G$ the homomorphism with $\ker(\pi) = K$. I want to show that there exist an homomorphism $\mu\colon G\to G'$ such that $\nu = \mu\circ \pi$. It is obvious that if such a function exists then $\mu(x_i)=a_i$.

Since $(1,0,0)\star(0,0,1) = (0,0,1)\star(1,0,0) = (1,0,1)$, $(1,0,0)\star(0,1,0) = (0,1,0)\star(1,0,0) = (1,1,0)$ and $(0,1,0)\star(0,0,1) = (1,0,0)\star(1,0,0)\star(0,1,0) = (1,1,1)$, it follow that the relations of $G$ are in $K'$ too. So, since $K$ is the smaller normal subgroup that contains them, we can conclude that $K\subseteq K'$.

By the third isomorphism theorem, $FG^{(3)}/K' \cong (FG^{(3)}/K)/(K/K')$ or in other word $\mu$ exists (actually it is also unique by the universal property of the free groups).

My last answer started from this point.

Since $\nu$ is surjective, $\mu$ have to be surjective too. In other words, $\mu^{-1}(a)$ contains at least an element for every $a\in G'$. An obvious choice is the elements $x_3^hx_1^mx_2^l$, in fact $\mu(x_3^hx_1^mx_2^l) = \mu(x_3)^h\star\mu(x_1)^m\star\mu(x_2)^l = a_3^h\star a_1^m\star a_2^l = (h,l,m)$.

Lets consider an element $w\in G$ and one decomposition $w = \prod x_i^{\varepsilon_i}$ as the product of elements of the set $\{x_1, x_2, x_3\}$. I want to show that there exists a product of the form $x_3^hx_1^mx_2^l$ such that $x_3^hx_1^mx_2^l = w$.

I do it considering the product $\prod x_i^{\varepsilon_i}$ as a succession of elements of $\{x_1, x_2, x_3\}$ and then I transform it in the wanted form in a finite number of steps. Lets define the transformations:

• Since, by the relations, $x_3$ and $x_3^{-1}$ commute with the other generators, the first transformation consists in move an $x_3$ or an $x_3^{-1}$ at the beginning of the succession.
• The second one, consists in deleting $x_ix_i^{-1}$ or $x_i^{-1}x_i$ from the succession.
• The third one consists in the application of the first relation $x_2x_1 = x_3x_1x_2$.
• The fourth transformation is simply $x_2^{-1}x_1^{-1} = x_3x_1^{-1}x_2^{-1}$ that is the inverse of the third rule.

It is obvious that these first four transformations transform a product in an equivalent one (in G).

Let's consider the product $x_2x_1^{-1}x_2^{-1}x_1$. If we apply the four transformation to it, we have the equation $x_2x_1^{-1}x_2^{-1}x_1 = x_3^{-1}$. So:

• The fifth one consists in the use of the equivalence $x_2x_1^{-1} = x_3^{-1}x_1^{-1}x_2$ that is a direct conseguence of $x_2x_1^{-1}x_2^{-1}x_1 = x_3^{-1}$.
• The last transformation is $x_2^{-1}x_1 = x_3^{-1}x_1x_2^{-1}$ that it is analog to the fifth one.

Like for the first four transformation, the last two send products in products with the same result.

Using them, we can transform every product in a product of the form $x_3^hx_1^mx_2^l$ with the same result. Since every such product has a different image in G', we conclude that $\mu$ define an isomorphism between the two groups.

Your original group $G$ is $FG^{(3)}/K$, where $K$ is the normal subgroup generated by the relations you mentioned. There is no need to show isomorphism between the two: you are given equality!

I think you have already shown that the elements of $G'$ are simply relabelings of elements in $G$. All of their multiplication properties and relations have been preserved by the maps.

Added: So you have constructed a surjection $f:FG^{(3)}\rightarrow G'$, and you have convinced yourself that the smallest normal subgroup containing the relations, call it $N$, is contained in $ker(f)$. We would like to show that $ker(f)=N$ in our case.

This is essentially trying to show that $G'$ does not have any more relations that we are unaware of. (If $G'$ had "more relations" than $G$, then $ker(f)\supsetneq N$.) I have been going back and forth on this with myself but I can't see how it is done :(

There must be some standard trick that a group theorist would use to finish this argument.