Presentation of the group $\mathbb Z^3$ with a special product (how to complete the proof?) Let $G$ be the structure with underlying set $\mathbb Z^3$ and product defined by:
$$(k_1,l_1,m_1)(k_2,l_2,m_2) = (k_1 + k_2 + l_1m_2, l_1+l_2,m_1+m_2)$$
It's easy to prove that $G$ is a group with neutral element $(0,0,0)$. I would like to prove that this group has the presentation:
$$x_2x_1 = x_3 x_1x_2, x_3x_1 = x_1 x_3, x_3x_2=x_2x_3$$
in the free group generated by $x_1,x_2,x_3$.
What I did was the following.
Let $e_3 = (1,0,0)$, $e_2 = (0,1,0)$ and $e_1 =(0,0,1)$ (the choice of these is because I'd like them to "match" the $x_i$'s). 
We have by easy verification that:
$$e_2e_1 = e_3e_1e_2, e_3e_1=e_1e_3, e_3e_2=e_2e_3 \tag1$$
Let $K$ be the normal subgroup of $FG^3$ generated by the elements:
$$x_2x_1x_2^{-1}x_1^{-1}x_3^{-1}, x_3x_1x_3^{-1}x_1^{-1}, x_3x_2x_3^{-1}x_2^{-1}$$
Let $f: FG^3 \to G$ be the homomorphism sending $x_i$ to $e_i$. It is surjective since the $e_i$ generate $G$; for we have:
$$(k,l,m) = (k,0,0)(0,l,m) = (k,0,0)(0,0,m)(0,l,0)=e_3^k e_1^me_2^l$$
Now $K \subset \ker f$, by $(1)$. I would like to prove that $\ker f \subset K$ to conclude.
How to do this?
I know that the question has been asked before here, but I don't understand the answer there.
 A: Let $H = FG^{3}/K$, and let $f_1,f_2,f_3$ be the images of the generators $x_1,x_2,x_3$. Note that there is a map $\phi:H \rightarrow G$ sending $f_i$ to $e_i$, and we are done if we can show $\phi$ has trivial kernel. 
In $H$ we have $f_1f_3 = f_3f_1$ and $f_2f_3 = f_3f_2$, so the set of all elements in $H$ commuting with $f_3$ is a subgroup of $H$ containing all generators, and therefore is all of $H$. 
We can also show that $f_2^{l}f_1 = f_3^{l}f_1f_2^{l}$ for all integers $l$. First we can prove it for positive integers by induction: the base case is trivial and for the inductive step we get 
$$f_2^{l+1}f_1 = f_2(f_2^{l}f_1) = f_2(f_3^{l}f_1f_2^{l}) = f_3^{l}(f_2f_1)f_2^{l} = f_3^{l}(f_3f_1f_2)f_2^{l} = f_3^{l+1}f_1f_2^{l+1}$$
We can multiply both sides of that equation by $f_2^{-l}$ on both sides to get $f_1f_2^{-l} = f_3^{l}f_2^{-l}f_1$, or $f_2^{-l}f_1 = f_3^{-l}f_1f_2^{-l}$, this covers it for negative integers. 
Starting with $f_2^{l}f_1 = f_3^{l}f_1f_2^{l}$ and multiplying it on both sides by $f_1^{-1}$ we get $f_1^{-1}f_2^{l} = f_3^{l}f_2^{l}f_1^{-1}$, or $f_2^{l}f_1^{-1} = f_3^{-l}f_1^{-1}f_2^{l}$. This gives the general equation $f_2^{l}f_1^{\pm 1} = f_3^{\pm l}f_1^{\pm 1}f_2^{l}$.
We are ready for the main claim: Every element in $H$ can be written as $f_3^{k}f_1^{m}f_2^{l}$ for some $k,m,l$. This will finish our proof, since $\phi(f_3^{k}f_1^{m}f_2^{l}) = (k,l,m)$.
To prove the claim, let $S$ be the subset of $H$ of all elements satisfying that condition, and $H_0$ the subgroup of $H$ defined by $\{g \in H \mid xg \in S \text{ and } xg^{-1} \in S \text{ for all } x \in S\}$. We are done if we can show that $H_0$ contains all the generators of $H$, which would imply $H_0 = H$ and hence $S = H$. 
So, let $x = f_3^{k}f_1^{m}f_2^{l} \in S$. We have, from the earlier computations:
$$xf_3^{\pm 1} = f_3^{k \pm 1}f_1^{m}f_2^{l}$$
$$xf_2^{\pm 1} = f_3^{k}f_1^{m}f_2^{l \pm 1}$$
$$xf_1^{\pm 1} = f_3^{k}f_1^{m}f_2^{l}f_1^{\pm 1} = f_3^{k}f_1^{m}f_3^{\pm l}f_1^{\pm 1}f_2^{l} = f_3^{k \pm l}f_1^{m \pm 1}f_2^{l}$$
And this finishes the argument.
