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From Jacobson's Basic Algebra I, page 70,

Let $G$ be the group defined by the following relations in $FG^{(3)}$: $$x_2x_1=x_3x_1x_2, \qquad x_3x_1=x_1x_3,\qquad x_3x_2=x_2x_3.$$ Show that $G$ is isomorphic to the group $G'$ defined to be the set of triples of integers $(k,l,m)$ with $$(k_1,l_1,m_1)(k_2,l_2,m_2)=(k_1+k_2+l_1m_2,l_1+l_2,m_1+m_2).$$

My thoughts: I was able to show that $G'$ is generated by $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, since $(h,l,m)=(1,0,0)^{h-lm}(0,1,0)^l(0,0,1)^m$. Letting $(0,0,1)=a_1$, $(0,1,0)=a_2$, and $(1,0,0)=a_3$, I calculate that $a_2a_1=a_3a_1a_2,a_3a_1=a_1a_3,a_3a_2,a_2a_3$. So they look like the satisfy the same relations as the $x_i$. (I'm not sure if this is necessary.)

So taking the set $X=\{x_1,x_2,x_3\}$, I have a map $x_i\mapsto a_i$, which gives a homomorphism of $FG^{(3)}$ into $G'$ such that $\bar{x}_i\mapsto a_i$, and this homomorphism is in fact an epimorphism since it maps onto a set of generators for $G'$. Thus $FG^{(3)}/K\simeq G'$ where $K$ is the kernel of the homomorphism. Since $G\simeq FG^{(3)}/K$, $G\simeq G'$.

I can't quite justify if $G\simeq FG^{(3)}/K$, from the comments, I understand why the generated normal subgroup $K$ is contained in the kernel $\ker\nu$ of the induced homomorphism $FG^{(3)}\to G'$, but I don't follow why $\ker\nu\subset K$. Why does $\ker\nu\subset K$? Thanks.

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It looks not only fine to me but pretty good, as defining those $a_i'$s and finding their relations is not trivial. – DonAntonio May 30 '12 at 23:28
@Adeal First deduce that the normal subgroup with the generators we discussed below is contained in the kernel. – rschwieb May 31 '12 at 1:28
@rschwieb Thanks! I denote $\nu\colon FG^{(3)}\to G'$ to be the induced homomorphism. Then for the generator $bab^{-1}a^{-1}c^{-1}$, $$\nu(bab^{-1}a^{-1}c^{-1})=\nu(b)\nu(a)\nu(b)^{-1}\nu(a)^{-1}\nu(c)^{-1}=a_2a_1‌​a_2^{-1}a_1^{-1}a_3^{-1}=1$$ based on the relation $a_2a_1=a_3a_1a_2$. The same follows for the other generators. So the kernel, which is normal as the kernel of a homomorphism, contains the generators of $K$ and thus $K$ itself. But how does it follow that $\ker\nu\subset K$? – Adelaide Dokras May 31 '12 at 2:29
@Adeal It's going to take a lot of text so I am appending it to my solution. – rschwieb May 31 '12 at 11:23
up vote 2 down vote accepted

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 three 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 homorphism 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.

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Thanks! I appreciate the detail in this answer. – Adelaide Dokras Jun 4 '12 at 23:07

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.

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Thanks rschwieb, why is the normal subgroup generated by the relations the same as the kernel of the induced homomorphism $FG^{(3)}\to G'$? – Adelaide Dokras May 30 '12 at 23:25
@Adeal You can get that if you work a bit on it. You can rewrite all your generating relations as (something)=1, and then the preimage of (something) has to be in the kernel of your map. Actually I think using an isomorphism theorem is a bit overkill. You have essentially already shown that $G'$ is the same quotient of a free group on some relations, the only difference is that the operation is superficially different. – rschwieb May 30 '12 at 23:34
The thing that confuses me is that the normal subgroup generated is supposed to the intersection of all normal subgroups containing the relations $S$, where $S$ is a subset of a group. But the relations are equations, not elements. What does that mean here? – Adelaide Dokras May 30 '12 at 23:38
@Adeal "How do you convert relations into elements of K?" Good question! Let's take $x_2x_1=x_3x_1x_2$ as an example. To save typing and remind you that these are cosets, I'm going to write it as $(bK)(aK)=(cK)(aK)(bK)$. We want this relation to hold in the quotient. To affect it, we just require that $bab^{-1}a^{-1}c^{-1}\in K$. This being the case, then the cosets obey the identity, because $bab^{-1}a^{-1}c^{-1}\in K$ iff $baK=cabK$ iff $(bK)(aK)=(cK)(aK)(bK)$. By requiring $cac^{-1}a^{-1}$ and $cbc^{-1}b^{-1}$ to be in $K$, we can guarantee the relations hold for the quotient. – rschwieb May 31 '12 at 1:20
@Adeal To summarize, $K$ is the smallest normal subgroup containing $bab^{-1}a^{-1}c^{-1}$, $cac^{-1}a^{-1}$ and $cbc^{-1}b^{-1}$. – rschwieb May 31 '12 at 1:21

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