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Let $FG^{(3)}$ denote the free group generated by $3$ elements. Let $Z=(\Bbb Z^3,\oplus,{\bf 0})$ denote the group with addition

$$(a_1,a_2,a_3)\oplus (b_1,b_2,b_3)=(a_1+b_1+a_2b_3,a_2+b_2,a_3+b_3)$$

Let $G$ be the group defined by the relations

$$x_2x_1=x_3x_1x_2\\x_1x_3=x_3x_1\\x_2x_3=x_3x_2$$

on $FG^{(3)}$.

Note the first equation can be written as $x_2x_1=x_1x_2x_3$.

I ought to show $G\simeq Z$.

For starters, note that for any element of $\Bbb Z^3$ we can write

$$(a,b,c)=ae_1\oplus ce_3\oplus be_2$$ so that $\{e_1,e_2,e_3\}$ genereate $Z$. Moreover, $e_1,e_2,e_3$ fulfill the relations with $x_1=e_3,x_2=e_2,x_3=e_1$. We thus consdier the natural (canonical) homomorphism $FG^{(3)}\to \Bbb Z^3$ by sending $x_1\to e_3$, $x_2\to e_2$ and $x_3\to e_1$.

How can I show that the kernel of this homomoprhism is the normal subgroup generated by the elements obtained when equating the relations to the unity? It is clear that the normal subgroup is contained in this kernel, but how does one show it is actually the kernel?

ADD This is all the author says about relations on free groups:

"A group $G$ is said to be finitely generated if it contains a finite group of generators $\{a_1;1\leq a_i\leq r\}$. Then we have the homomorphism $\eta$ of $FG^{(r)}$ sending $x_i\to a_i$. Since the $a_i$ generate $G$, this is an epimorphism and $G\simeq FG^{(r)}/K$ where $K$ is the kernel of $\eta$. The normal subgroup $K$ is called the set of relations connecting the generators $a_i$. If $S$ is a subset of a group, we can define the normal subgroup generated by $S$ to be the intersection of all normal subgroups of the group containing $S$. If $S$ is a subset of $FG^{(r)}$ we say that $G$ is defined by the relations $S$ if $G\simeq FG^{(r)}/K$ where $K$ is the normal subgroup generated by $S$. If $S$ is finite, then we say that $G$ is a finitely presented group."

The author gave an example on how to show two groups are isomorphic. It went something like this: Let $D_n$ be the dihedral group. We show that it is defined by the relations $$\tag 1 x^n,y^2,xyxy$$ on the free group generated by two elements. It is clear that $D_n$ is generated by the rotation $R$ by $2\pi /n$ and the reflection $S$ through the $x$-axis. Moreover $R^n=1$, $S^2=1$ and $SRS=R^{-1}$. Hence $D_n$ is homomorphic to $FG^{(2)}/K$ where $K$ is the normal subgroup generated by the elements in $(1)$.

To show that it is isomorphic, we show that $FG^{(2)}/K$ has order less than $2n$. Let $\bar x=xK$ and $\bar y =yK$. Then $\bar x^n=1$, $\bar y^2=1$ and $\bar x\bar y\bar x\bar y=1$. Moreover, $ \bar y\bar x =\bar x^{-1} \bar y$ from the above implies that $\bar x^k =\bar x^{-k}\bar y$. Consider the set $\{\bar x^k,\bar x^k \bar y:1\leq k\leq n-1\}$. Then the product of any two elements is one of these elements, it contains $1$ and is closed under inverses. Hence it is a subgroup, but since it contains the generators $\bar x$ and $\bar y$, it is all of $FG^{(2)}/K$. Thus $|FG^{(2)}/K|\leq 2n$, which implies $D_n\simeq FG^{(2)}/K$.

Questions

$(1)$ I understand the reasoning by after "Hence $D_n$ is homomorphic..." but I don't know if I understand why this is so. Let $\eta$ be the natural homomorphism that sends $R\to x$, $S\to y$, and let $\zeta$ be $x\mapsto xK=\bar x$. We then use the homomorphism "natural" homomoprhism $R\mapsto \bar x$ and $S\mapsto \bar y$?

$(2)$ Can a similar argument be used on the exercise I am given? I see that $\Bbb Z^3$ is not finite at all, so I don't see how this can be used.

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1 Answer 1

up vote 2 down vote accepted

Since $x_3 \in Z(G)$ (by the second and third of the relations), the first relation $x_2x_1 = x_1x_2x_3$ implies that $x_2^{-1}x_1 = x_1x_2^{-1}x_3^{-1}$, $x_2x_1^{-1}=x_1^{-1}x_2x_3^{-1}$ and $x_2^{-1}x_1^{-1}=x_1^{-1}x_2^{-1}x_3$.

By using all of these relations, you can write any element of $G$ in the form $x_1^ix_2^jx_3^k$, with $i,j,k \in {\mathbb Z}$. Now it is straightforward to show that your homomorphism from $G$ to $Z$ maps the elements $x_1^ix_2^jx_3^k$ onto distinct elements of $Z$. So it is an injective homomorphism and, since you have already proved that it is surjective, it is an isomorphism.

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I see. So we're also exploiting the relations as Jacobson does to obtain a structure on $G$. Could you tell me about what I ask in $(1)$? –  Pedro Tamaroff Feb 23 '13 at 18:34
    
Then it doesn't matter to us what the normal subgroup generated by those relations is? How did we use the fact that the generators we chose fulfill these relations? –  Pedro Tamaroff Feb 23 '13 at 19:01
    
Also note that $x_1^{k}x_2^{l}x_3^m$ gets mapped into $(m,l,k)$. –  Pedro Tamaroff Feb 23 '13 at 19:02
1  
You wrote in your post that it is clear that the normal subgroup $N$, say, of $F:=FG^{(3)}$ generated by the relators is contained in the kernel of the homomorphism from $F$. That is where you used the fact that the generators of $Z$ fulfil the relations. Since $N$ is in the kernel, we get an induced homomorphism from $G = F/N$ to $Z$. We proved that $N$ is the exact kernel indirectly, by proving that the this induced homomorphism is injective. –  Derek Holt Feb 23 '13 at 20:26
    
I see. Thanks, Derek. This helped a lot. –  Pedro Tamaroff Feb 23 '13 at 22:04

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