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"Check in detail that $\langle x,y \mid xyx^{-1}y^{-1} \rangle$ is a presentation for $\mathbb{Z} \times \mathbb{Z}$." Exercise 27.5 from "Groups and Symmetry" M.A.Armstrong.

This should be an easy exercise but I'm completely unable to answer it. I know the group would be equal to the quotient group $F(X)/N$ where $X$ is an alphabet and $N$ is the smallest normal subgroup of $F(X)$ that contains $xyx^{-1}y^{-1}$. However these are just words, I'm not sure how to proceed.

The free group $F(X)$ is an absolutely huge group; if I understand correctly it would contains elements like these:

$x^5y^2x^9y^4x^7yxyxy^{-8}x^{-8}y^4xy^7xyx^{-2}yx^{-9}y^{-4}$

Even if you would quotient out $xyx^{-1}y^{-1}$ you would still be left with absolutely huge words, and it's entirely unclear to me why this would be isomorphic to $\mathbb{Z} \times \mathbb{Z}$. An added difficulty would be how to prove in general that the group we quotient by is normal; in this specific case I think we'd use the fact that $xyx^{-1}y^{-1}$ denotes the commutator but I'd be lost in a more general case.

Any help would be much appreciated.

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Take into account that when you divide that huge group by its commutator subgroup, the elements $\,x,y\,$ commute, so that the elements you wrote is simply $\,x^6y^p\,$...not so terrible, uh? –  DonAntonio Jan 20 '13 at 9:30
    
You don't have to prove that what you quotient out is normal, since that is true by definition. The group $G = \langle X \mid R \rangle$ is defined to be $F/N$, where $F$ is free on $X$, and $N$ is the normal closure of $R$ in $F$, which is the smallest normal subgroup of $F$ containing $R$. –  Derek Holt Jan 20 '13 at 11:49
    
@DerekHolt: Can we deduce like this: "Cosider $H=\langle x\rangle$. According to $G$'s relation $H$ is normal in $G$ and the presentation of $G/H$ is $\langle y \mid y=1\rangle\cong\mathbb Z$ so $G$ is s semi direct product of $H\cong\mathbb Z$ by $\langle y\rangle\cong\mathbb Z$. Since two copies of $\mathbb Z$ here are normal in the group so $G\cong\mathbb Z\times\mathbb Z$". Thanks. –  B. S. Jan 20 '13 at 14:41
    
@DerekHolt: Sorry for the nice wrong in my comment. I meant $G/H=\langle y\rangle\cong\mathbb Z$. –  B. S. Jan 20 '13 at 19:53
    
@Babak: Yes you could do it by observing that this presentation defines a semidirect product, which is in fact a direct product. But it is very difficult to answer this type of question without being told what standard results the OP already knows about group presentations. –  Derek Holt Jan 21 '13 at 9:25

3 Answers 3

up vote 3 down vote accepted

In the definition of a presentation of a group by generators and relations, we take the quotient of the free group on the generators by the smallest normal subgroup generated by the relations. This has 2 consequences: (a) normality of the subgroup we quotient by is not a problem; (b) the subgroup we quotient by is huge.

This subgroup contains not just $xyx^{-1}y^{-1}$ and its powers, but (because of normality) it also contains anything of the form $wxyx^{-1}y^{-1}w^{-1}$ where $w$ is an arbitrary word in the free group on $x,y$, as well as any product of any finite number of words of that form (and their inverses), and conjugates of those products by any words, and any products of those things, and their conjugates, and so on.

So, the free group is huge, but so is the group we quotient by, so it's reasonable to expect that we get something "small" like $\mathbb{Z} \times \mathbb{Z}$.

In fact, the best way to think about groups presented by generators and relations is not to think explicitly about what we quotient by. Intuitively, the reason this quotient group is $\mathbb{Z} \times \mathbb{Z}$ is because the relation $xyx^{-1}y^{-1}=1$ implies $xy=yx$, which allows us to rearrange the letters in a word inside the free group into the form $x^a y^b$. The product becomes $(x^a y^b) (x^c y^d) = x^{a+c} y^{b+d}$ so the group is $\mathbb{Z} \times \mathbb{Z}$.

A precise argument use the following universal property of quotient groups: Let $H$ is a normal subgroup of $G$. Suppose $K$ is an arbitrary group for which we have a homomorphism $f : G \to K$ with kernel containing $H$. Then there exists a unique homomorphism $\overline{f}: G/H \to K$ such that the following composition is $f$: $$G \to G/H \to K.$$ Here the first map in the composition is the natural projection map from $G$ to $G/H$, and the second is $\overline{f}$.

Intuitively, this universal property is saying that $G/H$ is the "largest" group for which we have a surjective homomorphism from $G$ with kernel containing $H$, because any other group with this property "factors through" $G/H$, as in the 2-step composition above.

All of this means that in your example problem, we must show that $\mathbb{Z} \times \mathbb{Z}$ is the largest group which is generated by 2 elements $x', y'$ with the property $x' y' x'^{-1} y'^{-1} = 1$. To prove that this largest group is $\mathbb{Z} \times \mathbb{Z}$ means that:

(1) there is a surjection $F_2 \to \mathbb{Z} \times \mathbb{Z}$.

(2) for any other group $G$ has this property (with 2 elements $x'', y''$ such that $x'' y'' x''^{-1} y''^{-1} = 1$), there is a map $\overline{f}$ from $\mathbb{Z} \times \mathbb{Z}$ to $G$ such that this composition takes takes $x$ to $x''$ and $y$ to $y''$: $$F_2 \to \mathbb{Z} \times \mathbb{Z} \to G$$

(where the first map is the surjection in (1) and the second is $\overline{f}$.)

This is much easier because we don't have to think directly about all the words in the complicated normal subgroup generated by $xyx^{-1}y^{-1}$. To prove (1), we find 2 elements $x', y'$ that generate $\mathbb{Z} \times \mathbb{Z}$ and satisfy $x' y' x'^{-1} y'^{-1} = 1$. Then by the universal property of the free group, there is a unique map $F_2 \to \mathbb{Z} \times \mathbb{Z}$ taking $x \mapsto x'$ and $y \mapsto y'$. So we get our map (1). Then for (2), we have an arbitrary group $G$ with 2 elements $x'', y''$ satisfying our relation. By using that relation, we can reduce any element of $G$ to the form $(x'')^a (y'')^b$. Check that we must send $(a,b) \mapsto (x'')^a (y'')^b$ to get a homomorphism as in (2).

This was long and I've left out many details at the end, but I hope it was helpful.

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Thank you for your long and insightful post. I still have a few questions. Proof: define the function f such that it sends x to (1,0) and y to (0,1). It's obviously surjective. (1) done. Now for (2), consider a group G with x'' and y'' satisfying the equation. Now since $xyx^{-1}x^{-1}=e$ we find $xy=yx$ and the group is abelian. So it is "obvious" (obvious enough?) that any element can be reduced to (x'')^a (y'')^b. Because $x^ay^bx^cx^d=x^{a+c}y^{b+d}?$ For (2) there is a composition of function f' such that:$ f(x^ay^bx^cy^d)=(a+c,b+d)$ and $fbar(a+c,b+d)=x''^{a+c}y''^{b+d}.$ (cont.) –  Lee Wang Jan 20 '13 at 12:53
    
If I understand the idea correctly just showing that there is a composition of functions $F_2\mapsto Z \times Z \mapsto G$ is enough to prove that we don't quotient out too much. The devil is in the details however so I'm not sure. Also I'm not entirely sure how to extend the idea that commutativity reduces all words to $x''^ay''^b$ to words of any length. Finally is this technique applicable to all problems of a similar nature? That is checking if a particular presentation correctly represent a group? –  Lee Wang Jan 20 '13 at 12:57
    
Similar techniques work for checking any presentation. In general, you make a guess for the group (by playing around with the generators and relations) and check that you have a surjective map as in (1). To get the map in (2), usually you come up with some sort of "normal form" like $x^a y^b$ for all elements, and then map this to the corresponding element with the "same" form $x''^a y''^b$ in $G$. The tricky part in (2) is making sure that the normal form is really normal, i.e., you don't have 2 different $a$ and $b$ that equal the same element in your guessed group. (cont) –  Ted Jan 20 '13 at 18:43
    
Otherwise when you map $x^a y^b \mapsto x''^a y''^b$ you might be sending the same element to 2 different elements in $G$, which would be bad. For a more difficult problem of this sort where it's harder to guess the group, see: math.stackexchange.com/questions/103525/… –  Ted Jan 20 '13 at 18:47

Hint: $\mathbb{Z} \times \mathbb{Z}$ is the abelianization of $\mathbb{F}_2$, so consider the quotient $\mathbb{F}_2/D(\mathbb{F}_2)$.

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The main thing you need to show is that $\langle x,y|xyx^{-1}y^{-1}\rangle$ is abelian. From this you get an homomorphism into $\mathbb Z^2$ (send $x$ to $(1,0)$ and $y$ to $(0,1)$), and it is easy to show that this is an isomorphism. To show that $\langle x,y|xyx^{-1}y^{-1}\rangle$ is commutative, you want to show that all its commutators vanish. This is equivalent to all commutators of $\langle x,y\rangle$ being in the normal subgroup generated by $xyx^{-1}y^{-1}$. Try writing out the commutators of some elements and seeing if you can reduce these to something like $xyx^{-1}y^{-1}$ by conjugating. This should give you an idea of how to proceed.

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To avoid having to manipulate commutators and words,you can show that any group $G = \langle X \rangle$ in which the elements of $X$ commute is abelian, by using the facts that centralizers of elements and $Z(G)$ are subgroups of $G$. The centralizer of any $x \in X$ contains $X$ and hence contains $\langle X \rangle =G$, so $X \subseteq Z(G)$ and hence $G = \langle X \rangle \subseteq Z(G)$. –  Derek Holt Jan 20 '13 at 11:47

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