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In the group $G=\langle x,y,z\mid xy=yx,zx=x^2z,zy=yz\rangle$, how can I prove that the elements of the form $x^iy^jz^k$ are all distinct?

As Arturo is asking I'm editing this question: I want to understand what is a set of normal forms. At the beginning I thought that it was $x^iy^jz^k$, but they made me notice that is not so obvious how $z^{-1}x$ can be written in this form. Actually I think I proved that it cannot be, in fact: if $z^{-1}x=x^iy^jz^k$ then conjugating by $z$ we obtain $xz^{-1}=zx^iy^jz^{k-1}=x^{2i}y^jz^k$ and so $z^{-1}=x^{2i-1}y^jz^k$ and so $1=x^{2i-1}y^jz^{k+1}$, I also proved that if $x^iy^jz^k=1$ then $i=j=k=0$, and so we have $2i-1=0$ that has no solution in $\mathbb{Z}$. So could you tell me how can I compute a set of normal forms? (they don't have to be too complicated because I'm asked to draw the cayley graph)

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Use the same technique I described in the answer to your other question:… – Ted Jan 29 '12 at 8:55
@Ted: I tried, but I didn't find any group that contains all the relations and in which $x^iy^jz^k\neq1$. – John Jan 29 '12 at 21:05
Remember that you don't have to use the same group for all $(i,j,k)$. To get started, you can try the trick suggested in the comments to my answer: If you put x=1, then you are left with just the relation yz=zy. Does that help? – Ted Jan 29 '12 at 21:22
@Ted: So G has a quotient isomorphic to $\mathbb{Z}^2$ and so all the elements of the type $y^jz^k$ are distinct. So why it remains to prove is that $<y,z>\cap<x>=1$ but if I follows the technique in the comment to your answer this doesn't work, because if I put $y=z=1$ then this force $x=1$, any idea? – John Jan 29 '12 at 21:32
Here's a better approach for this problem, which amounts to "guess the group, then verify it". From the relations, we obviously have a subgroup $\langle x,y \rangle$, and $z$ acts by conjugation on this subgroup. This suggests constructing a semidirect product, and then verifying that in this semidirect product, all $x^i y^j z^k$ are distinct. – Ted Jan 29 '12 at 21:41
up vote 2 down vote accepted

Consider the semidirect product $$G' = \langle X^i,Y \rangle \rtimes_{\varphi} \langle Z \rangle $$ where $i$ ranges over $\mathbb{Z}[1/2]$, powers of $X$ multiply in the obvious way and they commute with $Y$ (which has infinite order). The element $Z$ also has infinite order, and the action of $Z$ is $X^i Y^j \to X^{2i} Y^j$ (here $i \in \mathbb{Z}[1/2], j \in \mathbb{Z})$. Then $X, Y, Z$ satisfy the given relations (with the map $x \mapsto X$, etc., of course), and by construction, all $X^i Y^j Z^k \ne 1$ for $(i,j,k) \ne (0,0,0)$, hence $x^i y^j z^k \ne 1$.

Edit: As Jack Schmidt points out in the comments, the original construction with $i \in \mathbb{Z}$ didn't work because the action of $Z$ is not an automorphism (only even powers of $X$ would be in the image).

Edit 2: As Jack Schmidt points out in the comment to Manolito's answer, this can be fixed by letting $i$ range over $\mathbb{Z}[1/2]$ (rationals whose denominator is a power of 2) instead of just $\mathbb{Z}$. The map $G \to G'$ is surjective because $x^z$ has image $X^{1/2}$, $x^{z^k}$ has image $x^{1/2^k}$, etc., so $G'$ is a quotient of $G$.

Edit 3: In fact, we can prove that $G \cong G'$. It is sufficient to show that every element of $G$ has a normal form $$(x^a)^{z^b} z^c y^d$$ where $a$ is an odd integer and $b, c, d$ are arbitrary integers. These elements are distinct because the image of the above element in $G'$ is $X^{a/2^b} Z^c Y^d$, and the decomposition of an element of $\mathbb{Z}[1/2]$ into the form $a/2^b$, where $a$ is odd, is unique.

Take any word in $x,y,z$ and their inverses. First of all, since $y$ commutes with everything, we can just pull all the $y$'s out to the right. From now on I'll assume we just have $x$'s and $z$'s.

To avoid a lot of double superscripts, I'm going to write $x^{[a]}$ for $x^{z^a}$. Notice that the original relation $zx = x^2z$ can be written as $$x^{[-1]} = x^2$$ and also that we have the law $$(x^{[a]})^{[b]} = x^{[a+b]}.$$ The above identities continue to hold if $x$ is replaced by any integer power of $x$.

So we have a word in $x$'s and $z$'s (and their inverses). By repeatedly applying the identity $$z^{-a} x = x^{[a]} z^{-a}$$ we can pull the $z$'s out to the right as well, and we are just left with a product of elements of the form $x^{[a]}$. We must prove that such a product has the form $(x^b)^{[a]}$.

First of all, we have $$(x^2)^{[a]} = (x^{[-1]})^{[a]} = x^{[a-1]}$$ and the above is still valid if $x$ is replaced by any integer power of $x$, hence by iterating $$(x^{2^k})^{[a]} = x^{[a-k]}$$ for integers $k \ge 0$. Now the general product $$x^{[a]} x^{[b]} = x^{[a]} x^{[a-(a-b)]} = x^{[a]} (x^{2^{a-b}})^{[a]} = (x^{1+2^{a-b}})^{[a]}$$ valid for $a-b \ge 0$, has the desired form. If $a < b$, we rewrite $[a] = [b-(b-a)]$ instead, and do a similar calculation. The even more general product $(x^m)^{[a]} (x^n)^{[b]}$ is similar.

So elements of $G$ have the normal form $(x^a)^{[b]} y^c z^d$, and we just need to show that we can make $a$ odd. This follows from the identity $$(x^{2a})^{[b]} = ((x^a)^2)^{[b]} = ((x^a)^{[-1]})^{[b]} = (x^a)^{[b-1]}$$ which we can repeatedly apply to remove all the powers of 2 from $a$.

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How does $Z^{-1}$ act? Notice that the proposed mapping $X^i Y^j\mapsto X^{2i} Y^j$ is not surjective, since $X$ is not in its image. You'll need to fix the base group a bit. – Jack Schmidt Jan 30 '12 at 0:45
Since $(x^z)^2 = x$ we have that $x^z$ maps onto $X^{1/2}$ and $x^{z^k}$ maps onto $X^{1/2^k}$. – Jack Schmidt Jan 30 '12 at 11:49
@Jack Schmidt Oh, right... fixed. – Ted Jan 31 '12 at 3:58
@Jack Schmidt Thanks for correcting all my errors... In fact, it looks we have $G \cong G'$. I've tried to write this out carefully above; hopefully I haven't messed up this time. This has been a fun question. – Ted Jan 31 '12 at 6:19
Looks good. Thanks for finishing it out. In case you have to work with more 2-group things, I'll point out an annoyance with "2": If a=b then $x^{[a]} x^{[b]}$ does not have normal form $(x^{1+2^{(a-b)}})^{[a]}$ since $1+2^{(a-b)}$ is then even, $1+1=2$. Of course you just get the obvious relation, $x^{[a]} x^{[a]} = x^{[a-1]}$, but in some cases with 2-groups you actually get different and proof-breaking results. Generally 1+p^a is not divisible by p, but (p=2,a=0) is irritating. – Jack Schmidt Jan 31 '12 at 17:22

let's do it the intuitive way:

If you take some time to look at the relations, you will notice that every word in $x, y, z$ can be put into the form $x^iy^jz^k$, as $x$ and $y$ commute, and so do $y$ and $z$. Though $x$ and $z$ do not commute, strictly speaking, the relator $zx=x^2z$ lets you move all the $x$ to the left, doubling their exponents. Technically speaking, every word has a unique normal form $x^iy^jz^k$, and obviously, two such normal forms with different exponents are distinct.

From the relators you can also deduce that the group is the direct product of the infinite free group $\langle y \rangle$ (isomorphic to $\mathbb{Z}$) with the group $\langle x, z \mid zx=x^2z \rangle$.

Hope this helps.

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you said that obviously two normal forms with different exponents are distinct, could you explain me why? – John Jan 29 '12 at 21:04
@John: Compare the images in the quotient $G/\langle x,z\rangle$ to get the exponents of $y$ are equal; compare the images in the quotieng $G/\langle y\rangle$ to get that the exponents of $x$ and of $z$ are the same (since the expression $x^iz^j$ in $\langle x,z\rangle$ is unique). – Arturo Magidin Jan 29 '12 at 22:07
@John: Because you can prove the exponent of $x$ is unique by projecting onto groups of the form $C_{2^{n}-1}\rtimes C_2$, with $x$ generating the cyclic group of order $2^n-1$ and $z$ acting as the automorphism $x\mapsto x^2$; and you can prove the exponent of $z$ is unique by adding the relation $x=1$ and mapping onto the infinite cyclic group. – Arturo Magidin Jan 29 '12 at 22:29
In summary: Manolito's answer is right, but it takes some work to see that it is right. – Gerry Myerson Jan 29 '12 at 22:35
I think a normal form would be $z^{-a}x^bz^c$ with $a,b\gt 0$, with some extra restrictions; it is clear any word in $x$ and $z$ can be written this way, but that it is not unique (since $z^{-1}x^{2k}z = x^k$). – Arturo Magidin Jan 30 '12 at 0:30

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