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Suppose $G$ is a group generated by elements $x$ and $y$ where $xy^2 = y^3x$ and $yx^3 = x^2y$ What can you prove about $G$?

I've just been playing around with the relations but I can't seem to get anywhere. It seems as if I'm going in circles. I don't know what to look for in terms of how the equations should look in order to give me an insight into the structure of the group.

So given a group presentation, are there any general strategies of what to look for in the relations (as in how to rearrange them) in order to obtain insight about the group?

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It's a pretty open-ended question, and therefore hard to answer. But, here are a couple of things to look at and how they apply in this example.

First, you should wonder whether the group is finite or not. If you've got more generators than relators (or relations) then your group is infinite. However, this does not apply here, since the numbers of generators and relations are both equal to two.

The next thing to look at is the abelianisation, $G/[G,G]$. For a finitely presented group, this is easy to do. First write your relations as relators: $$G = \langle x,y\mid x^{-1}y^{-3}xy^{2}, y^{-1}x^{-2}yx^{3}\rangle .$$ Now compute the relation matrix for the presentation. This is the matrix $(\sigma_{t}(w))$ of exponent sums of generators $t$ in each relator $w$ in the presentation. In this case, taking $w = x^{-1}y^{-3}xy^{2}$, for example, the exponent sum $\sigma_{x}(w)$ of $x$ in $w$ is $0$. (Just add up the exponents that appear on the letter $x$.) Similarly, $\sigma_{y}(w) = -1$. For the relator $w_{2} = y^{-1}x^{-2}yx^{3}$ we get $\sigma_{x}(w_{2}) = 1$ and $\sigma_{y}(w_{2}) = 0$. Therefore, the relator matrix is $$\begin{bmatrix} -1&0\\ 0&1\end{bmatrix}.$$ In general, you'd need to compute the Smith form for this to get the abelianised invariants, but in this case, there is nothing really to do, since it indicates that the abelianisation of $G$ is trivial. Therefore, you can conclude that $G$ is a perfect group (i.e., that $G = [G,G]$).

However, as soon as I see a finitely presented perfect group, I'm immediately suspicious that the group is itself trivial. (This need not be true, of course, but it's something to look into.) Therefore, you might try to show that the relations imply that the group is trivial.

Now, since it's me, I'd just go and ask a computer algebra system (such as Maple) to tell me.

> with(GroupTheory):                              
> G := < x,y | x.y^2 = y^3 . x, y.x^3 = x^2 . y >;
G :=

    < (x, y) |  (x^(-3).y^(-1).x^2.y, y^(-2).x^(-1).y^3.x)>

> GroupOrder( G );
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But you can try to work it out by hand also, and for this example, it's not too difficult. (In general, it can be very difficult!) From the relation $xy^2 = y^3x$ we get $xy = y^3xy^{-1}$, and from $yx^3 = x^2y$ we get $xy = x^{-1}yx^3$. Hence, $y^3xy^{-1} = x^{-1}yx^3$. From this last equality we get $xy^3x = yx^3y$, from which we obtain $x^2yx = x^2y^2$ by using the original relations twice. Now this implies that $x = y$ by cancelling $x^2y$ on the left. Putting this back into the original relation $xy^2 = y^3x$, we get $x^3 = x^4$, whence, $x = 1$. Therefore, $y = 1$, and the group is trivial.

[Full disclaimer: It helped a lot that I knew the answer before I started.]

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  • $\begingroup$ This is not my area and so this may be a basic question, but why is it that seeing you have a finitely presented perfect group makes you suspicious the group is trivial? $\endgroup$ – fourier1234 Jul 7 '19 at 19:34
  • $\begingroup$ @fourier1234 What I meant is that, once I had seen that the abelianisation is trivial, I would wonder whether it was because the group itself was trivial. Does that make more sense? $\endgroup$ – James Jul 10 '19 at 22:35
  • $\begingroup$ Yes it does, thanks for replying. $\endgroup$ – fourier1234 Jul 11 '19 at 8:54
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Here is something you can say. Any word in $x$ and $y$ built with positive exponents can be reduced to $$y^a(xy)^bx^c$$ with $a,b,c\geq0$. The first relation allows you to commute powers of $y$ that are greater than $2$ to the left at the cost of raising the exponent on $y$. The second relation allows you to do the same for $x$ but to the right. There is a similar statement to make for words built with negative exponents.

I'm not sure how helpful this is, especially since it is not applicable to most of the words in $x$ in $y$ (the ones that involve mixes of positive and negative exponents) but the question was open-ended.


You can also say that $$xy^3x=yx^3y$$

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