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This comes from Artin Second Edition, page 219. Artin defined $G = \langle x,y\mid x^3, y^3, yxyxy\rangle$, and uses the Todd-Coxeter Algorithm to show that the subgroup $H = \langle y\rangle$ has index 1, and therefore $G = H$ is the cyclic group of order 3.

That being the case, $x$ cannot be either $y$ or $y^2$, for then the third relation would not be satisfied. So the relation $x=1$ must follow from the given relations. Is there another way of seeing this besides from the Todd-Coxeter algorithm?

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I think you mean $G/H$ rather than $G=H$? –  Matthew Kahle Jan 11 '12 at 22:34
    
I mean that $H$ is defined as a subgroup of $G$, and in fact, $G = H$. Was that unclear? –  Dustan Levenstein Jan 11 '12 at 22:38
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4 Answers 4

up vote 17 down vote accepted

Let's see. We have $yxyxy=1$, so (multiplying by $y$ on the left) $y=y^2xyxy$, so (cancelling $y$ on the right) $y^2xy=x^{-1}$.

Also, $yxyxy=1$, so $yxyxy^2=y$, or $xyxy^2=1$. So $yxy^2=x^{-1}$.

It follows that $y^2xy=yxy^2$, or $yx=xy$. (So the group is Abelian.)

But then $1=yxyxy=x^2y^3=x^2$. Since $x^3=1$ as well, we finally conclude $x=1$.

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yeah, I definitely knew it would be sufficient to show that $x$ and $y$ commute. Nicely done! –  Dustan Levenstein Jan 11 '12 at 23:24
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Here's one way -- quite a bit ad hoc.

The basic idea is write all the elements in terms of $z := yx$. From the third relation, we can see that $z^2 y = 1$, or $z^2 = y^{-1} = y^2$. Therefore, $y = y^4 = z^4$. Now, we can also write $x$ in terms of $z$: $$x = y^{-1} z = y^2 \cdot z = z^8 \cdot z = z^9 .$$ Now $x$ and $y$ commute, both being powers of $z$. (It is a simple exercise to show that $x = 1$ from this. The last line in Andres' answer explains this.)


Here's an alternative approach, which is what I originally followed. Armed with these two identities, we can rewrite all three given relations entirely in terms of $z$:

  • $(z^{9})^3 = 1$;
  • $(z^{4})^3 = 1$; and
  • $z^2 \cdot z^4 = 1$.

From these observations, since $\gcd(27, 12, 6)=3$, we get that $z^3 = 1$. Finally, plugging this back, we obtain $x = z^9 = 1$ and $y = zx^{-1} = z$.

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In other words, assuming that $x^3=y^3=yxyxy=e$, the goal is to prove that $x=e$.

Note that $xyx=y^2(yxyxy)y^2=y^4=y$ hence $xy=(xyx)x^2=yx^2$ $(*)$.

Imagine one wants to carry every $y$ in $x=xy^3$ to the leftmost end of the product. Using $(*)$ twice, one first gets $$ x=xy^3=(xy)y^2=(yx^2)y^2=yx(xy)y=yx(yx^2)y=y(xy)x(xy), $$ and, again using $(*)$ twice, $$ x=y(yx^2)x(yx^2)=y^2x^3yx^2=y^3x^2=x^2. $$ Thus, $x=x^2$ and $x=e$.

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I computed the powers of $z = xy$:

$xy$, $y^2$, $x$, $y$, $xy^2$, $x^2$, $xy$, $y^2$, $1$.

This in particular gives $z^7 = z$ so $z^6 = x^2 = 1$, and $x^3 = x^2 = 1$ gives $x = 1$.

All you need from this though is

$z^2 = y^2$
$z^3 = x$

So $1 = y^6 = (z^2)^3 = (z^3)^2 = x^2$, then the last step of $x^3 = x^2 = 1 $ implies $x = 1$.

I posted this solution since the idea is just simplying that "playing around" with powers of $xy$ is algebraically easy and lets you do stuff. So I think computing up to its "period" of 9 is more insightful than just listing the final result.

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