Why does $x$ vanish in the group $\langle x,y\mid x^3, y^3, yxyxy\rangle$? 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?
 A: 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 $y = zx^{-1} = z$. 
A: 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.
A: 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 $x=e$.
A: Applying the way Artin used, you see from the following TC algorithm that $[G:H]=1$.

A: 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 $x=1$.
A: $e=y^{-1}(yxyxy)y=xyxy^{-1}$ so $y=xyx$ and $xy=yx^{-1}$; therefore $e=y^3=(xyx)^3=xyx^{-1}yx^{-1}yx=xxyxyyx=x^{-1}yxy^{-1}x$.  Cancelling on the left and right yields $x=e$ and the result follows.
