Show that $\langle x,y\mid xyx=yxy,x^3=y^2\rangle\cong\{e\}$.

So we are trying to show that $\langle x,y\mid xyx=yxy,x^3=y^2\rangle$ is isomorphic to the trivial group which contains only the identity.

The only approach we can use by manipulating the relations $xyx=yxy,x^3=y^2$. For example, from $xyx=yxy$ we can get $xyxy=yx$. But I am clueless on how can we relate it to the trivial group?

My questions are:
1. How can we show the isomorphism?
2. I googled this statement and the Andrews-Curtis Conjecture appears quiet often, how does it relate to the conjecture? Is it potential counterexample to the conjecture?

Just to note: I have just learned some topics related to group generators and relations, but not the Todd-Coxeter Algorithm yet, so I might not be able to use some advanced tricks.

Helps are greatly appreciated. Many thanks!


$xyxy = yxy^2 = yx^4 = y^3 x$ by post-multiplying $xyx$ by $y$.

$xyxy = x^2 yx$ by pre-multiplying $yxy$ by $x$.

Therefore $y^3 x = x^2 y x$, so (by cancelling $x$ from RHS) $y^3 = x^2 y$, so (by cancelling $y$ from RHS) $y^2 = x^2$.

Therefore $\underbrace{y^2 = x^3}_{\text{relation}} = x^2$, so $x = e$ and hence $y=e$.


It can be shown that this is a representation of the trivial group by some brute force manipulations: \begin{eqnarray*} xyxy&=&(xyx)y=(yxy)y=yxy^2=yxx^3=yx^4=y^3x,\\ xyxy&=&x(yxy)=x(xyx)=x^2yx,\\ \end{eqnarray*} so $y^3x=x^2yx$. Right-multiplying both sides by $(yx)^{-1}$ shows that $x^2=y^2$. Hence $x^3=y^2=x^2$, from which it is immediate that $x=e$ and hence $y=e$, and we are done.

  • 1
    $\begingroup$ Ninja edit there ;) $\endgroup$ – Patrick Stevens Aug 27 '15 at 6:36

Multiply on the right by $x$ to get $yxyx=xyx^2$. Rewrite $yxyx=yyxy=x^4y=xy^3$ Therefore cancelling $xy$, from the relations above, to get $x^2=y^2=x^3$ so $x=1$, so $y^2=y$ and thus $y=1$.

  • $\begingroup$ Isn’t this just the same as the answers above (albeit 3+ years later?) $\endgroup$ – JavaMan Oct 21 '18 at 15:47

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