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The following was an exercise to check if a group is trivial or finite. The group is given by $$G=\langle x,y : yxy^{-1}=x^2, xyx^{-1}=y^2\rangle.$$ Question: Is $G$ a trivial group? or finite group?

My attempt We must have that $x$ and $y$ should be of finite and odd order. For this, we first show that $\langle x\rangle=\langle x^2\rangle$. If this is not true, then $\langle x^2\rangle$ is a normal subgroup of $G$, being normalized by both $x$ and $y$. Hence if $\bar{G}$ denotes $G/\langle x^2\rangle$, then in $\bar{G}$, we have generators $\bar{x}$ and $\bar{y}$ with relation $\bar{y}\bar{x}\bar{y}^{-1}=1.$ Hence $\bar{x}=1$. This means $x\langle x^2\rangle=\langle x^2\rangle$. Hence $x.x^2=x^{2k}$ which implies that $x$ is of finite and odd order.

Similarly $y$ is of odd order.

How to proceed to get whether $G$ is trivial finite or infinite?

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  • $\begingroup$ Doesn't a similar argument show that $\langle x\rangle$ itself is a normal subgroup, and that $G/\langle x\rangle$ is trivial, so that $y$ must be a power of $x$? $\endgroup$ – Greg Martin Oct 4 '16 at 7:01
  • $\begingroup$ The relations imply that $\langle x\rangle$ is normal in $G$. Show that the quotient $G/\langle x\rangle$ is trivial, so $G=\langle x\rangle$. Then show that, since $y$ is a power of $x$, the relation $yxy^{-1} = x^2$ implies that $x=1$, so $G$ is trivial. $\endgroup$ – James Oct 4 '16 at 7:36
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Observe that

$$\color{red}{x^2}=yxy^{-1}=y(xy^{-1}x^{-1})x=y(xyx^{-1})^{-1}x=yy^{-2}x=\color{red}{y^{-1}x}\implies x=y^{-1}$$

and from here that the group is the trivial one, since for example:

$$x^2=yxy^{-1}=yx^2\implies y=1\implies x=1$$

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Sadly, it's possible that even if x and y have finite order, the the generated group has infinite order (for example, in the group $G=⟨a,b : a^2 = b^2 = 1⟩$.)

But you have a little more information if you play with your given relations. More specifically, given any word w in your group, you can prove that there exists some integers k, l, such that $$w = x^ky^l,$$ where k is bounded by the order of x and l is bounded by the order of y.

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  • $\begingroup$ But here, $aba^{-1}$ is not $b^2$, isn't it? $\endgroup$ – p Groups Oct 4 '16 at 8:40
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Note that $$ x^4 = (yxy^{-1})^2 = y(x^2)y^{-1} = yyxy^{-1}y^{-1} \\ = (y^2)x(y^{-2})=xyx^{-1}xxy^{-1}x^{-1} = x(yxy^{-1})x^{-1} = xx^2x^{-1} = x^2 $$ so $x^2=e$ hence $yxy^{-1}=e$ and so $x=e$ and $y=e$. Hence the group is trivial.

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