Cyclic Group Presentation

Show that the the group with presentation $$\langle x, y\ \mid\ x^2=y^2x^2y,\ (xy^2)^2=yx^2, \ yx^{-1}y^2=x^7\rangle$$ is cyclic of order 24.

This presentation was obtained using the Todd-Coxeter process for a subgroup of index 2 in the group presented in problem 476854.

• Hint: try drawing the Cayley graph with $x,y$ as generators. – msteve Jul 21 '14 at 19:12
• I think it might be easier to let $z=x*y^{-1}$ and eliminate $x$. – i. m. soloveichik Jul 21 '14 at 20:27
• @mesel If the question was interesting enough to attract the efforts of several people, it seems a likely candidate to remain open. – James Jul 22 '14 at 21:30

$$x^2=y^2x^2y$$

$$x=x^{-1}y^2x^2y=y^{-1}(yx^{-1}y^2)x^2y$$ $$x=y^{-1}x^9y$$

then we can say that;

$$y^2x^2y=x^2=y^{-1}x^{18}y$$ $$y^2x^2=y^{-1}x^{18}$$ $$y^3=x^{16}$$

Now, $$(xy^2)^2=yx^2$$ $$xy^2xy^2=yx^2$$ $$xy^2x(y^3)=yx^2y=y^3x^2y^2=x^{18}y^2$$ $$y^2x^{17}=x^{17}y^2$$ $$y^5x=xy^{5}$$

As a last step;

$$x^2=y^2x^2y$$ $$x^{18}=x^2y^3=y^3x^2=(y^5x^2)y=x^2y^6$$ $$x^2y^3=x^2y^6$$ $$e=y^3$$

Thus, we can say that $x,y^2$ will commute with each other so will $x,y^4=y$. And $x=y^{-1}x^9y=x^9\implies x^8=e$. From that point you can easily conclude that $G$ is an cyclic group of order $24$.

N.Q.E.D

• In the antepenultimate equality, I think you mean $x^{18}$ instead of $x^{14}$? Also, how to you get that $x$ and $y^2$ commute? The 1st identity says that $y$ and $x^2$ commute... – Nishant Jul 22 '14 at 3:14
• "N.Q.E.D."?${}$ – Najib Idrissi Jul 22 '14 at 8:26
• @Nishant: yeah it is $x^{18}$ before last step we have found that $x,y^5$ commute with each other and in last step we found that $y^3=e$ that is why $y^2,x$ commute each other. – mesel Jul 22 '14 at 10:04
• @NajibIdrissi: 'Not quite easily done' – mesel Jul 22 '14 at 10:10

First we manipulate the relations to show that the group $G$ defined by the presentation is the direct product $\langle x\rangle\times\langle y\rangle$, with $x$ of order a divisor of $8$ and $y$ of order a divisor of $3$. Then, to show that $x$ and $y$ have orders $8$ and $3$ exactly, we exhibit a homomorphism from $G$ onto a group in which the images of $x$ and $y$ have those respective orders.

Let's label the relations for easier reference, thus: $$x^2 = y^2x^2y;\tag{e1}\label{e1}$$ $$yx^{-1}y^2 = x^7;\tag{e2}\label{e2}$$ $$(xy^2)^2 = yx^2.\tag{e3}\label{e3}$$ From \eqref{e2} we get $$y^2 = xy^{-1}x^7.\tag{e4}\label{e4}$$ Using \eqref{e1} and \eqref{e4} we can derive the following. \begin{align} x^2 &\stackrel{\eqref{e1}}= y^2x^2y \\ &\stackrel{\eqref{e4}}= xy^{-1}x^7x^2y \\ &= xy^{-1}x^9y, \end{align} which implies $$x = y^{-1}x^9y \;\;\text{and}\;\; x^{-1} = y^{-1}x^{-9}y.\tag{e5}\label{e5}$$ Using \eqref{e5} in \eqref{e2}, we obtain $$x^7 \stackrel{\eqref{e2}}= yx^{-1}y^2 \stackrel{\eqref{e5}}= y(y^{-1}x^{-9}y)y^2 = x^{-9}y^3.\tag{e6}\label{e6}$$ From this follows the crucial relation $$x^{16} = y^3.\tag{e7}\label{e7}$$ In particular, $y^3$ commutes with $x$ (and hence, with its powers).

Next, we use \eqref{e2} and \eqref{e7} as follows: \begin{align}\tag{e8}\label{e8} x^7y &\stackrel{\eqref{e2}}= yx^{-1}y^3 \\ &\stackrel{\eqref{e7}}= yx^{-1}x^{16} \\ &= yx^{15}. \end{align}

Now, because $y^3$ commutes with $x$, we can write $$x^2y^2 \stackrel{\eqref{e1}}= y^2x^2y^3 = y^5x^2,$$ and then, using that $y^3$ commutes with $x^{-1}$ (and \eqref{e2}), we get $$x^7y^{-5} \stackrel{\eqref{e2}}= yx^{-1}y^3 = y^{-2}x^{-1} = (xy^2)^{-1}.$$ Taking inverses, this leads to $$xy^2 = y^5x^{-7}.\tag{e9}\label{e9}$$ But, \begin{align} yx^2 &\stackrel{\eqref{e3}}= (xy^2)^2 \\ &\stackrel{\eqref{e9}}= y^5x^{-7}y^5x^{-7} \\ &\stackrel{\eqref{e2}}= y^5(y^{-2}xy^{-1})y^5(y^{-2}xy^{-1}) \\ &=y^3xy^2xy^{-1}; \end{align} whence $$x^2 = y^2xy^2xy^{-1} = (y^2x)^2y^{-1}.\tag{e10}\label{e10}$$ Now, \begin{align} yx^2 &\stackrel{\eqref{e3}}= (xy^2)^2 \\ &\stackrel{\eqref{e9}}= y^5x^{-7}y^5x^{-7} \\ &\stackrel{\eqref{e2}}= y^5(y^{-2}xy^{-1})y^5(y^{-2}xy^{-1}) \\ &= y^3xy^2xy^{-1}, \end{align} so that $$x^2 = y^2xy^2xy^{-1} = (y^2x)^2y^{-1}.\tag{e11}\label{e11}$$ Therefore, $$y^2x^2y \stackrel{\eqref{e1}}= x^2 \stackrel{\eqref{e11}}= (y^2x)^2y^{-1},$$ which, upon cancelling $y^2x$ on the left leaves us with $xy = y^2xy^{-1}$ or, equivalently, $$xy^2 = y^2x.$$

Recall, however, that also $y^3$ commutes with $x$ so, in fact, $y$ itself commutes with $x$.

We're almost done. Using the knowledge that $x$ and $y$ commute, the first relation \eqref{e1} gives us that $x^2 = x^2y^3$, or $y^3=1$. Then, from $y^3 = 1$ and \eqref{e2} we get $$x^7 = yx^{-1}y^2 = x^{-1}y^3 = x^{-1},$$ so that $x^8 = 1$.

This shows that the group $G$ defined by your presentation is abelian, generated by $x$ of order dividing $8$ and $y$ of order dividing $3$. Since the orders of $x$ and $y$ are coprime, we have $$\langle x\rangle\cap\langle y\rangle = 1$$ in $G$, and so $G = \langle x\rangle\times\langle y\rangle = \langle xy\rangle$ is cyclic. We also know that the order of $G$, which is equal to $\mid\langle x\rangle\mid\cdot\mid\langle y\rangle\mid$, is a divisor of $24 = 8\cdot 3$.

Finally, to show that the orders of $x$ and $y$ are indeed $8$ and $3$, respectively, I'll leave it as an exercise to show that the map $$G\to\langle (1,2,3,4,5,6,7,8)(9,10,11)\rangle$$ defined on the generators by $$x\mapsto (1,2,3,4,5,6,7,8)\;\;\text{and}\;\; y\mapsto (9,10,11)$$ does indeed define a homomorphism. (Just check that the permutation images of $x$ and $y$ satisfy the relations.)

• +100k Very good James. As I commented above, I did these tedious manipulations on 7 sheet of papers but couldn't find the right way. Awful relations. (-: – mrs Jul 22 '14 at 12:17