5

Problem says:

Let $G=\langle x,y|x^4=y^5=1,x^{-1}yx=y^2\rangle$. Show that $G$ is nonabelian group of order 20.

To show it, I tried to turn $x^ny^m$ into $y^kx^l$ for some $k,l$. Since I have $yx=xy^2$, $xy^m=(xy^2)y^{m-2}=\cdots =y^kxy^{m-2k}$. So, if $m$ is even, $xy^m=y^{m/2}x$. If $m$ is odd, then $xy^m=y^{(m-1)/2}xy$. So, I have 12 distinct elements of $G$, $1,x,x^2,x^3,y,y^2,y^3,y^4,xy,xy^3,yx,y^2x$. How could I do find more here? And I want to the general way to solve this type of problem.

Darae-Uri
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3 Answers3

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The third relation can be rewritten as $yx=xy^2$ and allows us to transform any word into an equivalent word of the form $x^ny^m$, and per the first relations, $0\le n\le 3$ and $0\le m\le 4$. This shows us that $G$ has at most twenty elements. To show that these are actually pairwise distinct, you could in principle spell out the full multiplication table. But you might also exhibit the group $C_5\rtimes C_4$ that can has generators obeying the given relations. Ultimately, once we know that our twenty suspected elements are really distinct, we see that $yx=xy^2\ne xy$, which shows that $G$ is not abelian.

Hagen von Eitzen
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4

A slightly more concrete presentation of the argument from Hagen von Eitzen's answer.

Consider the group (verify that it is subgroup of a suitable matrix group) $$ H=\left\{\left(\begin{array}{cc}a&b\\0&1\end{array}\right)\bigg\vert\ a,b\in\Bbb{Z}_5,a\neq0\right\}. $$ It has 20 elements. The matrix $$ Y=\left(\begin{array}{cc}1&1\\0&1\end{array}\right) $$ has order five. The matrix $$ X=\left(\begin{array}{cc}3&0\\0&1\end{array}\right) $$ has order four. I am leaving it to you to verify these and the remaining relation $X^{-1}YX=Y^2$ (observe that $3=2^{-1}$). Anyway, the relations are there meaning that we have a homomorphism $f$ from your group $G$ to this group $H$ such that $f(x)=X$ and $f(y)=Y$. To complete the proof you need to show that $X$ and $Y$ generate a group of order $20$, i.e. all of $H$. Then you can polish off:

  • $G$ has at most $20$ elements as per the calculations using the relations.
  • $G$ has at least $20$ elements because it has a homomorphic image $H$ with $20$ elements.
  • Ergo, $G$ has exactly $20$ elements and is also isomorphic to $H$ (in particular as $H$ is non-abelian, so is $G$).
Jyrki Lahtonen
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  • Here $Y$ generates a subgroup $\simeq C_5$ that is $\unlhd H$. The diagonal matrices form another subgroup $\simeq C_4$. The whole group $H$ is their semidirect product bringing us back Hagen's answer. Some readers may be unfamiliar with the semidirect product. That's why I chose to rewrite it using matrices. – Jyrki Lahtonen Apr 25 '16 at 14:31
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Given that $yx=y^2x$ every element can be written $x^ny^m$ with $n\leq 3$ and $m\leq 4$. That's exactly $20$ elements so it's enough to show that different elements have different writings.

karmalu
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