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I tried to show that the following group is abelian by manipulation the relations but they didn't work. Please show me the right way. The group is $$G:=\left<x,y \mid xyxy^2=yxyx^2=1\right>$$

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Thank you for the edit. –  Nancy Rutkowskie Mar 2 '13 at 17:45
You're welcome! –  Andreas Caranti Mar 2 '13 at 18:25
Yea abelian @NancyR –  B11b Mar 3 '13 at 1:16
Does the notation above mean 'for every pair, is is true that ...' or 'for every pair of distinct elements ...' ? –  josinalvo Mar 4 '13 at 1:35

3 Answers 3

up vote 20 down vote accepted

HINT: From $xyxy^2 = 1$, you get $xyx = y^{-2}$. Try substituting this into $yxyx^2 = 1$.

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Excellent hint/answer. +1 –  DonAntonio Mar 2 '13 at 17:41

Sorry for this kind of answer. @Tara's hint is enough but mine is base on Van Kampen diagram.

enter image description here

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Rats, also this one is nice though not so elementary as Tara's...+1 –  DonAntonio Mar 2 '13 at 17:54
@DonAntonio: Thanks Don. I really ouldn't draw so I arranged it on a paper and this is one of my paintings. Thanks again. :) –  B. S. Mar 2 '13 at 17:57
Very nice! ++++ –  amWhy Mar 3 '13 at 0:04

Hint: You can identify $xyxy^2$ as a subword of $yxyx^2$. In details:

$$\begin{array}{ll} yxyx^2=1 & \Rightarrow xyxyx^2=x \\ & \Rightarrow (xyxy^2)y^{-1}x^2=x \\ & \Rightarrow y^{-1}x^2=x \\ & \Rightarrow y^{-1}x=1 \\ & \Rightarrow x=y \end{array}$$

So $G \simeq \langle x \mid x^5=1 \rangle \simeq \mathbb{Z}_5$.

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Noooo! Please please please don't provide full answers to questions where someone (especially me =] ) has given a nice little hint, and the answer with the hint has been accepted. It spoils all the fun. –  Tara B Mar 2 '13 at 17:52
OK, now I don't mind. =] Thanks! –  Tara B Mar 2 '13 at 17:55
it's not so bad, Tara. In fact, your hint is almost a complete answer as it follows from it at once that the group is cyclic (and generated by $\,x\,$)... –  DonAntonio Mar 2 '13 at 17:55
@DonAntonio: Yeah, I know. I've just had this happen quite a few times, sometimes when there was a little more to do. This wouldn't have bothered me much if hadn't been for those other times. –  Tara B Mar 2 '13 at 17:56
@TaraB: I always like this kind of answer. I remembered I saw it for the first time done by Brian. It really help the OP reflect about hints. But as always, Mouse can't wait not to move on gray area and... ;-) –  B. S. Mar 2 '13 at 18:00

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