Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$G:=\left<x,y \mid xyxy^2=yxyx^2=1\right>$$

share|improve this question
    
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
add comment

3 Answers

up vote 20 down vote accepted

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

share|improve this answer
    
Excellent hint/answer. +1 –  DonAntonio Mar 2 '13 at 17:41
add comment

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

enter image description here

share|improve this answer
1  
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
add comment

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

share|improve this answer
2  
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
    
+1..................... –  DonAntonio Mar 2 '13 at 17:54
    
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
show 9 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.