Is this group abelian?

Question

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

Does the notation above mean 'for every pair, is is true that ...' or 'for every pair of distinct elements ...' ? — josinalvo, Mar 04 '13 at 01:35

score 23 · Accepted Answer · answered Mar 02 '13 at 17:34

23

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

answered Mar 02 '13 at 17:34

Tara B

5,341

Excellent hint/answer. +1 – DonAntonio Mar 02 '13 at 17:41

score 14 · Answer 2 · answered Mar 02 '13 at 17:44

14

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

enter image description here

answered Mar 02 '13 at 17:44

Mikasa

67,374

1

Rats, also this one is nice though not so elementary as Tara's...+1 – DonAntonio Mar 02 '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. :) – Mikasa Mar 02 '13 at 17:57
1

Very nice! ++++ – amWhy Mar 03 '13 at 00:04

Seirios · Answer 3 · 2013-03-02T20:12:33.083

7

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

edited Mar 02 '13 at 20:12

answered Mar 02 '13 at 17:51

Seirios

33,157

3

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 02 '13 at 17:52
OK, now I don't mind. =] Thanks! – Tara B Mar 02 '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 02 '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 02 '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... ;-) – Mikasa Mar 02 '13 at 18:00
@BabakS.: You mean answers with a 'spoiler' box? Seirios added that after I complained. I think it's good too. I only found out you could do that when someone suggested I use it in an answer on the computer science site. It doesn't seem to be used as much here. – Tara B Mar 02 '13 at 18:04
@Seirios: How about you make the spoiler box not include the first line of your answer? That way it can be partially a 'hint' answer, too. – Tara B Mar 02 '13 at 18:06
@TaraB: Yes. I saw that before and after you did. Nice job Seirios. +1 – Mikasa Mar 02 '13 at 18:07
@Seirios: Sorry, far too many comments here now, but I just noticed that you seem to have multiplied both sides by $x$ for no particular reason. You don't actually ever use that extra $x$. So you could shorten your proof by two lines without it. – Tara B Mar 02 '13 at 18:11
@TaraB: Thank for your comments, I improved my answer. – Seirios Mar 02 '13 at 18:17
@Seirios: Oh, I'm sorry. I thought I was probably being stupid (I'm quite tired today), and now that you've edited your answer, I see where you used the extra $x$ before and I actually like the old version better. This one is only shorter because you 'skipped' a bit at the end. – Tara B Mar 02 '13 at 18:21
@TaraB: I have no preference, so I edited to write my old argument. – Seirios Mar 02 '13 at 20:14
Sorry for all the comment mess. Also +1. =] – Tara B Mar 02 '13 at 20:49

Is this group abelian?

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