You didn't learn that notation yet? (12) is the permutation that switches 1 and 2 and does nothing to 3. (123) is 1 goes to 2, 2 goes to 3, 3 goes to 1. Etc. Each element goes to the place where the thing to its right was, until you get to the end of the parenthesis and then it cycles back to the beginning. So, in the group of four elements, (12)(34) would switch 1 and 2 and switch 3 and 4.

Ack! I got to the end of the thread and it's all mathy!! Eek!

Yes. The husband type person is watching the Yankees/Angels game. I seek refuge here, only to be mathed at.

Also, my pun got no love, and now I'm worried that it wasn't clear what the "stinks" was in reference to, and that I've offended everyone. It was supposed to be P-C saying it stinks that he missed the swag.

Now I'm all over explainy.

But still? I math not.

Huh. Nope. Or, rather, maybe, but like I said I missed the class. So I get that I can find a noncommuting pair for any particular set, but how do I go about proving the claim for all n > 2?

Tell me if this makes any sense: Dn must be a subset of Sn (right?). Then there's some permutation (reflection or diagonal flip) such that vertex x goes to itself, but all the other vertices change, and also some premutation (rotation) such that all vertices change by 1. Then I can show these don't commute. Would that suffice?

Cindy, I liked the pun.

I'm off to bed. I know--it's ridiculously early. But last night I went to bed early and felt human this morning so I'm going to try for it again tomorrow.

Ack! I got to the end of the thread and it's all mathy!! Eek!

Sorry. It's just me, and, well... let's just say it gets increasingly hard to get homework help as I take more classes. This is the last problem, I promise (well, other than the two I have unsatisfactory answers for).

Look! Turtles!

(Oh, and I totally got that the stinkiness referred to the missing stuff, not the missed stuff.)

Cindy, I liked the pun.

Thank you. Of course, it was a limerick, but we knew that. We were just testing those others.

Also, my pun got no love, and now I'm worried that it wasn't clear what the "stinks" was in reference to, and that I've offended everyone. It was supposed to be P-C saying it stinks that he missed the swag.

I got it! Although at first, I read it as Pete saying that.

Hmm. I guess what I was saying is essentially the same as (123) and (23), in that (123) composed with (23) gives (132), but (23) composed with (123) gives, er... (213).

(Which looks enticingly like an area code, but I don't know for where. 212 I would know, but that's not a valid permutation. Mind you, 617 is, proving once again its supremacy! Go Sox!)

Tell me if this makes any sense: Dn must be a subset of Sn (right?). Then there's some permutation (reflection or diagonal flip) such that vertex x goes to itself, but all the other vertices change, and also some premutation (rotation) such that all vertices change by 1. Then I can show these don't commute. Would that suffice?

The thing about all vertices but one changing would only work if n is odd. If you're doing it that way, then you'd probably have to do two cases, one for even n and one for odd n.

So I get that I can find a noncommuting pair for any particular set, but how do I go about proving the claim for all n > 2?

The way I was going about it was finding the smallest group which would be a subgroup of any Sn with n greater than 2, i.e. S3, then showing that that's nonabelian, and therefore any group that contains it is nonabelian.

All the above just goes to prove that you never know what kind of conversation you're going to wander into around here.