Spike's Bitches 26: Damn right I'm impure!
[NAFDA] Spike-centric discussion. Lusty, lewd (only occasionally crude), risque (and frisque), bawdy (Oh, lawdy!), flirty ('cuz we're purty), raunchy talk inside. Caveat lector.
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.
she was in the middle of telling me she may have cancer
Oh hell, Emily. I'll cross my fingers on the cat scratch??? And maybe some math~ma because it sounds funny in my head.
You file for unemployment online in California. It's pretty easy.
Oh good. My nap has made me just feel edgy and cranky, so good news is basically good right now. And I think I have a headache from crying earlier. Clotted cream seems like it would help.
they are perky because they are not large enough to be used shelfishly.
Don't be shelfish! Share with the rest of the 'fistas...
