Take jobs as they come -- and we'll never be under the heel of nobody ever again. No matter how long the arm of the Alliance might get, we'll just get ourselves a little further.

Mal ,'Out Of Gas'


Spike's Bitches 27: I'm Embarrassed for Our Kind.  

[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.


Emily - Nov 07, 2005 3:05:06 pm PST #3349 of 10003
"In the equation E = mc⬧, c⬧ is a pretty big honking number." - Scola

Yay! Okay, here are my two remaining issues:

If a finite group has only one subgroup of a particular order, how do I know that subgroup is normal? Also, I need a group which has no elements of finite order greater than one, but which has a factor group with all finite elements. Any takers?

(Not that I think any subgroup is abnormal. I would never want to put that kind of pressure on a subgroup, which after all is constantly living in the shadow of the Big G, which is having its own issues about the lingering possibility of being called simple.)


Trudy Booth - Nov 07, 2005 3:13:46 pm PST #3350 of 10003
Greece's financial crisis threatens to take down all of Western civilization - a civilization they themselves founded. A rather tragic irony - which is something they also invented. - Jon Stewart

The nice thing about comfort food is that it's actually comforting.

Chow down, ladies.


Gris - Nov 07, 2005 3:18:23 pm PST #3351 of 10003
Hey. New board.

Okay. For the first one, you have a subgoup N of G that is of order k. Take g in G and n in N such that g^-1*n*g is not in N. Then the group defined by g^-1*n2* g for each n2 in N will be a group of order k (I'm 97% sure - showing it is your job) that isn't N, a nice contradiction.

For the second one, um, I dunno. I'd guess there's some way to show that (Z, +) works, but I'm not sure. Sorry, I'm not so great with this stuff.


Emily - Nov 07, 2005 3:32:50 pm PST #3352 of 10003
"In the equation E = mc⬧, c⬧ is a pretty big honking number." - Scola

I was heading in that direction, but hadn't quite managed to make it into a subgroup of its own. Thank you! Now the second one... rrrrgh. Part of the problem is that I keep getting messed up by the word element -- I understand an element of G, but is an element of the factor group G/H a coset? So that would translate to finding an infinite group with a finite subgroup -- I can do that, I think, but I want to make sure what's being measured here.


Gris - Nov 07, 2005 3:35:23 pm PST #3353 of 10003
Hey. New board.

Hmm. I'll think on it and get back to you if I tink of something. Maybe do some reviewing of terms. For now, walking back to my apartment, so it'll be a few.


SuziQ - Nov 07, 2005 3:38:48 pm PST #3354 of 10003
Back tattoos of the mother is that you are absolutely right - Ame

True tales of insanity.

My mom had some medical equipment delivered today, including an IV pole. It came in a few pieces. Mom asked K-Bug to unpack it for her.

K-Bug stood there, looking at the pieces and asked where the ivy was.

Oy!!!!


Cashmere - Nov 07, 2005 3:40:46 pm PST #3355 of 10003
Now tagless for your comfort.

K-Bug stood there, looking at the pieces and asked where the ivy was.

That's funny!


Emily - Nov 07, 2005 3:55:05 pm PST #3356 of 10003
"In the equation E = mc⬧, c⬧ is a pretty big honking number." - Scola

Just in case you're wondering D., I've decided to go with the factor group G/{e}, because the question says G can't have any elements of finite order >1, but the factor group has to have elements of finite order. No ">1" in that part of the sentence. So I feel like it's looking for something else, but that's what I've come up with.

Now off to spin some bullshit about the history of math in India.


Emily - Nov 07, 2005 3:59:29 pm PST #3357 of 10003
"In the equation E = mc⬧, c⬧ is a pretty big honking number." - Scola

Oh hell. I forgot I hadn't finished another problem. I've gotten myself in a bind over showing that the inner automorphisms of G form a normal subgroup of the group of automorphisms of G under function composition. Damn.


Connie Neil - Nov 07, 2005 4:09:09 pm PST #3358 of 10003
brillig

I've gotten myself in a bind over showing that the inner automorphisms of G form a normal subgroup of the group of automorphisms of G under function composition.

Current leader of my list of English Sentences That Make No Sense To Me. All I know is that it sounds bad.