Well, it's just good to know that when the chips are down and things look grim you'll feed off the girl who loves you to save your own ass!

Xander ,'Chosen'


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.


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.


Emily - Nov 07, 2005 4:12:17 pm PST #3359 of 10003
"In the equation E = mc⬧, c⬧ is a pretty big honking number." - Scola

It is. It's very bad. Cause, dammit, f(gf'(x)g') is not like gxg', no matter how hard I try to make it be!