If G and G' are finite groups, and there's a homomorphism from G to G', I need to show that the cardinality of the image of G is a divisor of the cardinality of G. My thinking (about the case where it's not one-to-one or trivial) so far has been this: there's an equivalence relation, where a and b in G are in the same cell if they map to the same element of G', and then (and here's the fuzzy part) you can pick one representative of each cell in such a way that they make up a subgroup H which has the same cardinality as the image of G. And then... er, I thought I had something which came next relating |H| to |G|, but now I'm not sure. In any case, am I even on the right track?
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.
Hmm. I'm not sure, but that does seem like it's probably the right track.
Well, does the cardinality of a finite subgroup have to be the divisor of the cardinality of the finite group it's in, or is that only for cyclic groups?
I have this sneaking suspicion that the answer should have something to do with cosets, but honestly, my brain's not taking to cosets very well.
Hil, probably the contact was too dry, and it tried to take your eyeball with it. Y'know, or something. If it still feels bad tomorrow, after you have your eyes closed for sleeping a decent amount (assuming/hoping you're sleeping a decent amount), then you probably want to do something about it...
My friend came over and talked to me while I cleaned, and put clothes on hangers, and in the hour she was here, I got more done than I have in WEEKS. See? I just need someone to come guilt/entertain me. Otherwise I end up on the internet...
Well, does the cardinality of a finite subgroup have to be the divisor of the cardinality of the finite group it's in,
Yes. At least, I'm almost positive it does. It's been awhile since I've really looked at algebra, but I'm just about certain that's true.
Lillian pictures up: [link]
Back from trick or treating. Best. Halloween. Ever.
Mainly because Emmett was totally committed to his Harpo performance, never broke character when he whistled "thank you" and gurned on command. He was an especial hit with grandpas who declared things like, "Harpo! You get EXTRA candy!"
Cutiebel!
Love the kitty costume. Everyone has such cute kids!
Ellie looks so wary in her peapod. And a certain young lady looks just beautiful in her black dress. Then we have Em, FEAR HER EVIL PINK CUTENESS!
Annabel approves of the Lilybean.
Me: Annabel, there's Lillian.
Annabel: (pointing to the screen) Hi!
Yes. At least, I'm almost positive it does. It's been awhile since I've really looked at algebra, but I'm just about certain that's true.
All right. Well, I'll put what I have, and he'll likely give me some credit for it, anyway. It's not even that big a deal (especially since I didn't hand in a whole chapter's worth of homework), but I don't like to leave it blank. Thank you! Also, good luck with the eye. Sounds like you probably just irritated it a bunch.