EEEE!!!! overdose of cutons!!! [link]
'Jaynestown'
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.
Eeee! An adorable Kitty!Bel!
My turn, my turn. Here are my big cutie heads.
eta - have fixed the access. I will switch it back to friends only in a bit since it is pics of the kids - but it is public for now.
Taunter!
(IOW, getting a 403, you don't have permission page)
Everyone's babies (big and little) are so cute!
I'm going to stagger over to my bed and keel over from a massive lethal cuteness overload.
Hil, that sounds awful. Do you have a medical advice place you can call? My insurance used to have a 24-hour nurse line, for medical questions where you didn't think you needed to go to the emergency room, but weren't sure.
Er, and I have a math question, but I feel like maybe you should be resting your eye.
Ahem. And now that I thought to look, apparently that was an hour ago. Oh well.
K-Bug looks really, really dead, there. Hope she scared lots of people! CJ is just too cute for words. You make pretty (big) babies, Maidengurl.
Skippity Skim...my sister's dial up SUCKS! It sucks for viewing pictures, dammit!
Owen is the boy you want Em to marry, but Leif is the boy she ends up bringing home.
I love Gud.
Halloween pictures will be uploaded when we return home tomorrow night. Then there will be catching up. Maybe.
Emily, I do have a medical advice line, but I lost the number. My eye is feeling a bit better now -- still sort of a "something isn't right here" feeling, but not really hurting anymore.
What's your math question?
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?