#exult@irc.freenode.net logs for 24 Mar 2004 (GMT)

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[19:21:20] <wjp> hi
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[21:50:05] <Fingolfin> wjp: hm... you don't happen to know this, do you...? Assume you have a finite group G, and a normal (=invariante) subgroup N of G.... consider the quotien (=factor) group G/N ... when is G/N isomorphic to a subgroup to G ?
[21:51:09] * Fingolfin wonders if #math is any good for such questions
[21:55:42] <wjp> hm, G is finite? *thinks*
[21:56:21] <Fingolfin> I've been thinking about it all day, after a 1st year student asked me, and for the hell of it, I can't think of the answer (i.e. besides guessing, of course <g>)
[21:56:23] <wjp> any idea how explicit you want the answer?
[21:56:29] <Fingolfin> and I haven't found any good answers either
[21:56:32] <Fingolfin> well...
[21:56:53] <wjp> it's equivalent to \exists M \subset G: G = N \oplus M, I guess
[21:56:58] <Fingolfin> I don't want a full proof... but for now, I have no good idea to tackle this... of course it's true for simple groups <g>
[21:57:03] <Fingolfin> indeed
[21:57:06] <Fingolfin> I was that far, too :-)
[21:57:22] <Fingolfin> an abelian group can be decomposed into a sum of cyclic groups
[21:57:25] <Fingolfin> which probably helps
[21:57:43] <wjp> abelian is fairly easy to classify fully
[21:57:46] <Fingolfin> but no idea where to start for a non-abelian group
[21:57:49] <Fingolfin> yup, indeed
[21:58:15] <wjp> hm, let me look something up
[21:58:24] <Fingolfin> I tried some simple small groups, it was possible to ginf such a subgroup M for all of the cases I tried (including non-abelian ones)...
[21:58:29] <Fingolfin> err
[21:58:41] <Fingolfin> s/simple// (did mean "small" not "simple" in the math sense :-)
[21:58:55] <wjp> C_4 / C_2
[21:59:30] <wjp> hm, maybe I'm misinterpreting the question
[21:59:49] <wjp> you just want an isomorphism? not necessarily a splitting of G -> N -> 0 ?
[22:00:33] <wjp> uh, G -> G/N -> 0
[22:02:43] <Fingolfin> hum... what is "O" here? in any case, indeed "just" an isomorphism between G/N and another subgroup M of G
[22:02:50] <wjp> (note that $C_4 / C_2 \cong C_2$, but $C_4 \ne C_2 \oplus C_2$ )
[22:03:04] <wjp> 0 is the group of one element
[22:03:29] <wjp> are you familiar with the exact sequence notation, btw?
[22:03:41] <wjp> (and what it means for an exact sequence to split?)
[22:06:24] <Fingolfin> right, regarding the \oplus :-). And I know what an exact sequence is (you have a sequence of maps, between a sequence of spaces, and the images of one map match the kernel of the next, IIRC ?
[22:06:46] <wjp> yes
[22:06:46] <Fingolfin> dunno about the split. Maybe I forgot, maybe I never heard it :-)
[22:06:59] * Fingolfin looks it up on mathworld
[22:07:18] <Fingolfin> (anyway, no direct sum is needed, "just" an isomorphism)
[22:07:45] <Fingolfin> so in this case, C_4/C_2 \cong C_2 is fine
[22:08:34] <Fingolfin> I am wondering about a) the answer, b) if the answer not "always", then at least one counter example, and c) ideally of course a proof of the answer
[22:08:48] <Fingolfin> of course I am not expecting all that from you, but I thought that you maybe have an idea or have heard of this before...
[22:09:57] <Fingolfin> (well, I *am* expecting it all of you, of course, you should know it all ;-)! But i don't expect you to do my work... just so we don't have a misunderstanding :-)
[22:12:04] <Fingolfin> ahh, reading the entry about splitting is still enlightening. While it doesn't answer the original problem, it's very useful to know :-)
[23:04:43] <Fingolfin> well at least now I know the problem is not completly trivial. Or wjp is just mean and keeps the answer to himself. But that doesn't fit in my view of wjp (which is entierly to positive :-)
[23:05:43] <wjp> hm, no, I can't really think of anything sensible (and non-obvious) to say about it
[23:14:00] <wjp> anyway, time to go
[23:14:01] <wjp> g'night
[23:14:10] <wjp> I'll let you know if I think of anything :-)
[23:22:48] <Fingolfin> hehe thanks
[23:22:50] <Fingolfin> and sleep well
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