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Thu 30 Sep, 2004 04:35 pm
I'm not quite sure this forum is up to this question, because I haven't seen any questions about advanced topics in mathematics. Oh well, here goes nothing:
Let G be a group such that it's abelization (G/G') is a finitely generated group. Let G/G' be the direct sum of the following cyclic groups: <x1>,...,<xn>. We also ask that for every i there is a representitive yi of xi (yi+G'=xi) such that ord(xi)=ord(yi). Let H be the free product of <y1>,...,<yn>. Why is the following true?
Gamma(n,H)/Gamma(n+1,H) (lower central series) is an abelian group (obvious) whose torsion subgroup is finite of order dividing a power of the order of the torsion subgroup of G/G' (not so obvious).
Can you help me with this?
I suggest you look here:
http://www.ocf.berkeley.edu/~wwu/contents.shtml
Check the mathematics discussions in the "medium" and "hard" sections of the riddles forum.
It makes me wonder. The mathematical experts of you - just what is your mathematical education? I'll start: I'm halfway through my B. Sc. in mathematics, but I'm also taking very advanced courses in Abstract Algebra (the mathematical topic I like the most).
My Modern algebra is rusty so bear with me
G and its inverse G' are isomorphic abelian groups then all gubgroups of G and G' are abelian and isomorphic. (Actually I seem to remember this falling out of the requirements of groups). Therefore all of the cyclic supgroups <xi> are abelian and isomorphic
Moreover, if G/G' are inverse isomorphic groups then |G|=|G'| and the order of all equivalent cyclic subgroups are equal, e.g. |<xi>|=|<yi>|, <yi>=<xi>', consequently the product of cyclic groups is the identity, e.g. <xi><yi>=e and the free product is the product ∏<xi><yi>=e for all n. Since |e|=1 then everything boils down to 1
As for the former question Γ(n,H)/ Γ(n+1,H) are both Γ(n+1,e) which is abelian by definition, and the order of the tortional subgroup is 1.
So I've shown the trivial case it true(?) but then I could very well be wrong, with Modern Algebras I often am.
When confronted with braintwisters like this I frequent [URL=http://]mathworld[/URL] for definitions and properties, cruise here [URL=http://]Dr Math-[/URL] for basic proofs, then I pull out an old Dover Modern Algebra Text and start suffering until in absolute frustration I call an old college chum who was better than I at modern algebra
and let him start suffering too.
BTW I'm an Engineer. That means I'm estatic when I find an equation that matches my world.
Rap
Rara, while I don't consider myself an expert, I'm an undergraduate engineering student, so I've taken up through calc 4. Raprap, I like finding equations to model the real world also (as well as algorithms) although I wouldn't say I am exstatic. Although I must admit I am not very interested in discrete math, proofs, and math for maths sake like this. What kind of engineer are you?
Re: My Modern algebra is rusty so bear with me
raprap wrote:G and its inverse G' are isomorphic abelian groups then all gubgroups of G and G' are abelian and isomorphic. (Actually I seem to remember this falling out of the requirements of groups). Therefore all of the cyclic supgroups <xi> are abelian and isomorphic
Moreover, if G/G' are inverse isomorphic groups then |G|=|G'| and the order of all equivalent cyclic subgroups are equal, e.g. |<xi>|=|<yi>|, <yi>=<xi>', consequently the product of cyclic groups is the identity, e.g. <xi><yi>=e and the free product is the product ∏<xi><yi>=e for all n. Since |e|=1 then everything boils down to 1
As for the former question Γ(n,H)/ Γ(n+1,H) are both Γ(n+1,e) which is abelian by definition, and the order of the tortional subgroup is 1.
So I've shown the trivial case it true(?) but then I could very well be wrong, with Modern Algebras I often am.
When confronted with braintwisters like this I frequent [URL=http://]mathworld[/URL] for definitions and properties, cruise here [URL=http://]Dr Math-[/URL] for basic proofs, then I pull out an old Dover Modern Algebra Text and start suffering until in absolute frustration I call an old college chum who was better than I at modern algebra
and let him start suffering too.
BTW I'm an Engineer. That means I'm estatic when I find an equation that matches my world.
Rap
"G and its inverse G'"? G' is the commutator subgroup of G.
"G and its inverse G' are isomorphic abelian groups"?!?!?!? Whos said G was abelian? I just said that G's abelization (G/G') is a finitely generated (abelian) group.
The rest is just too bizzare for me to respond to.
P.S.
I asked my prof. this question and he said it has something to do with Lie algebras (which I currently know nothing about).
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