mm-363
===
Subject:
: Algebra ProblemCan anyone give me a direction
here.A composite (positive integer) is a product ab with a and
b not necesserailydistinct integers in {2,3,4,...}. Show that
every composite is expressibleas xy+xz+yz+1, with x, y, z
being positive integers.Thanks.
===
Subject:
: Re: Algebra
Problem===> Can anyone give me a direction here.> A composite
(positive integer) is a product ab with a and b not
necesseraily> distinct integers in {2,3,4,...}. Show that
every composite is expressible> as xy+xz+yz+1, with x, y, z
being positive integers.1x + 1x + 1*1 + 1 = 2x + 2 = 2(x+1)1x
+ 2x + 1*2 + 1 = 3x + 3 = 3(x+1)1x + (p-1)x + 1*(p-1) + 1 = px
+ p = p(x+1)
===
Subject:
: Re: Is union of subgroups also a subgroup
?===> Today, I studied group theory.> but, I don't know if A,B
are subgroups of some group G,> then wheather AUB is also a
subgroup of G or not.> If somebody know this, please post
reply.> the following is my insufficient proof about this,>
please point out wrong side.> pf>> Since A,B are subgroups of
some group G,> AUB is not empty. now let a,b are in AUB.> if a
in A, b in A then ab^-1 in A =>ab^-1 in AUB.> if a in B, b in B
then ab^-1 in B =>ab^-1 in AUB.> if a in A, b in B then => ???
(How can I show ab^-1 in AUB ?)The smallest subgroup
containing A and B is the subgroup genera by Aand B. In Z+Z,
let A = Z+{0}, B = {0}+Z A / B does not contain (1,1)but any
subgroup containing A and B contains (1,0) + (0,1) = (1,1)so
A/B can't be a subgroup as it's not closed w.r.t +.Exercise:
show the intersection of any number of subgroups is a
subgroup.===
===
Subject:
: Effective help for students, provided by
a group of scientistsDear Students! If you need a high-quality
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===
Subject:
: Call for Book
Proposal - Series on Statistical Science and Applied
ProbabilityOriginator: bergv@math.uiuc.edu (Maarten
Bergvelt)Dear Colleagues,The Series on Statistical Science and
Applied Probability invites bookproposals for this series.More
about the series
athttp://www.wspc.com/books/series/asssap_series.shtmlThis
Series comprises accounts, by leading experts, from those
areasof statistical science and applied probability where
advancedmathematical tools are essential. Both research
monographs andtextbooks will be included, and the series
promotes purely theoreticalworks as well as studies that
combine theory and applications.To contribute to this book
series, contact editor@worldscientific.comSome recently
published titles under the series include:Statistical
Experiments and DecisionsAsymptotic Theoryby A N Shiryaev & V
G SpokoinyNon-Gaussian MertonBlackScholes Theoryby Svetlana I
Boyarchenko & Sergei Z LevendorskiiSincerely,Ole E
Barndorff-Nielsen (University of Aarhus, Denmark)Series
Editor===
===
Subject:
: Non-separable Hilbert spacesOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)I was wondering if
anyone knows some good references to techniques in
non-separable Hilbert spaces. Stephen===
===
Subject:
: Re:
RHomOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)Thank
you for your answer!> I think so.> It suffices to show
Ext_T^j(X,Y)=Ext_A^j(X,Y) for any two X,Y in T.Why? (We want
to prove RHom_T(X,Y)=RHom_A(X,Y) withoutboundedness
assumptions on the complexes X and Y.)> Fix X and consider
both sides as delta-functors in Y. The first> clearly is
universal.> To see that the second is universal, one has to
show that for each Y> in T there is> an injective module I
which is torsion and has Y as a submodule.> There clearly is
an injective I with 0 -> Y -> I exact. Since the ring> is
commutative,> I contains the torsion submodule I_tors and
clearly Y lies in it.> Injectivity is equivalent to full
divisibility (Cohn, Algebra vol3)> and this shows> that I_tors
again is injective.> So the right hand side delta functor is
erasable, hence universal and> thus coincides> with the
first.===
===
Subject:
: Re: combinatorial group theroyOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)> Two independent
questions.> 1) Does there exist an infinite, finitely presen,
simple group G,> and a two-transitive action of G?> 2) Does
there exist a residually finite, finitely genera group,>
satisfying a nontrivial relation, but not virtually solvable?>
Thanks in advance,> --> Yves de Cornulier2). Yes. By the
Neumann-Neumann proof of the Higman-Neumann-Neumannembedding
theorem, a finite p-group G can be embedded in a
finite2-generator group H, such that H is a subgroup of a
repea wreathproduct GwrCwrD, with C and D cyclic. Then the 2nd
derived subgroup H''of H has exponent p. If p ge 5, we may
choose G to have arbitrarilylarge derived length. Let F be a
free group of rank 2, and let N be theintersection of all
normal subgroups K of F such that F/K is finite andhas 2nd
derived subgroup of exponent p. By the above, F/N is
infinite,2-genera, and insoluble, but each finite factor group
of F/N issoluble, therefore F/N is not virtually
soluble.Avinoam Mann===
===
Subject:
: The analytic domain in the
implicit function theoremOriginator: bergv@math.uiuc.edu
(Maarten Bergvelt)Greetings!I'm looking for references about
the ``size'' of theanalytic domain in the implicit function
theorem, that is,if F(x, y) = c defines implicitly y = f(x) in
a neighborhood Bof x0 (with F(x0, y0) = c and y0 = f(x0)),
which is the sizeof B?I've found only two references:CHANG, H.
C., HE, W. PRABHU, N. ``The Analytic Domain in theImplicit
Function Theorem'', Journal of Inequalities in
Pure(http://jipam.vu.edu.au/v4n1/061_02.html)andSCPARPELLO, G.
M.; RITELLI, D. ``A Historical Outline of theTheorem of
Implicit Functions'', Divulgaciones Matematicas,vol. 10, no.
2, 2002, pp. 171-180. (It says Dini was the firstperson to
formalize the implicit function theorem in twovariables.
Moreover, Dini gave an estimative for the size ofthe
neighborhood
B).(http://www.emis.de/journals/DM/vX2/art6.pdf)Are there best
estimatives? Any other reference?Thanks in advance,
Humberto.===
===
Subject:
: Re: The analytic domain in the implicit
function theoremOriginator: bergv@math.uiuc.edu (Maarten
Bergvelt)> I'm looking for references about the ``size'' of
the> analytic domain in the implicit function theorem, that
is,> if F(x, y) = c defines implicitly y = f(x) in a
neighborhood B> of x0 (with F(x0, y0) = c and y0 = f(x0)),
which is the size> of B?> CHANG, H. C., HE, W. PRABHU, N.
``The Analytic Domain in the> Implicit Function Theorem'',
Journal of Inequalities in Pure>
(http://jipam.vu.edu.au/v4n1/061_02.html)They write in their
abstract, '' To the best of our knowledge, our result is the
first bound on the domain of validity of the Implicit Function
Theorem.''which only means that their best knowledge is poor.
See, for example,A. Neumaier,Existence regions and error
bounds for implicit and inverse functions,Z. Angew. Math.
Mech. 65 (1985), 49-55.scanned
text:http://www.mat.univie.ac.at/~neum/publist.htmland the
references given there. (Well, this is about C^1 equations,not
analytic ones. But the results extend immediately.)For
constructive methods, see also Chapter 5 of my book A.
Neumaier Interval Methods for Systems of Equations Cambridge
Univ. Press, Cambridge 1990which gives a number of further
references.In case you'd like to buy the book: It is currently
out of print,but will be reprin in a few months.Arnold
Neumaier===
===
Subject:
: Re: combinatorial group theroyOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)>1) Does there exist an
infinite, finitely presen, simple group G,...> I recall
Kenneth S. Brown mentioned to me in the (first half of) the>
eighties that he discovered such a group (probably published
in> Inventiones)...and a two-transitive action of
G?*******************************Hi:You may want to check
Hirsch's traduction to Kurosh's book The TheoryofGroups,
Chelsea Pub. Company, N.Y., 2nd english edition, 1960,
VolumeII, and the references there.In Appendix G there's a
very nice example of an infinite f.g. (4generators,
4relations) group that either it or one of its quotient groups
issimple (thisexample is actually G.Higgman's: A f.g. infinite
simple group, J.Lon. Math.Soc., vol 26 (1951), pp. 61-64).
This is the simpler example of such agroupthat I know.Good
luck!Tonio===
===
Subject:
: Re: combinatorial group
theroyOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> Does
there exist a residually finite, finitely genera group,>
satisfying a nontrivial relation, but not virtually solvable?>
Yes. By the Neumann-Neumann proof of the
Higman-Neumann-Neumann> embedding theorem, a finite p-group
GYou mean: a group of exponent p?> can be embedded in a
finite> 2-generator group HCan H be taken to be a p-group (=
of order a power of p)? This would givea much stronger result,
replacing residually finite by residuallyp-finite., such that H
is a subgroup of a repea wreath> product GwrCwrD, with C and D
cyclic. Then the 2nd derived subgroup H''> of H has exponent
p. If p ge 5, we may choose G to have arbitrarily> large
derived length.Thank you very much. Your answer permits to
construct a finitelygenera profinite group (and perhaps
pro-p),that contains no free nonabelian subgroup, but with no
open solvablesubgroup, answering negatively a question of
Breuillard and Gelander.Yves de Cornulier===
===
Subject:
:
eigenvalues with positive imaginary partOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)Does anyone know a
reference/proof for the following?S is a symmetric matrix.D is
a diagonal matrix with positive values on the diagonal.i is
sqrt(-1)then the eigenvalues of S + i D all have positive
imaginary part.