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Jim
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PostSubject: Exam question   Fri Nov 13, 2009 4:50 pm

Hi. This is Jim. This is the problem that I showed Bruno last night from an old Concordia Ph. D. Comprehensive Examination. Question 1.(b)

Prove: If f(x) -> + infinity and f'(x) -> 0 as x -> infinity then f is not a quotient of two polynomials.

The solution I have seems too simple so I'd like to see if there is a proof using the logarithmic derivative of f that Bruno suggested.
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PostSubject: Re: Exam question   Fri Nov 13, 2009 5:37 pm

Hello Jim!

The solution I suggested does not work after all. Indeed the log. derivative of a quotient of polynomial goes to 0 at infinity, so we can't use that!

I would go like this : suppose f(x)=p(x)/q(x) is a quotient of polynomials. Then deg(p(x))>deg(q(x)), because of the first assumption. But f'(x)=(p'(x)q(x)-p(x)q'(x))/q(x)^2 is also a quotient of polynomials. The only way for this to converge to 0 at infinity is if we have : deg(p'(x)q(x)-p(x)q'(x))<deg(2q(x)). It's easy to check that deg(p'(x)q(x)-p(x)q'(x))=deg(p(x))+deg(q(x))-1. So we must have deg(p(x))-1<deg(q(x)), i.e. deg(p(x)) <= deg(q(x)), which contradicts deg(p(x))>deg(q(x)). cat

What's your solution?
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PostSubject: Re: Exam question   Fri Nov 13, 2009 7:48 pm

Here is my solution to #2a. Gave me some trouble!

Let C be the set of conjugate subgroups of H, and N be the normalizer of H in G. Then we can rephrase the problem by saying that if G is the union of the elements of C, then H=G.

If |C|>1 then we must have |C| > [G : H], because every element of C has order |H|, but since |C|>1 the union is not disjoint - the elements of C have the identity in common.

A well known theorem states that |C| = [G : N]. Therefore [G : N] > [G : H], i.e. |G|/|N| > |G|/|H|, or |H|>|N|. This is absurd, because H is contained in N. Therefore |C|=1, H is normal in G and H=G.
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Jim
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PostSubject: Re: Exam question   Sun Nov 15, 2009 2:05 pm

Yes, that is exactly my solution. I haven't tried the group theory question yet though. So I will try it and see if I get the same answer.
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PostSubject: Re: Exam question   Fri Feb 05, 2010 5:26 pm

#2a is exercise 24, page 131 in Abstract Algebra Dummit & Foote.

I'd say : if a subgroup contains a representative for each conjugacy class, it must be the entire group.
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