Recent content by symbol0

I don't follow you mathwonk. If the curve is going counterclockwise, the winding number is greater or equal to 0. And how is that related to the question?

To jgens: you say 1 leq |f|. How do we know f is not a function like f(z)=z which has a minimum in D ? To office shredder: to get that contradiction don't we need to know that integrals are continuous ? is that a fact?

I want to understand how do you conclude |f| = 1 on D. I know f cannot have a maximum in D. So |f|<1 on D. ...what else?

Thank you jgens. About your approach, you say f doesn't have a min in D. How do we know 0 is not a min ?

Do you mean that if I integrate f on circles with radii approaching 1, I get 0 = int f approaches int 1/z =2ipi ? but does the continuity of f imply the continuity of the integral like that?

Show that there is no holomorphic function f in the unit disc D that extends continuously to |z|=1 such that f(z) =1/z for |z|=1 Some thoughts that might not be relevant: If such f existed then, I can see that f maps the unit circle to the unit circle and the unit disc onto the unit disc...

Thank you Compuchip

Let M and N be normal subgroups of G such that G=MN. Prove that G/(M\bigcapN)\cong(G/M)x(G/N). I tried coming up with an isomorphism from G to (G/M)x(G/N) such that the kernel is M\bigcapN, so that I can use the fundamental homomorphism theorem. I tried f(a) = (aM, aN). It is an...

Thank you micromass, I got it.

The order of hN in G/N is the smallest integer k such that h^k = n for some n in N. We know that k divides |G/N|. I don't know if k divides |H|. We know that h^k is in the intersection of H and N.

Let G be a finite group, let H be a subgroup of G and let N be a normal subgroup of G. Show that if |H| and |G:N| are relatively prime then H is a subgroup of N. I have tried using the fact that since N is normal, HN is a subgroup of G. Suposing that H is not contained in N, I tried finding...

Jarle, In one of your responses to my question, you sent me this link:http://mathforum.org/library/drmath/view/51638.html". Can I say that in general: An infinite set of prime roots of prime numbers are linearly independent over the rationals?

No confusion mathwonk. I get the difference between the uses of the word "finite" Thank you so much guys

So I guess adjoining an algebraic element to a field is what you are calling an algebraic extension. In my book the only thing defined is algebraic extension over a field K: a field in which every element is algebraic over K. My book has that adjoining a finite amount of algebraic elements of...

What's the difference between being an algebraic extension of a field and being algebraic over a field?