{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 03 - MAT 322 Algebra with Galois Theory Professor Claus...

This preview shows pages 1–3. Sign up to view the full content.

MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 3 Rik Sengupta October 9, 2009 1. Show that Q / Z is isomorphic to μ ( C ) , the group of z C * such that z n = 1 for some positive integer n . Is it a finitely generated abelian group? What is the torsion subgroup of Q / Z ? Solution. We define a homomorphism from Q to μ ( C ) by ϕ : Q μ ( C ), with the following mapping: p 7→ e 2 πip . This is obviously a ho- momorphism, because ϕ ( p + q ) = e 2 πi ( p + q ) = e 2 πip e 2 πiq = ϕ ( p ) ϕ ( q ). In fact, the image space of the homomorphism has been chosen because we can simply raise the image of the homomorphism to the denominator of the rational number in question, and then we will get 1, as follows: [ ϕ ( p q )] q = e 2 πi p q q = e 2 πip = 1 Thus the image space is well-defined, in the sense that all the images of the map lie in μ ( C ). Note that the kernel of this homomorphism is the set of points p Q for which ϕ ( p ) = 1, i.e. e 2 πip = 1. But this happens exactly when p Z . So, the kernel of the homomorphism is simply Z . Furthermore, the map is surjective, because given any element z μ ( C ), we know z n = 1 for some n , and hence, z is one of the n th roots of unity. So we can simply write z as e 2 πik for some rational k . Then, k is in Q , so it is a preimage in the homomorphism, i.e. our map is surjective. So then, from the First Isomorphism Theorem, we immediately get Q / Z is isomorphic to μ ( C ). It is not finitely generated. Suppose it is in fact a finitely generated group, generated by p 1 Z , p 2 Z , . . . p r Z for finitely many rationals p i . Then we can simply take the denominators of all the p i and find some natural number that is coprime to them all. Then, take any rational s whose denominator in its lowest form is the natural number we just found. Then, s Z is not 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
in the group generated by the finitely many elements, even though it is in Q / Z . Hence, Q / Z is not a finitely generated abelian group. Every element in the group Q / Z has some finite order that is the denom- inator of the rational number. This means that since all the elements in the group have finite order, all the elements are also present in the tor- sion subgroup. Furthermore, the torsion subgroup, being a subgroup, is contained within the actual group. This is possible only when the torsion subgroup equals the group itself. Hence, ( Q / Z ) tor = Q / Z . 2. Let G be a group, and let G der be the subgroup generated by all commuta- tors [ x, y ] = xyx - 1 y - 1 , for x, y G . Show that G der is a normal subgroup of G , and that the corresponding quotient group G ab is abelian. Describe the group D ab n up to isomorphism, for all integers n 3 . Solution. We will show that for each x G der , gxg - 1 is also in G der whenever g G . Since G der is the subgroup generated by all commutators in G , for each x G der we have x = c 1 c 2 . . . c m – a word of commutators – so c i = [ a i , b i ] for all i .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}