Abstract Algebra: Review Problems on Groups and Galois by Paul T. Bateman

By Paul T. Bateman

My aim is to supply a few assist in reviewing Chapters 7 and eight of our ebook summary Algebra. i've got integrated summaries of almost all these sections, including a few basic reviews. The evaluation difficulties are meant to have really brief solutions, and to be extra standard of examination questions than of ordinary textbook exercises.By assuming that it is a evaluate. i've been capable make a few minor alterations within the order of presentation. the 1st part covers a number of examples of teams. In featuring those examples, i've got brought a few recommendations that aren't studied until eventually later within the textual content. i feel it truly is valuable to have the examples accrued in a single spot, that you should seek advice from them as you review.A whole record of the definitions and theorems within the textual content are available on the net web site wu. math. niu. edu/^beachy/aaol/ . This website additionally has a few team multiplication tables that are not within the textual content. I may still word minor alterations in notation-I've used 1 to indicate the id component to a bunch (instead of e). and i have used the abbreviation "iff" for "if and purely if".Abstract Algebra starts off on the undergraduate point, yet Chapters 7-9 are written at a degree that we examine acceptable for a pupil who has spent the higher a part of a 12 months studying summary algebra. even though it is extra sharply concentrated than the normal graduate point textbooks, and doesn't move into as a lot generality. i am hoping that its good points make it an excellent position to profit approximately teams and Galois concept, or to study the fundamental definitions and theorems.Finally, i need to gratefully recognize the help of Northern Illinois collage whereas penning this evaluation. As a part of the popularity as a "Presidential instructing Professor. i used to be given go away in Spring 2000 to paintings on tasks regarding instructing.

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Let P be a Sylow 5-subgroup and let Q be a Sylow 7-subgroup. At least one of these subgroups must be normal, since otherwise we would have 21 · 4 elements of order 5 and 15·6 elements of order 7. Therefore P Q is a subgroup, and it must be normal since its index is the smallest prime divisor of |G|, so we can apply Review Problem 11. Since P Q is normal and contains a Sylow 5-subgroup, we can reduce to the number 35 when considering the number of Sylow 5-subgroups, and thus n5 (G) = n5 (P Q) = 1.

Describe G3 , up to isomorphism. Solution: The order of G3 must be a common divisor of 24 and 30, so it must be a divisor of 6. Since any group of order less than 6 is abelian, G3 must be isomorphic to the symmetric group S3 . 2. Prove that a finite group whose only automorphism is the identity map must have order at most two. Solution: All inner automorphisms are trivial, so G is abelian. Then α(x) = x−1 is an automorphism, so it is trivial, forcing x = x−1 for all x ∈ G. If G is written additively, then G has a vector space structure over the field Z2 .

Therefore the splitting field must have subfields of degree 3 and of degree 2, so the degree of the splitting field over Z2 must be 6. 4. Let F be an extension field of K. Show that the set of all elements of F that are algebraic over K is a subfield of F . 7, but it is worth repeating. Whatever you do, don’t try to start with two elements and work with their respective minimal polynomials. If u, v are algebraic elements of F , then K(u, v) is a finite extension of K. Since u + v, u − v, and uv all belong to the finite extension K(u, v), these elements are algebraic.

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