A First Course in Abstract Algebra - Instructor's Solutions by John B. Fraleigh

By John B. Fraleigh

Instructor's ideas guide to a primary direction in summary Algebra that is uploaded right here: http://bibliotik.org/torrents/27184 contains solutions to all problems.

Considered a vintage via many, a primary direction in summary Algebra is an in-depth creation to summary algebra. all for teams, jewelry and fields, this article supplies scholars an organization beginning for extra really good paintings through emphasizing an figuring out of the character of algebraic structures.

* This classical method of summary algebra makes a speciality of applications.

* The textual content is aimed at high-level classes at colleges with powerful arithmetic programs.

* available pedagogy comprises old notes written by way of Victor Katz, an expert at the historical past of math.

* through starting with a examine of team concept, this article presents scholars with a simple transition to axiomatic arithmetic.

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Identity: We are given that e ∈ H. Inverses: For all a ∈ H, the equation a2 = e means that a−1 = a ∈ H. Thus H is a subgroup. 48. Yes, H is a subgroup for order 3, by essentially the same proof as in the preceding exercise. No, H is not a subgroup for order 4, because the square of an element of order 4 has order 2, so H is not closed under the operation. For prime positive integers, H will be a subgroup. 49. S3 is a counterexample. 50. a. (h, k) = (h, e)(e, k) b. (h, e)(e, k) = (h, k) = (e, k)(h, e).

The definition is correct. 19. T T T F T F T T F T (i) See the last sentence in this section. 10. Cosets and the Theorem of Lagrange 35 20. This is impossible. For a subgroup H of an abelian group G, we have a + H = H + a for all a ∈ G. 21. For any group G, just take the subgroup H = G. 22. The subgroup {0} of Z6 . 23. This is impossible. Because the cells are disjoint and nonempty, their number cannot exceed the order of the group. 24. This is impossible. The number of cells must divide the order of the group, and 4 does not divide 6.

We claim that no lower power of ab is equal to e, for suppose that (ab)n = an bn = e. Then an = b−n = c must be an element of both H and K, and thus generates a subgroup of H of order dividing r which must also be a subgroup of K of order dividing s. Because r and s are relatively prime, we see that we must have c = e, so an = bn = e. But then n is divisible by both r and s, and because r and s are relatively prime, we have n = rs. Thus ab generates the desired cyclic subgroup of G of order rs.

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