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# Abstract Algebra with Applications: Rings and Fields by Karlheinz Spindler

By Karlheinz Spindler

A accomplished presentation of summary algebra and an in-depth therapy of the purposes of algebraic innovations and the connection of algebra to different disciplines, akin to quantity idea, combinatorics, geometry, topology, differential equations, and Markov chains.

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Additional resources for Abstract Algebra with Applications: Rings and Fields

Example text

Exercises 1. If G has at least two subgroups of prime order p; and p does not divide jGj, then no subgroup of order p is normal. ) Find an analogous assertion for higher powers of p. 2. If N / G and g 2 G has order n, then Ng (viewed as an element of G=N ) has order dividing n. Give an example where equality does not hold. 2 2 Sn and An There are many proofs of Theorem 21 in the literature. Here are two good alternative approaches; Exercise 4 contains the fastest proof that I know, but is a bit tricky 3.

Fermat's Little Theorem) If p is a prime number and (a; p) = 1; then ap 1 1 (mod p). Proof. Recall Euler(p) = f1; : : : ; p 1g is a group under multiplication and contains a, so it su ces to prove o(a)jp 1. But this is true by Corollary 13, since jEuler(p)j = p 1. Fermat's Little Theorem also helps test whether a given large number is prime, since it turns out that ap 1 is usually not congruent to 1 mod p when 13 p is not prime. Thus, given a large number p, if we compute ap 1 for twenty random values of a < p and always get 1 (mod p), we can be virtually certain that p is prime.

Since (i a) = (a i) we may assume a never appears on the right side in a transposition. Clearly a appears at least twice. ) 5. The only nontrivial normal subgroup of A is the Klein group. 1 1 1 ( ) 4 38 6. If (1) 6= N / An for n 5 then N = An . (Hint: Mimic the proof of Theorem 24. ) 7. If H Sn and H contains every transposition, then H = Sn . 8. If H is a subgroup of Sn containing = (1 2 : : : n) and = (1 2); then H = Sn . (Hint: i i = (i i + 1) 2 H: But then one has (i i + 1)(i 1 i)(i i + 1) = (i 1 i + 1) 2 H .