By Thomas C. Craven
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Extra info for Abstract algebra [Lecture notes]
Let a ∈ G. We claim an = 1. If not, then there exists some prime p in the factorization of |a| that appears to a higher power than it does in n, say n = ps n1 , |a| = pr m and r > s, where n1 and m are relatively prime to p. 8), hence am cp has order pr n1 > ps n1 = n (by Exercise 31, p. 180—assigned as homework), a contradiction. But now we have seen that every element of G is a root of the equation xn − 1, which has at most n roots in the field. Since c generates a subgroup of order n, it must actually generate all of G and therefore G is cyclic.
2) f (a)−1 = f (a)−1 eH = f (a)−1 f (eG ) = f (a)−1 f (aa−1 ) = f (a)−1 f (a)f (a−1 ) = f (a−1 ). (3) eH ∈ Im f by (1). For f (a), f (b) ∈ Im f , f (a)f (b)−1 = f (ab−1 ) ∈ Im f by (2). (4) f is gives a bijection of G and Im f , hence is an isomorphism. The real complexity of the symmetric groups Sn can be seen by the next result that every finite group is isomorphic to a subgroup of some Sn . In fact, this can be generalized to infinite groups as well. 20 (Cayley’s Theorem). Every group G is isomorphic to a group of permutations.
We use this to show that arithmetic works “modulo I”. 5. Let I be an ideal of a ring R. If a ≡ b (mod I) and c ≡ d (mod I), then (1) a + c ≡ b + d (mod I); (2) ac ≡ bd (mod I). Proof. (1) (a + c) − (b + d) = (a − b) + (c − d). Since a − b ∈ I and c − d ∈ I, so is (a + c) − (b + d), hence a + c ≡ b + d (mod I). (2) ac − bd = ac − bc + bc − bd = (a − b)c + b(c − d) ∈ I since I is closed under multiplication on both sides. Therefore ac ≡ bd (mod I). Looking at this proof, we see that it is multiplication that fails if we have only a left or right ideal that is not 2-sided.