By Joseph J. Rotman
A person who has studied summary algebra and linear algebra as an undergraduate can comprehend this ebook. the 1st six chapters supply fabric for a primary direction, whereas the remainder of the publication covers extra complicated issues. This revised version keeps the readability of presentation that used to be the hallmark of the former variations. From the studies: "Rotman has given us a really readable and worthy textual content, and has proven us many attractive vistas alongside his selected route." --MATHEMATICAL experiences
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Additional info for An Introduction to the Theory of Groups, 4th Edition
We omit proofs of standard results. 1. 1. Deﬁnition. Let G be a set on which a binary operation is deﬁned. Denote the image of (a, b) ∈ G × G under the operation by ab. Then G is a group with identity element 1 ∈ G if for all a, b, c ∈ G the following hold. • Associativity: a(bc) = (ab)c. • Identity: 1a = a = a1. • Invertibility: there is an inverse a−1 ∈ G such that aa−1 = 1 = a−1 a. The trivial group, denoted 1, consists only of its identity. The group G is abelian if for all a, b ∈ G we have • Commutativity: ab = ba.
18. The Frattini subgroup. The Frattini subgroup Frat(G) of a ﬁnite group G is the intersection of its maximal subgroups (the proper subgroups of G not properly contained in any proper subgroup). Clearly Frat(G) is characteristic, so any automorphism of G induces an automorphism of G/Frat(G). Suppose that G is a ﬁnite p-group. Then Frat(G) = G P where P is the subgroup of G generated by the pth powers of all elements of G. Consequently G/Frat(G) is an elementary abelian p-group. In this case, if |G/Frat(G)| = pr then r is the minimum possible size of a generating set for G.
Proof. We induct on |G|. Clearly the theorem is true for cyclic G. Suppose that G is non-cyclic and the theorem is true for all groups of order less than |G|. Since G is solvable, it has a proper normal subgroup N such that G/N is cyclic. The exponent of G/N divides the exponent of G, so is square-free. Therefore, we can assume that |G : N | = q, q prime. Select an element g of G \ N , so that q is the maximum power of q dividing |g|. Say |g| = qm where q and m are coprime. Then g = g m has order q and is not in N : if it were, then for integers x, y such N.
An Introduction to the Theory of Groups, 4th Edition by Joseph J. Rotman