By Joseph J. Rotman

A person who has studied summary algebra and linear algebra as an undergraduate can comprehend this e-book. the 1st six chapters supply fabric for a primary path, whereas the remainder of the publication covers extra complicated subject matters. 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 helpful textual content, and has proven us many attractive vistas alongside his selected route." --MATHEMATICAL REVIEWS

Retail PDF from Springer; a number of chapters mixed into one dossier; scanned, searchable.

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J. Rotman, An Introduction to the Theory of Groups © Springer-Verlag New York, Inc. 1995 Subgroups 21 as a subset of a group G*, then it is much simpler to show that G is a subgroup of G* than to verify all the group axioms for G. For example, the four permutations of the 4-group V form a group because they constitute a subgroup of S4. 2. A subset S of a group G is a subgroup s, t E Simply SCl E S. if and only if Proof. If s E S, then 1s- 1 = S-1 E S, and if s, t E S, then S(C l )-1 converse is also easy.

If p is a prime, prove that Zp2 *' X Zp Zn. (Hint. 64. ,. is a homomorphism if and only if G is abelian. 65. Let A be an abelian group, and let IX: H -> A and p: K -> A be homomorphisms. Prove that there exists a unique homomorphism 1': H x K -> A with y(h, 1) = lX(h) for all hE Hand 1'(1, k) = P(k) for all k E K. Show that this may be false if A is not abelian. We now take another point of view. It is easy to multiply two polynomials together; it is harder to factor a given polynomial. We have just seen how to multiply two groups together; can one factor a given group?

If Sand T are subgroups of a finite group G, then ISTIIS n TI = lSI I n Remark. The subset ST need not be a subgroup. Proof. Define a function cp: S x T --+ ST by (s, t) H st. Since cp is a surjection, it suffices to show that if x E ST, then Icp-1(x)1 = IS n We show that cp-1(X) = {(sd,d- 1t): dES n T}. It is clear that cp-1(X) contains the right side. For the reverse inclusion, let (s, t), (a, f) E cp-1(X); that is, s, a E S, t, f E T, and st = x = af. Thus, s-la = tf- 1 E S n T; let d = s-la = tf- 1 denote their common value.