Algebraic and analytic aspects of operator algebras by I. Kaplansky

By I. Kaplansky

An algebraic prelude Continuity of automorphisms and derivations $C^*$-algebra axiomatics and simple effects Derivations of $C^*$-algebras Homogeneous $C^*$-algebras CCR-algebras $W^*$ and $AW^*$-algebras Miscellany Mappings keeping invertible parts Nonassociativity Bibliography

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This contradicts the fact that R is an integral 0 domain, so n must be prime. 9) Definition. A subset S of a ring R is a subring if S, under the operations of multiplication and addition on R, is a ring. Thus S is a subring of R if and only if S is an additive subgroup of R that is closed under multiplication. We will now present a number of examples of rings. Many of the math- ematical systems with which you are already familiar are rings. Thus the integers Z are an integral domain, while the rational numbers Q, the real numbers R, and the complex numbers C are fields.

Let d = (m,n), the greatest common divisor of m and n. Then n I (n/d)m, so (n/d)m = 0 in Z. , m E Z. Also, any m E Zn determines an element 0m E Aut(Zn) by the formula ¢,n(r) = rm. To see this we need to check that Om is an automorphism of Zn. But if On(r) _ 46,n(s) then rm = sm in Zn, which implies that (r - s)m = 0 E Zn. , r = s in Z. Therefore, we have a one-to-one correspondence of sets Aut(Zn) Z,, given by 0m~ 'm. 4 Permutation Representations and the Sylow Theorems If X is any set, then the set Sx = {one-to-one correspondences f : X -+ X) is a group under functional composition.

By hypothesis, G = NH, so existence of the factorization is clear. Suppose a = n1h1 = n2h2. Then n2'n1 = h2hi 1 E Nf1H = {e}. Therefore, 0 n1=n2 and h1=h2. According to this lemma, G is set theoretically the cartesian product set N x H, but the group structures are different. 12) shows that H = H/(H n N) (HN)/N = (NH)/N = GIN. Thus, H is determined once we have N. A natural question is then, given all groups C such that G is the semidirect product groups N and H, H. As one answer to N and H of subgroups N and H where N this problem, we will present a construction showing how to produce all semidirect products.

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