New PDF release: Algebra & Number Theory By Andrew Baker

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Deduce that σr is multiplicative. c) Show that [r] ∗ [r] satisfies [r] ∗ [r](n) = nr τ (n). d) Find a general formula for [r] ∗ [s](n) when s < r. 3-4. For n ∈ Z+ , prove the following formulæ, where the functions are defined in the text or in earlier questions. (a) µ(d)σ(n/d) = n; d|n (b) µ(d)τ (n/d) = 1; d|n σr (d)µ(n/d) = nr . (c) d|n CHAPTER 4 Finite and infinite sets, cardinality and countability The natural numbers originally arose from counting elements in sets. There are two very different possible ‘sizes’ for sets, namely finite and infinite, and in this section we discuss these concepts in detail.

GROUPS AND GROUP ACTIONS 2. 4 and consider the standard set with n elements n = {1, 2, . . , n}. The Sn = Perm(n) is called the symmetric group on n objects or the symmetric group of degree n or the permutation group on n objects. 5. Sn has order |Sn | = n!. Proof. Defining an element σ ∈ Sn is equivalent to specifying the list σ(1), σ(2), . . , σ(n) consisting of the n numbers 1, 2, . . , n taken in some order with no repetitions. To do this we have • n choices for σ(1), • n − 1 choices for σ(2) (taken from the remaining n − 1 elements), • and so on.

Proof. Suppose that f : X −→ Y is a surjection. Choose any element P ∈ P(X) and define the function h : X −→ P(X); h(x) = g(f (x)) if f (x) ∈ Z, P if f (x) ∈ / Z. We easily see that h is a surjection, contradicting Russell’s Paradox. Thus no such surjection can exist. 5. 16 (Cantor). , there is no bijection N0 −→ R. Proof. Suppose that R is countable and therefore the obviously infinite subset (0, 1] ⊆ R is countable. Then we can list the elements of (0, 1]: q0 , q1 , . . , qn , . . qn,1 qn,2 · · · qn,k · · · , where for each k, qn,k = 0, 1, .