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1 is strictly multiplicative. An important example is the M¨ obius function µ : Z+ −→ R defined as follows. 19, we have the prime power factorization n = pr11 pr22 · · · prt t , where for each j, pj is a prime, 1 rj and 2 p1 < p2 < · · · < pt . We set 0 if any rj > 1, µ(n) = µ(pr11 pr22 · · · prt t ) = t (−1) if all rj = 1. So for example, if n = p is a prime, µ(p) = −1, while µ(p2 ) = 0. Also, µ(60) = µ(22 × 3 × 5) = 0. 2. The M¨ obius function µ is multiplicative. Proof. This follows from the definition and the fact that the prime power factorizations of two coprime natural numbers m, n have no common prime factors.

By Induction on r, the number of prime factors in the prime power factorization of n = pr11 · · · prt t , so r = r1 + · · · + rt . If r = 1, then n = p is prime and µ(p) = −1, hence µ(d) = 1 − 1 = 0. d|p Assume that whenever r < k. Then if r = k, let n = mprt t where pt is a prime factor of n. Then µ(n) = µ(m)µ(prt t ) and so µ(d) = d|n (µ(d) + µ(dpt )) = d|m (µ(d) + µ(d)µ(pt )) = d|m µ(d)(1 − 1) = 0. d|m This gives the Inductive Step. 2. Convolution and M¨ obius Inversion Let θ, ψ : Z+ −→ R (or C) be arithmetic functions.

Decompose σ= 1 2 3 4 5 2 5 3 1 4 ∈ S5 into a product of transpositions. Solution. We have σ = (3)(1 2 5 4) = (1 2 5 4) = (1 4)(1 5)(1 2). Some alternative decompositions are σ = (2 1)(2 4)(2 5) = (5 2)(5 1)(5 4). 5. Symmetry groups Let S be a set of points in Rn , where n = 1, 2, 3, . .. , |ϕ(u) − ϕ(v)| = |u − v| (u, v ∈ S). 13. Let ϕ be a symmetry of S ⊆ Rn . Then a) ϕ is a bijection and ϕ−1 is also a symmetry of S; b) ϕ preserves distances between points and angles between lines joining points.