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29. Let G Sn and let X = n. For σ ∈ G and k ∈ n let σk = σ(k). This defines an action of (G, ◦) on n. 7. 30. Let X ⊆ Rn and let G Sym(X) be a subgroup of the symmetry group of X. For ϕ ∈ G and x ∈ X, let ϕx = ϕ(x). This defines an action of (G, ◦) on X. Suppose we have an action of a group (G, ∗) on a set X. For x ∈ X, the stabilizer of x is StabG (x) = {g ∈ G : gx = x} ⊆ G, and the orbit of x is OrbG (x) = {gx : g ∈ G} ⊆ X. Notice that x = ιx, so x ∈ OrbG (x) and ι ∈ StabG (x). Thus StabG (x) = ∅ and OrbG (x) = ∅.

8. Show that the function σ = σ1 satisfies µ(d)σ(n/d) = n (n ∈ Z+ ). d|n Solution. By definition, σ(n) = d, d|n hence σ = id ∗η. By M¨obius Inversion, id = id ∗δ = id ∗(η ∗ µ) = (id ∗η) ∗ µ = σ ∗ µ = µ ∗ σ, so for n ∈ Z+ , n= µ(d)σ(n/d) = d|n σ(d)µ(n/d). 9. If θ, ψ are multiplicative arithmetic functions, then θ ∗ψ is multiplicative. ¨ 2. CONVOLUTION AND MOBIUS INVERSION 51 Proof. If m, n be coprime positive integers, θ ∗ ψ(mn) = θ(d)ψ(mn/d) d|mn = θ(rs)ψ(mn/rs) r|m s|n = θ(r)θ(s)ψ((m/r)(n/s)) r|m s|n θ(r)θ(s)ψ(m/r)ψ(n/s) = r|m s|n = θ(s)ψ(n/s) θ(r)ψ(m/r) r|m s|n = θ ∗ ψ(m)θ ∗ ψ(n).

These conditions actually characterise finite sets. In the next section we investigate how to recognise infinite sets. 3. COUNTABLE SETS 55 2. 6. Let X be a set. a) X is infinite if and only if there is an injection X −→ P where P ⊆ X is a proper subset. b) X is infinite if and only if there is a surjection Q −→ X where Q ⊆ X is a proper subset. c) X is infinite if and only if there is an injection N0 −→ X. d) X is infinite if and only if there is a subset T ⊆ X and an injection N0 −→ T . 7. The set of all natural numbers N0 = {0, 1, 2, .

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Algebra and number theory, U Glasgow notes by Baker.

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