By Andrew Baker

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The sign of σ is the number sgn σ = (−1)cσ = +1 if cσ is even, −1 if cσ is odd. Then sgn : Sn −→ {+1, −1}. Notice that {+1, −1} is actually a group under multiplication. 7. The function sgn : Sn −→ {+1, −1} satisfies sgn(τ σ) = sgn(τ ) sgn(σ) (τ, σ ∈ Sn ). Proof. By considering the arrow diagram for τ σ obtained by joining the diagrams for σ and τ , we see that the total number of crossings is cσ + cτ . If we straighten out the paths starting at each number in the top row, so that we change the total number of crossings by 2 each time.

Vn ) the symmetry group is the dihedral group of order 2n D2n , with elements ι, α, α2 , . . , αn−1 , τ, ατ, α2 τ, . . , αn−1 τ where αk is an anticlockwise rotation through 2πk/n about the centre and τ is a reflection in the line through V1 and the centre. Moreover we have |α| = n, |τ | = 2, τ ατ = αn−1 = α−1 . 6. SUBGROUPS AND LAGRANGE’S THEOREM 35 In permutation notation this becomes α = (V1 V2 · · · Vn ), but τ is more complicated to describe. For example, if n = 6 we have α = (V1 V2 V3 V4 V5 V6 ), τ = (V2 V6 )(V3 V5 ), while if n = 7 α = (V1 V2 V3 V4 V5 V6 V7 ), τ = (V2 V7 )(V3 V6 )(V4 V5 ).

FINITE AND INFINITE SETS, CARDINALITY AND COUNTABILITY (e) The set of all positive rational numbers Q+ = a : a, b ∈ N0 , a, b > 0 . b Solution. (a) Since S is infinite it cannot be empty. Let S0 = S. By WOP, S0 has a least element s0 say. Now consider the set S1 = S − {s0 }; this is not empty since otherwise S would be finite. Again WOP ensures that there is a least element s1 ∈ S1 . Continuing, we can construct a sequence s0 , s1 , . . , sn , . . of elements in S with sn the least element of Sn = S − {s0 , s1 , .