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If xH = yH then xH ∩ yH = ∅. Conversely, suppose that xH ∩ yH = ∅. If yh ∈ xH for some h ∈ H, then x−1 yh ∈ H. For k ∈ H, x−1 yk = (x−1 yh)(h−1 k), which is in H since x−1 yh, h−1 k ∈ H and H is a subgroup of G. Hence yH ⊆ xH. Repeating this argument with x and y interchanged we also see that xH ⊆ yH. Combining these inclusions we obtain xH = yH. ii) For each g ∈ G, |gH| = |H|. If gh = gk for h, k ∈ H then g −1 (gh) = g −1 (gk) and so h = k. Thus there is a bijection θ : H −→ gH; θ(h) = gh, which implies that the sets H and gH have the same number of elements.

If A = [a0 ; a1 , . . , an ] with A > 1, show that 1/A = [0; a0 , a1 , . . , an ]. Let x > 1 be a real number. Show that the n th convergent of the continued fraction representation of x agrees with the (n−1) th convergent of the continued fraction representation of 1/x. 1 1 1-18. Find the continued fraction expansions of √ and √ . Determine as many conver5 5−1 gents as you can. √ 1 1-19. Investigate the continued fraction expansions of 6 and √ . Determine as many con6 vergents as you can. 1-20.

9. The elements of A3 are ι= 1 2 3 , 1 2 3 1 2 3 , 2 3 1 1 2 3 . /2. 4. The cycle type of a permutation Suppose σ ∈ Sn . Now carry out the following steps. • Form the sequence 1 → σ(1) → σ 2 (1) → · · · → σ r1 −1 (1) → σ r1 (1) = 1 where σ k (j) = σ(σ k−1 (j)) and r1 is the smallest positive power for which this is true. • Take the smallest number k2 = 1, 2, . . , n for which k2 = σ t (1) for every t. Form the sequence 1 → σ(k2 ) → σ 2 (k2 ) → · · · → σ r2 −1 (k2 ) → σ r2 (k2 ) = 1 where r2 is the smallest positive power for which this is true.

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