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47 Consider for m, n, a, b ∈ Z>0 the element am, bn ∈ Zm × Zn in the group (Zm × Zn, +). Show that the order of this element can be computed as follows o am, bn = lcm o am , o bn = lcm lcm(a, m) lcm(b, n) , a b and that it is a divisor of lcm(m, n). In particular, the group Zm × Zn is not cyclic if m and n share a common divisor apart from one.. 48 Compute the order of 621, 933 ∈ Z21 × Z33. 49 Let σ and π be two disjoint cycles in Sn of length k respectively l. Show that o(σ ◦ π) = lcm(k, l). D) The Homomorphism Theorem If there exists a group isomorphism from one group G to another group H then these two groups are the same from the point of view of group theory.

N} = · i=1 ai1. 35 Set ki = |ai1| and aij = σj−1(ai1) then due to (16) we get t {1, . . , n} = · (17) {ai1, ai2, . . , aiki }. i=1 It remains to show that σ = σ1 ◦ · · · ◦ σt where σi = (ai1 · · · aiki ) is a ki-cycle. For this let b ∈ {1, . . , n} so that b = aij = σj−1(ai1) for some 1 ≤ i ≤ t and some 1 ≤ j ≤ ki. We now apply σ to b and get σ(b) = σ(aij) = σj(ai1) = aij+1, ai1, if j < ki, if j = ki = σi(b). Since the decomposition in (17) is disjoint and since b as well as σi(b) are contained in {ai1, .

G. 1 2 3 2 1 3 = 2 1 3 1 2 3 . In order to keep things well-arranged we advise, however, to write the numbers in the first row always in an ascending order. 14 This is the content of the Theorem of Cayley. 60. 3 The above representation of a permutation has the nice side effect that we can very easily invert it by simply exchanging the two rows of the scheme. e. if σ= 1 2 ... n σ(1) σ(2) . . σ(n) ∈ Sn then its inverse σ−1 is given by σ−1 = σ(1) σ(2) . . σ(n) 1 2 ... n . Mathematicians are lazy people, or maybe one should better say that they are efficient.

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