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A course in abstract algebra by Nicholas Jackson

By Nicholas Jackson

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55 Let σ = ( x1 x2 . . xk ) be a finite cyclic permutation in some (possibly infinite) symmetric group Sym( X ). Then σ has length or periodicity k, which we denote by l (σ ) = k. This is equal to the order |σ | of σ in the group Sym( X ). 56 Let σ = ( x1 x2 ) be a period–2 cyclic permutation in some (possibly infinite) symmetric group Sym( X ). Then we call σ a transposition. We noted earlier that σ3 = ι and τ 6 = ι. More formally, the order |σ | of σ in S5 is 3, while the order |τ | of τ in S5 is 6.

Conversely, suppose that n is such that li |n for 1 π n = σin . . σkn = ι . . ι = ι. i k. Then Therefore, π n = ι if and only if li |n for 1 i k. So in particular, n must be the smallest positive integer which divides each of the li , which is to say that it must be the lowest common multiple lcm(li , . . , lk ). So |σ | = l (σ ) = 3, but |τ | = lcm(2, 3) = 6. Now let’s look again at the permutation σ = (2 3 5). This has length 3, but is it possible to express it as a product of even shorter (but possibly not disjoint) cyclic permutations?

Next we look at 4, the smallest number not yet seen, which is again left unchanged by σ. This gives us another trivial cycle (4). And now we’ve dealt with all of the numbers in {1, 2, 3, 4, 5}, so that’s the end, and our permutation can therefore be written as (1)(2 3 5)(4). Except that we’re really only interested in the nontrivial bits of the permutation, which in this case is the 3–cycle (2 3 5); the 1–cycles (1) and (4) don’t tell us anything nontrivial about the permutation, so we discard them, leaving us with σ = (2 3 5).

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