24 Chapter 1. Groups I (ii) Each βi has order ri, by (i). Suppose that αM = (1). Since the βi commute, (1) = αM = (β1 · · · βt)M = β1 M · · · βt M . By Exercise 1.21 on page 16, disjoint- ness of the β’s implies that βi M = (1) for each i, so that Proposition 1.26 gives ri | M for all i that is, M is a common multiple of r1,...,rt. On the other hand, if m = lcm{r1,...,rt}, then it is easy to see that αm = (1). Therefore, α has order m. (iii) Write α as a product of disjoint cycles and use (ii). For example, a permutation in Sn has order 2 if and only if it is a product of disjoint transpositions. Example 1.28. Suppose a deck of cards is shuffled, so that the order of the cards has changed from 1, 2, 3, 4,..., 52 to 2, 1, 4, 3,..., 52, 51. If we shuffle again in the same way, then the cards return to their original order. But a similar thing happens for any permutation α of the 52 cards: if one repeats α sufficiently often, the deck is eventually restored to its original order. One way to see this uses our knowledge of permutations. Write α as a product of disjoint cycles, say, α = β1β2 · · · βt, where βi is an ri-cycle (our original shuffle is a product of disjoint transpositions). By Proposition 1.27, α has order k, where k is the least common multiple of the ri. Therefore, αk = (1). Here is a more general result with a simpler proof (abstract Algebra can be easier than Algebra): we show that if G is a finite group and a G, then ak = 1 for some k 1. Consider the list 1,a,a2,...,an,.... Since G is finite, there must be a repetition occurring on this infinite list: there are integers m n with am = an, and hence 1 = ama−n = am−n. We have shown that there is some positive power of a equal to 1. [Our original argument that αk = (1) for a permutation α of 52 cards is still worthwhile, because it gives an algorithm computing k.] Let us state formally what we proved in Example 1.28. Proposition 1.29. If G is a finite group, then every x G has finite order. Table 3 for S5 augments Table 2 on page 10. Cycle Structure Number Order Parity (1) 1 1 Even (1 2) 10 2 Odd (1 2 3) 20 3 Even (1 2 3 4) 30 4 Odd (1 2 3 4 5) 24 5 Even (1 2)(3 4 5) 20 6 Odd (1 2)(3 4) 15 2 Even 120 Table 3. Permutations in S5.
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