MATH 413 – ADDITIONAL TOPICS in GROUP THEORY 1. Order in Abelian Groups 1.1. Order of a Product in an Abelian Group. the First
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MATH 413 { ADDITIONAL TOPICS IN GROUP THEORY ALLAN YASHINSKI 1. Order in Abelian Groups 1.1. Order of a product in an abelian group. The first issue we shall address is the order of a product of two elements of finite order. Suppose G is a group and a; b 2 G have orders m = jaj and n = jbj. What can be said about jabj? Let's consider some abelian examples first. The following lemma will be used throughout. Lemma 1.1. Let G be an abelian group and a; b 2 G. Then for any n 2 Z, (ab)n = anbn: Exercise 1.A. Prove Lemma 1.1. (Hint: Prove it for n = 0 first. Then handle the case where n > 0 using induction. Lastly prove the case where n < 0 by using inverses to reduce to the case where n > 0.) Example 1. Let G = Z3 × Z4, a = (1; 0), and b = (0; 1). Then m = jaj = 3 and n = jbj = 4. Notice that ja + bj = j(1; 1)j = 12. Example 2. Let G = Z2 × Z4, a = (1; 0), and b = (0; 1). Then m = jaj = 2 and n = jbj = 4. Notice that ja + bj = j(1; 1)j = 4. Example 3. Let G = Z6 × Z4, a = (1; 0), and b = (0; 1). Then m = jaj = 6 and n = jbj = 4. Notice that ja + bj = j(1; 1)j = 12. Here is the general fact uniting these examples. Exercise 1.B. Let G = Zm × Zn and let a = (1; 0), b = (0; 1), so that m = jaj and n = jbj. Prove that ja + bj = lcm(m; n), the least common multiple1 of m and n. One may start to conjecture that if jaj = m and jbj = n, then jabj = lcm(m; n). However, this need not be the case, as the following examples show. Example 4. Suppose G is any group and a 2 G has order m. Then a−1 has order m also, and the product aa−1 = e has order 1, which is not equal to lcm(m; m) = m. Example 5. In Z12, consider a = 2 and b = 4. Then jaj = 6 and jbj = 3, and lcm(6; 3) = 6. However a + b = 6 has order 2 6= 6. Now we'll prove a result that is consistent with all of the above. Proposition 1.2. Let G be an abelian group and let a; b 2 G. Let m = jaj; n = jbj; and r = jabj. Then r j lcm(m; n). 1That is, lcm(m; n) is the smallest positive integer t such that m j t and n j t. It is a fact that mn lcm(m; n) = : gcd(m; n) 1 2 ALLAN YASHINSKI Proof. Let t = lcm(m; n). Since m j t and n j t, we have t = mk and t = n` for integers k; `. Using Lemma 1.1, (ab)t = atbt = amkbn` = (am)k(bn)` = ekb` = e: From Theorem 7.9 of the text, this implies r j t, as desired. This narrows down the possibilities for the order of a product ab in an abelian group. The answer will depend on more than just the numbers m = jaj and n = jbj. It will also depend on where a and b sit in the group in relation to each other2. We would like to identify situations where the order of ab is maximal, that is, equal to the least common multiple of jaj and jbj. Theorem 1.3. Let a; b be elements of an abelian group G and let m = jaj and n = jbj. If hai \ hbi = feg, then jabj = lcm(m; n). Proof. Let r = jabj and t = lcm(m; n). From Proposition 1.2, we have r j t, so r ≤ t. By definition of order, we have e = (ab)r = arbr, using Lemma 1.1. It follows that ar = b−r 2 hbi. Thus ar 2 hai \ hbi = feg, and so ar = e. Hence the order of a, which is m, must divide r. We also have br = (b−r)−1 = (ar)−1 = e−1 = e: So n j r as well. Hence r is a positive common multiple of m and n. By definition of least common multiple, t = lcm(m; n) ≤ r: We've shown r ≤ t and t ≤ r, so we've shown r = t. Compare the result of Theorem 1.3 with Example 4 and Example 5. In those examples, hai \ hbi 6= feg. Corollary 1.4. Let a; b be elements of an abelian group G and let m = jaj and n = jbj. If m and n are relatively prime, then jabj = mn. Exercise 1.C. Use Theorem 1.3 to prove Corollary 1.4. (Hint: hai \ hbi is a subgroup of both hai and hbi. Apply Lagrange's theorem to determine the order of hai \ hbi.) 1.2. Order of a product in nonabelian groups. For nonabelian groups, basi- cally nothing can be said about how to relate jabj to jaj and jbj. Example 6. Let G = S3, and consider a = (12) and b = (23). Then jaj = jbj = 2. However ab = (12)(23) = (123) is an element of order 3. Notice 3 has seemingly nothing to do with lcm(2; 2) = 2. Exercise 1.D. Consider the elements 0 −1 0 1 a = ; b = 1 0 −1 −1 of the nonabelian group GL(2; Z). Show that jaj = 4, jbj = 3, and yet ab has infinite order. 2A vague statement indeed. For example, if there is overlap between the cyclic subgroups hai and hbi, as in some of the above examples, the order of ab turns out to be less than lcm(jaj; jbj). MATH 413 { ADDITIONAL TOPICS IN GROUP THEORY 3 These examples stand in stark contrast to our above results for abelian groups. It gets much worse. Here is a somewhat difficult theorem, which we shall neither use nor prove3. Theorem 1.5. For any integers m; n; r > 1, there exists a finite group G and elements a; b 2 G such that jaj = m, jbj = n, and jabj = r. This theorem says that there is no relationship between jaj; jbj; and jabj in general nonabelian groups. Thus we should appreciate the results we have above for abelian groups. 1.3. Maximal order in finite abelian groups. We return to studying abelian groups. Let G be a finite abelian group. Since every element of G has finite order, it makes sense to discuss the largest order M of an element of G. Notice that M divides jGj by Lagrange's theorem, so M ≤ jGj. We also have M = jGj if and only if G is cyclic. Thus for non-cyclic abelian groups, M < jGj. We shall prove the following theorem. Theorem 1.6. Let G be a finite abelian group, and let M be the largest order of an element of G. Then for any a 2 G, the order of a divides M. In particular, aM = e for all a 2 G. Since M ≤ jGj, this strengthens the results on orders of elements obtained from Lagrange's theorem. Let's consider some examples before proving the theorem. Example 7. If G is cyclic, then G contains an element of order jGj by definition of cyclic. So M = jGj. In this case, the results of Theorem 1.6 coincide with Corollary 8.6 from the text. Example 8. Consider the abelian group G = Z4 × Z6. Notice that jGj = 24. The element (1; 1) has order 12, and this is the maximal order of an element of G because G is not cyclic. So M = 12 < 24 = jGj. Theorem 1.6 says that the order of every element of G divides 12, hence is 1; 2; 3; 4; 6; or 12. Further every (k; `) 2 Z4 × Z6 satisfies 12(k; `) = (0; 0). Example 9. The last example can be generalized to Zm × Zn. We claim that the maximal order M = lcm(m; n). Let t = lcm(m; n). Since m j t and n j t, we have t = mr and t = ns for some r; s 2 Z. Notice that for any (k; `) 2 Zm × Zn, t(k; `) = (tk; t`) = (mrk; ns`) = (0; 0) in Zm × Zn: By definition of order, we have j(k; `)j ≤ t. Since (k; `) 2 Zm × Zn was arbitrary and M is the order of some element of Zm × Zn, we see M ≤ t. On the other hand, the element (1; 1) has order t by Exercise 1.B, which shows M ≥ t. Thus M = t = lcm(m; n). In particular, we see Zm × Zn is cyclic if and only if lcm(m; n) = mn, which happens if and only if gcd(m; n) = 1. Example 10. One can consider products of cyclic groups with more factors. For example, the maximal order of an element of Z2 × Z2 × Z2 × Z2 is M = 2. The maximal order of an element of Z2 × Z3 × Z6 × Z8 is M = 24. As we shall see later, every finite abelian group is a product of cyclic groups. So these types of examples are the only examples to consider. Theorem 1.6 really is an abelian theorem. It fails badly for nonabelian groups. 3A proof can be found of page 28 of http://www.jmilne.org/math/CourseNotes/GT.pdf 4 ALLAN YASHINSKI Example 11. The nonabelian group G = S3 contains elements of order 1; 2; and 3 3. The largest order is 3, but 2 - 3, and certainly a does not hold for all a 2 S3. We will prove Theorem 1.6 using the following lemmas.