1. Finite Groups & Subgroups

2. Order of a group Definition: The number of elements of a group (finite or infinite) is called its order. Notation: We will use |G| to denote the order of group G.

3. Examples |D 4 | = |D n | = | | = |Z n | = |U(8)| = |U(11)| = |Z| = 8 2n 4 n 4 10 ∞

4. Order of an element Definition: The order of an element g in a group G is the smallest positive integer n such that g n = e (In additive notation, ng = 0). If no such integer exists, we say g has infinite order. Notation: The order of g is denoted |g|.

5. Examples In D 4, |R 90 | = In D 4, |H| = In Z 10, |4| = In Z 11, |4| = In U(8), |5| = In U(9), |5| = In Z, |1| = 4 ( R 4 90 = R 0 ) 2 ( H 2 = R 0 ) 5 (54 mod 10 = 0) 11 (114 mod 11 = 0) 2 (5 2 mod 8 = 1) 6 {5, 7, 8, 4, 2, 1} ∞ (n1 ≠ 0 for n>0)

6. Group G (mod 35) |G| = e = |5| = |10| = |15| = |20| = |30| = 6 15 6 6 1 2 3 51015202530 525155302010 15301025520 1551015202530 2030252015105 2520525103015 3010203051525

7. Subgroups Definition: If a subset H of a group G is itself a group under the operation of G, then we say that H is a subgroup of G.

8. Notation: We write H ≤ G to mean H is a subgroup of G. If H is not equal to G, we write H < G. We say H is a proper subgroup of G. {e} is called the trivial subgroup. All other subgroups are nontrivial.

9. R0R0 R 90 R 180 R 270 HVDD' R0R0 R0R0 R 90 R 180 R 270 HVDD' R 90 R 180 R 270 R0R0 D'DHV R 180 R 270 R0R0 R 90 VHD'D R 270 R0R0 R 90 R 180 DD'VH HHDV R0R0 R 180 R 90 R 270 VVD'HDR 180 R0R0 R 270 R 90 DDVD'HR 270 R 90 R0R0 R 180 D' HDVR 90 R 270 R 180 R0R0

10. R0R0 R 90 R 180 R 270 HVDD' R0R0 R0R0 R 90 R 180 R 270 HVDD' R 90 R 180 R 270 R0R0 D'DHV R 180 R 270 R0R0 R 90 VHD'D R 270 R0R0 R 90 R 180 DD'VH HHDV R0R0 R 180 R 90 R 270 VVD'HDR 180 R0R0 R 270 R 90 DDVD'HR 270 R 90 R0R0 R 180 D' HDVR 90 R 270 R 180 R0R0

11. R0R0 R 90 R 180 R 270 HVDD' R0R0 R0R0 R 90 R 180 R 270 HVDD' R 90 R 180 R 270 R0R0 D'DHV R 180 R 270 R0R0 R 90 VHD'D R 270 R0R0 R 90 R 180 DD'VH HHDV R0R0 R 180 R 90 R 270 VVD'HDR 180 R0R0 R 270 R 90 DDVD'HR 270 R 90 R0R0 R 180 D' HDVR 90 R 270 R 180 R0R0

12. R0R0 R 90 R 180 R 270 HVDD' R0R0 R0R0 R 90 R 180 R 270 HVDD' R 90 R 180 R 270 R0R0 D'DHV R 180 R 270 R0R0 R 90 VHD'D R 270 R0R0 R 90 R 180 DD'VH HHDV R0R0 R 180 R 90 R 270 VVD'HDR 180 R0R0 R 270 R 90 DDVD'HR 270 R 90 R0R0 R 180 D' HDVR 90 R 270 R 180 R0R0

13. R0R0 R 90 R 180 R 270 HVDD' R0R0 R0R0 R 90 R 180 R 270 HVDD' R 90 R 180 R 270 R0R0 D'DHV R 180 R 270 R0R0 R 90 VHD'D R 270 R0R0 R 90 R 180 DD'VH HHDV R0R0 R 180 R 90 R 270 VVD'HDR 180 R0R0 R 270 R 90 DDVD'HR 270 R 90 R0R0 R 180 D' HDVR 90 R 270 R 180 R0R0

14. R0R0 R 90 R 180 R 270 HVDD' R0R0 R0R0 R 90 R 180 R 270 HVDD' R 90 R 180 R 270 R0R0 D'DHV R 180 R 270 R0R0 R 90 VHD'D R 270 R0R0 R 90 R 180 DD'VH HHDV R0R0 R 180 R 90 R 270 VVD'HDR 180 R0R0 R 270 R 90 DDVD'HR 270 R 90 R0R0 R 180 D' HDVR 90 R 270 R 180 R0R0

15. Subgroup tests Three important tests tell us if a nonempty subset of a group G is a subgroup of G. One-Step Subgroup Test Two-Step Subgroup Test Finite Subgroup Test

16. One-Step Test Let H be a nonempty subset of a group G. If ab -1 belongs to H whenever a and b belong to H, then H is a subgroup of G. (In additive groups: If a–b belongs to H whenever a and b belong to H, then H ≤ G.)

17. Proof of One Step Test. Let G be a group, and H a nonempty subset of G. Suppose ab -1 is in H whenever a and b are in H. (*) We must show: 1.In H, multiplication is associative 2.The group identity e is in H 3.H has inverses 4.H is closed under the group multiplication. Then, H must be a subgroup of G.

18. (1) Multiplication is Associative: Choose any elements a, b, c in H. Since H is a subset of G, these elements are also in the group G, so (ab)c = a(bc) as required.

19. (2) H contains e Choose any x in H. (Since H is nonempty there has to be some element x in H) Let a = x and b = x. Then a and b are in H, so by (*) ab -1 = xx -1 = e is in H, as required.

20. (3) H has inverses Choose any x in H. Let a = e and b = x. Since a and b are in H, ab -1 = ex -1 = x -1 must be in H as well.

21. (4) H is closed Choose any x and y in H. Let a = x and b = y -1. Since a and b are in H, ab -1 = x(y -1 ) -1 = xy is also in H. We have shown that H is closed under the multiplication in G, and that H is associative, contains the identity, and has inverses. Therefore, H is a subgroup of G.

22. To use the One-Step Test 1.Identify the defining property P that distinguishes elements of H. 2.Prove the identity has property P. 3.Assume that two elements have property P 4.Show that ab -1 has property P. Then by the one-step test, H ≤ G. H≠  a,b in H ab -1 in H

23. Example: One Step Prove: Let G be an Abelian group with identity e. Let H = {x |x 2 = e}. Then H ≤ G. Proof: e 2 = e, so that H is nonempty. Assume a, b in H. Then (ab -1 ) 2 = a(b -1 a)b -1 = aab -1 b -1 (G is Abelian) = a 2 b -2 = a 2 (b 2 ) -1 = ee -1 (since a and b in H) = e. By the one-step test, H ≤ G.

24. Example One-Step Prove: The set 3Z = {3n | n in Z} (i.e. the integer multiples of 3) under the usual addition is a subgroup of Z. Proof: 0 = 30, so 3Z is not empty. Assume 3a and 3b are in 3Z. Then 3a – 3b = 3(a–b) is in 3Z. By the One-Step test, 3Z ≤ Z.

25. Terminology Let H be a nonempty subset of a group G with operation *. We say "H is closed under *" or "H is closed" when we mean "ab is in H whenever a and b are in H" We say "H is closed under inverses" when we mean "a -1 is in H whenever a is in H"

26. Two Step Test Let H be a nonempty subset of group G with operation *. If (1) H is closed under * and (2) H is closed under inverses, then H ≤ G Proof: Assume a and b are in H. By (2), b -1 is in H. By (1) ab -1 is in H. By the one-step test, H ≤ G.

27. Finite Subgroup Test Let H be a nonempty finite subset of a group G. If H is closed under the operation of G, then H ≤ G. Proof. Choose any a in H. By the two step test, it only remains to show that a -1 is in H.

28. To show a -1 is in H If a = e, then a -1 (= e) is in H. If a ≠ e, consider the sequence a,a 2,a 3 … Since H is closed, all are in H. Since H is finite, not all are unique. Say a i = a j where i < j. Cancel a i to get e = a j-i. Since a ≠ e, j-i > 1. Let b = a j-i-1. Then ab = a 1 a j-i-1 = a j-i = e So b = a -1 and b belongs to H.

29. Definition Let a be an element of a group G. The cyclic group generated by a, denoted is the set of all powers of a. That is, = {a n | n is an integer} In additive groups, = {na | n is a integer}

30. is a subgroup Let G be group, and let a be any element of G. Then is a subgroup of G. Proof: a is in, so is not empty. Choose any x = a m and y = a n in. xy –1 = a m (a n ) -1 = a m-n which belongs to since m–n is an integer. By the one-step test, is a subgroup of G.

31. Example = 51015202530 525155302010 15301025520 1551015202530 2030252015105 2520525103015 3010203051525

32. Example = {25, 30, 15} 51015202530 525155302010 15301025520 1551015202530 2030252015105 2520525103015 3010203051525