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Finite Groups & Subgroups 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. Examples |D4| = |Dn| = |<R90>| = |Zn| = |U(8)| = |U(11)| = |Z| = 8 2n 4 n 4 10 ∞ Order of an element • Definition: The order of an element g in a group G is the smallest positive integer n such that gn = 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|. Examples In D4, |R90| = In D4, |H| = In Z10, |4| = In Z11, |4| = In U(8), |5| = In U(9), |5| = In Z, |1| = 4 ( R490 = R0) 2 ( H2 = R 0 ) 5 (5•4 mod 10 = 0) 11 (11•4 mod 11 = 0) 2 (52 mod 8 = 1) 6 {5, 7, 8, 4, 2, 1} ∞ (n•1 ≠ 0 for n>0) Group G (•mod 35) • 5 10 15 20 25 30 5 25 15 5 30 20 10 10 15 30 10 25 5 20 15 5 10 15 20 25 30 20 30 25 20 15 10 5 25 20 5 25 10 30 15 30 10 20 30 5 15 25 |G| = e = |5| = |10| = |15| = |20| = |30| = 6 15 6 6 1 2 3 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. 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. R90 D4 • R0 R90 R180 R270 H V D D' R0 R0 R90 R180 R270 H V D D' R90 R90 R180 R270 R0 D' D H V R180 R180 R270 R0 R90 V H D' D R270 R270 R0 R90 R180 D D' V H H H D V D' R0 R180 R90 R270 V V D' H D R180 R0 R270 R90 D D V D' H R270 R90 R0 R180 D' D' H D V R90 R270 R180 R0 R180 R90 D4 • R0 R90 R180 R270 R0 R0 R90 R90 R90 R180 R180 R180 R270 R270 R270 R0 H H D V V D' D D V D' D' H H V D D' R180 R270 H V D D' R270 R0 D' D H V R0 R90 V H D' D R90 R180 D D' V H V D' R0 R180 R90 R270 H D R180 R0 R270 R90 D' H R270 R90 R0 R180 D V R90 R270 R180 R0 R0 R180 R90 D4 • R0 R90 R180 R270 H V D D' R0 R0 R90 R180 R270 H V D D' R90 R90 R180 R270 R0 D' D H V R180 R180 R270 R0 R90 V H D' D R270 R270 R0 R90 R180 D D' V H H H D V D' R0 R180 R90 R270 V V D' H D R180 R0 R270 R90 D D V D' H R270 R90 R0 R180 D' D' H D V R90 R270 R180 R0 {R0,R180,H,V} D4 • R0 R90 R180 R270 H V D D' R0 R0 R90 R180 R270 H V D D' R90 R90 R R0 D' D H V 180 R270 R180 R180 R270 R0 R90 V H D' D R270 R270 R0 R90 R180 D D' V H H H D V D' R0 R180 R90 R270 V V D' H D R180 R0 R270 R90 D D V D' H R270 R90 R0 R180 D' D' H D V R90 R270 R180 R0 {R0,R180,D,D'} D4 • R0 R90 R180 R270 H V D D' R0 R0 R90 R180 R270 H V D D' R90 R90 R R0 D' D H V 180 R270 R180 R180 R270 R0 R90 V H D' D R270 R270 R0 R90 R180 D D' V H H H D V D' R0 R180 R90 R270 V V D' H D R180 R0 R270 R90 D D V D' H R270 R90 R0 R180 D' D' H D V R90 R270 R180 R0 {R0,H,V} not a subgroup of D4 • R0 R90 R180 R270 H V D D' R0 R0 R90 R180 R270 H V D D' R90 R90 R180 R270 R0 D' D H V R180 R180 R270 R0 R90 V H D' D R270 R270 R0 R90 R180 D D' V H H H D V D' R0 R180 R90 R270 V V D' H D R180 R0 R270 R90 D D V D' H R270 R90 R0 R180 D' D' H D V R90 R270 R180 R0 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 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.) 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. (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. (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. (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. (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. To use the One-Step Test 1. Identify the defining property P that distinguishes elements of H. H≠ 2. Prove the identity has property P. 3. Assume that two elements have a,b in H property P -1 in H -1 ab 4. Show that ab has property P. Then by the one-step test, H ≤ G. Example: One Step • Prove: Let G be an Abelian group with identity e. Let H = {x |x2 = e}. Then H ≤ G. • Proof: e2 = e, so that H is nonempty. Assume a, b in H. Then (ab-1)2 = a(b-1a)b-1 = aab-1b-1 (G is Abelian) = a2b-2 = a2(b2)-1 = ee-1 (since a and b in H) = e. By the one-step test, H ≤ G. 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 = 3•0, 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. 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" 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. 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. To show a-1 is in H If a = e, then a-1 (= e) is in H. If a ≠ e, consider the sequence a,a2,a3… Since H is closed, all are in H. Since H is finite, not all are unique. Say ai = aj where i < j. Cancel ai to get e = aj-i. Since a ≠ e, j-i > 1. Let b = aj-i-1. Then ab = a1 aj-i-1 = aj-i = e So b = a-1 and b belongs to H. Definition • Let a be an element of a group G. The cyclic group generated by a, denoted <a> is the set of all powers of a. That is, <a> = {an | n is an integer} • In additive groups, <a> = {na | n is a integer} <a> is a subgroup • Let G be group, and let a be any element of G. Then <a> is a subgroup of G. • Proof: a is in <a>, so <a> is not empty. Choose any x = am and y = an in <a>. xy–1= am(an)-1 = am-n which belongs to <a> since m–n is an integer. By the one-step test, <a> is a subgroup of G. Example • 5 10 15 20 25 30 5 25 15 5 30 20 10 10 15 30 10 25 5 20 15 5 10 15 20 25 30 20 30 25 20 15 10 5 25 20 5 25 10 30 15 30 10 20 30 5 15 25 • <25> = Example • 5 10 15 20 25 30 5 25 15 5 30 20 10 10 15 30 10 25 5 20 15 5 10 15 20 25 30 20 30 25 20 15 10 5 25 20 5 25 10 30 15 30 10 20 30 5 15 25 • <25> = {25, 30, 15}