Finite Groups Subgroups Order of a group Definition
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 |D 4| = |Dn| = |<R 90>| = |Zn| = |U(8)| = |U(11)| = |Z| = 8 2 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 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 490 = R 0) 2 ( H 2 = 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.
• R 0 R 90 R 180 R 270 H V D D' R 90 R 180 R 270 R 0 D' D H V R 180 R 270 R 90 V H D' D R 270 R 0 R 90 R 180 D D' V H H H D V D' R 0 R 180 R 90 R 270 V V D' H D R 180 R 270 R 90 D D V D' H R 270 R 90 R 180 D' D' H D V R 90 R 270 R 180 R 0
• R 0 R 90 R 180 R 270 H V D D' R 90 R 180 R 270 R 0 D' D H V R 180 R 270 R 90 V H D' D R 270 R 0 R 90 R 180 D D' V H H H D V D' R 0 R 180 R 90 R 270 V V D' H D R 180 R 270 R 90 D D V D' H R 270 R 90 R 180 D' D' H D V R 90 R 270 R 180 R 0
• R 0 R 90 R 180 R 270 H V D D' R 90 R 180 R 270 R 0 D' D H V R 180 R 270 R 90 V H D' D R 270 R 0 R 90 R 180 D D' V H H H D V D' R 0 R 180 R 90 R 270 V V D' H D R 180 R 270 R 90 D D V D' H R 270 R 90 R 180 D' D' H D V R 90 R 270 R 180 R 0
• R 0 R 90 R 180 R 270 H V D D' R 90 R 180 R 270 R 0 D' D H V R 180 R 270 R 90 V H D' D R 270 R 0 R 90 R 180 D D' V H H H D V D' R 0 R 180 R 90 R 270 V V D' H D R 180 R 270 R 90 D D V D' H R 270 R 90 R 180 D' D' H D V R 90 R 270 R 180 R 0
• R 0 R 90 R 180 R 270 H V D D' R 90 R 180 R 270 R 0 D' D H V R 180 R 270 R 90 V H D' D R 270 R 0 R 90 R 180 D D' V H H H D V D' R 0 R 180 R 90 R 270 V V D' H D R 180 R 270 R 90 D D V D' H R 270 R 90 R 180 D' D' H D V R 90 R 270 R 180 R 0
• R 0 R 90 R 180 R 270 H V D D' R 90 R 180 R 270 R 0 D' D H V R 180 R 270 R 90 V H D' D R 270 R 0 R 90 R 180 D D' V H H H D V D' R 0 R 180 R 90 R 270 V V D' H D R 180 R 270 R 90 D D V D' H R 270 R 90 R 180 D' D' H D V R 90 R 270 R 180 R 0
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 |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.
Example One-Step • Prove: The set 3 Z = {3 n | n in Z} (i. e. the integer multiples of 3) under the usual addition is a subgroup of Z. • Proof: 0 = 3 • 0, so 3 Z is not empty. • Assume 3 a and 3 b are in 3 Z. • Then 3 a – 3 b = 3(a–b) is in 3 Z. • By the One-Step test, 3 Z ≤ 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, a 2, a 3… 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 = a 1 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}
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