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Subgroups Definition: a Subset H of a Group G Is a Subgroup of G If H Is

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Subgroups Definition: a Subset H of a Group G Is a Subgroup of G If H Is Subgroups Definition: A subset H of a group G is a subgroup of G if H is itself a group under the operation in G. Note: Every group G has at least two subgroups: G itself and the subgroup {e}, containing only the identity element. All other subgroups are said to be proper subgroups. Examples 1. GL(n,R), the set of invertible n ¥ n matrices with real entries is a group under matrix multiplication. We denote by SL(n,R) the set of n ¥ n matrices with real entries whose determinant is equal to 1. SL(n,R) is a proper subgroup of GL(n,R) . (GL(n,R), is called the general linear group and SL(n,R) the special linear group.) 2 3 2. In the group D4 , the group of symmetries of the square, the subset {e ,r,r ,r } forms a p proper subgroup, where r is the transformation defined by rotating units about the z- 2 axis. 3. In Z9 under the operation +, the subset {0, 3, 6} forms a proper subgroup. Problem 1: Find two different proper subgroups of S3. We will prove the following two theorems in class: Theorem: Let H be a nonempty subset of a group G. H is a subgroup of G iff (i) H is closed under the operation in G and (ii) every element in H has an inverse in H. For finite subsets, the situation is even simpler: Theorem: Let H be a nonempty finite subset of a group G. H is a subgroup of G iff H is closed under the operation in G . Problem 2: Let H and K be subgroups of a group G. (a) Prove that H « K is a subgroup of G. (b) Show that H » K need not be a subgroup † † Example: Let Z be the group of integers under addition. Define Hn to be the set of all multiples of n. It is easy to check that Hn is a subgroup of Z. Can you identify the subgroup Hn « H m ? Try it for H6 « H9 . † † † † † † † † † † † Note that the proof of part (a) of Problem 2 can be extended to prove that the intersection of any number of subgroups of G, finite or infinite, is again a subgroup. Cyclic Groups and Subgroups We can always construct a subset of a group G as follows: Choose any element a in G. Define n , i.e. a is the set consisting of all a ={a | n Œ Z} powers of a. Problem 3: Prove that a is a subgroup of G. Definition: a is called the cyclic subgroup generated by a. If a = G, then we say that G is a cyclic group. It is clear that cyclic groups are abelian. For the next result, we need to recall that two integers a and n are relatively prime if and only if gcd(a, n)=1. We have proved that if gcd(a, n)=1, then there are integers x and y such that ax + by =1. The converse of this statement is also true: Theorem: Let a and n be integers. Then gcd(a, n)=1 if and only if there are integers x and y such that ax + by =1. Problem 4: (a) Let U {a Z | gcd(a,n)=1}. Prove that U is a group under n = Œ n n multiplication modulo n. (Un is called the group of units in Zn .) (b) Determine whether or not Un is cyclic for n= 7, 8, 9, 15. We will prove the following in class. Theorem: Let G be a group and a Œ G. (1) If a has infinite order, then a is an infinite subgroup consisting of the distinct elements ak with k Œ Z . (2) If a has finite order n, then a is a subgroup of order n and 0 1 2 n-1 . a ={e = a ,a ,a ,...,a } † † Theorem: Every subgroup of a cyclic group is cyclic. Problem 5: Find all subgroups of U18. † Note: When the group operation is addition, we write the inverse of a by -a rather than a-1 , the identity by 0 rather than e, and ak by ka. For example, in the group of integers † † † under addition, the subgroup generated by 2 is 2 ={2k| k Œ Z}. Problem 6: Show that the additive group Z2 ¥ Z3 is cyclic, but Z2 ¥ Z2 is not. † † † † † † † † † † † † † † † † † † † Problem 7: Let G be a group of order n. Prove that G is cyclic if and only if G contains an element of order n. The notion of cyclic group can be generalized as follows. : Let S be a nonempty subset of a group G. Let S be the set of all possible products, in every order, of elements of S and their inverses. We will prove the following theorem in class. Theorem: Let S be a nonempty subset of a group G. (1) S is a subgroup of G that contains S. (2) If H is a subgroup of G that contains S, then H contains S . (3) S is the intersection of all subgroups of G that contain S. The second part of this last theorem states that S is the smallest subgroup of G that contains S . The group S is called the subgroup of G generated by S. Note that when S = {a}, S is just the cyclic subgroup generated by a. In the case when S =G, we say that G is generated by S, and the elements of S are called generators of G. k Example: Recall that we showed that every element in D4 could be represented by r or p ark for k=0, 1, 2, 3, where r is the transformation defined by rotating units about the 2 z-axis, and a is rotation p units about the line y=x in the x-y plane. Thus D4 is generated by S ={a, r}. Problem 8: Show that U15 is generated by {2, 13}. † † † † † † † † † † † † † † † † 10-1-2019 Cyclic Groups Cyclic groups are groups in which every element is a power of some fixed element. (If the group is abelian and I’m using + as the operation, then I should say instead that every element is a multiple of some fixed element.) Here are the relevant definitions. Definition. Let G be a group, g ∈ G. The order of g is the smallest positive integer n such that gn = 1. If there is no positive integer n such that gn = 1, then g has infinite order. In the case of an abelian group with + as the operation and 0 as the identity, the order of g is the smallest positive integer n such that ng = 0. Definition. If G is a group and g ∈ G, then the subgroup generated by g is hgi = {gn | n ∈ Z}. If the group is abelian and I’m using + as the operation, then hgi = {ng | n ∈ Z}. Definition. A group G is cyclic if G = hgi for some g ∈ G. g is a generator of hgi. If a generator g has order n, G = hgi is cyclic of order n. If a generator g has infinite order, G = hgi is infinite cyclic. Example. (The integers and the integers mod n are cyclic) Show that Z and Zn for n> 0 are cyclic. Z is an infinite cyclic group, because every element is a multipleof1(orof −1). For instance, 117 = 117·1. (Remember that “117 · 1” is really shorthand for 1+1+ ··· +1 — 1 added to itself 117 times.) In fact, it is the only infinite cyclic group up to isomorphism. Notice that a cyclic group can have more than one generator. If n is a positive integer, Zn is a cyclic group of order n generated by 1. For example, 1 generates Z7, since 1+1=2 1+1+1=3 1+1+1+1=4 1+1+1+1+1=5 1+1+1+1+1+1=6 1+1+1+1+1+1+1=0 1 In other words, if you add 1 to itself repeatedly, you eventually cycle back to 0. 0 6 1 5 2 4 3 acyclicgroupoforder7 Notice that 3 also generates Z7: 3+3=6 3+3+3=2 3+3+3+3=5 3+3+3+3+3=1 3+3+3+3+3+3=4 3+3+3+3+3+3+3=0 The “same” group can be written using multiplicative notation this way: 2 3 4 5 6 Z7 = {1,a,a ,a ,a ,a ,a }. In this form, a is a generator of Z7. It turns out that in Z7 = {0, 1, 2, 3, 4, 5, 6}, every nonzero element generates the group. On the other hand, in Z6 = {0, 1, 2, 3, 4, 5}, only 1 and 5 generate. Lemma. Let G = hgi be a finite cyclic group, where g has order n. Then the powers {1,g,...,gn−1} are distinct. Proof. Since g has order n, g, g2, ...gn−1 are all different from 1. Now I’ll show that the powers {1,g,...,gn−1} are distinct. Suppose gi = gj where 0 ≤ j<i<n. Then 0 <i − j < n and gi−j = 1, contrary to the preceding observation. Therefore, the powers {1,g,...,gn−1} are distinct. Lemma. Let G = hgi be infinite cyclic. If m and n are integers and m 6= n, then gm 6= gn. Proof. One of m, n is larger — suppose without loss of generality that m > n. I want to show that gm 6= gn; suppose this is false, so gm = gn.
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