Math 546, Exam 2 Information.


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1 Math 546, Exam 2 Information. 10/21/09, LC 303B, 10:1011:00. Exam 2 will be based on: Sections 3.2, 3.3, 3.4, 3.5; The corresponding assigned homework problems (see boylan/sccourses/546fa09/546.html) At minimum, you need to understand how to do the homework problems. Lecture notes: 9/1410/16. Topic List (not necessarily comprehensive): You will need to know: theorems, results, and definitions from class. 3.2: Subgroups. Let H be a subset of a group G. H is a subgroup if and only if H has inverses and is closed under the operation in G. If H is finite, then it is a subgroup if and only if it is closed under the operation in G. If H and K are subgroups of G, then H K is a subgroup of G, while it is generally not true that H K is a subgroup of G. Let a G. Then the cyclic subgroup of G generated by a is < a >= {x G : x = a n for some n Z}. In particular, G is cyclic if a G for which G =< a >. Cyclic groups include, for example, Z and Z n. For all n 3, S n is noncyclic (in fact, nonabelian); Z n is cyclic if and only if n {2, 4, p a, 2p a } where p is an odd prime. Let a G. Then the order of a, denoted by o(a) is the least positive integer n for which a n = e. If such an n N exists, a has finite order, o(a) = n. If not, then a has infinite order. Proposition and Corollary give important properties of order. An important example of the corollary is Euler s Theorem, which says that if gcd(a, n) = 1, then a φ(n) 1 (mod n). Lagrange s Theorem: If H is a subgroup of G, then we have H G. The key ingredient in the proof is an equivalence relation defined as follows: Let H be a subgroup of G. Then we have a b ab 1 H a Hb = {hb : h H}.
2 The equivalence classes are right cosets of H in G: if a G, the right coset of H in G represented by a is [a] = Ha = {ha : h H}. The coset represented by the identity element e G is He = H. All cosets have the same size, namely H. Further useful facts: 1. If G is cyclic, then G is abelian. 2. If G = p is prime, then G is cyclic. 3. Let a, b G. Then we have: o(a) = o(a 1 ); o(ab) = o(ba); if ab = ba and gcd(o(a), o(b)) = 1, then o(ab) = o(a)o(b). 3.3: Constructing examples. 1. Suppose that G = 4. Then, up to isomorphism, there are two possibilities for G: G = Z 4 or G = Z 2 Z Suppose that G = 6. Then, up to isomorphism, there are two possibilities for G: G = Z 6 or G = S 3. We studied two types of products: Let H, K G. Then the product HK is given by HK = {hk : h H, k K}. Now, suppose that H, K are subgroups of G. Then, we have Further, we proved that HK is a subgroup HK = KH (as sets). HK = H K H K. If G is abelian, then the condition HK = KH is automatically satisfied. For example, if a, b Z, then az + bz = gcd(a, b)z. Let G 1 and G 2 be groups. Then the direct product G 1 G 2 is given by G 1 G 2 = {(g 1, g 2 ) : g 1 G 1, g 2 G 2 }. The direct product G 1 G 2 is a group. If (a 1, a 2 ), (b 1, b 2 ) G 1 G 2, then (a 1, b 1 ) (a 2, b 2 ) = (a 1 b 1, a 2 b 2 ). The operation in the first entry is the operation in G 1 ; the operation in the second entry is the operation in G 2. The order of G 1 G 2 is G 1 G 2. The direct product can be generalized to any (finite) number of factors. The order of (a, b) G 1 G 2 is given by o((a, b)) = lcm(o(a), o(b)). 2
3 Now, let F be a field. We studied the matrix group GL n (F ), the general linear group of degree n over F. When p is prime, the group Z p is actually a field, in which case GL n (Z p ) is a group. 3.4: Isomorphisms. Two groups G 1 and G 2 are isomorphic (we write G 1 = G2 ) if and only if there is a map φ : G 1 G 2 (an isomorphism) which is: a bijection (it is oneone and onto); a homomorphism: a, b G 1, we have φ(ab) = φ(a)φ(b). If G 1 and G 2 are isomorphic, they are essentially the same groups; the elements are relabeled by φ. As such, we must have: 1. φ(a 1 a m ) = φ(a 1 ) φ(a m ); 2. φ(a 1 ) = φ(a) 1 ; 3. φ(e) = e (the identity in G 1 maps to the identity in G 2 ); 4. n Z, we have φ(a n ) = φ(a) n. Suppose that G 1 = G2 via an isomorphism φ. Further properties (Propositions 3.4.2, 3.4.3): Isomorphism is an equivalence relation. If o(a) = n in G 1, then o(φ(a)) = n in G 2. If G 1 is abelian, then so is G 2 ; if G 1 is cyclic, then so is G 2. In order to determine that G 1 = G2 (if this is, in fact, the case), some strategies are: Show that G 1 G 2. Show that one of G 1, G 2 is abelian; the other is not. Show that one of G 1, G 2 is cyclic; the other is not. Show that G 1 and G 2 have a different number of elements of a fixed order n. Lastly, we defined the kernel of a homomorphism φ : G 1 G 2 : ker φ = {x G 1 : φ(x) = e, the identity in G 2 } G 1. It turns out that ker φ is a subgroup of G 1. A homomorphism φ is oneone ker φ = {e} (Prop ). As an application (Prop ), we proved: If m, n N with gcd(m, n) = 1, then we have Z mn = Zm Z n ; Z mn = Z m Z n. 3.5: Cyclic groups. Let G be a cyclic group. Theorem 3.5.2, says: 1. if G is infinite, then G = Z; 2. if G is finite of order n, then G = Z n. 3
4 We seek to classify the subgroups of a cyclic group G. The first step in this direction is Theorem 3.5.1: Every subgroup H of a cyclic group G is cyclic. Consequences include: G infinite. If G =< a > is infinite, then the subgroups of G are exactly Moreover, we have {H =< a m >: m N}. < a m 1 > < a m 2 > m 2 m 1. Noting that G = Z in this case, the subgroups of Z are {H = mz : m N}, and we have m 1 Z m 2 Z m 2 m 1. G finite of order n. Let G =< a > have order n; thus, o(a) = n. Let k Z and let d = gcd(n, k) (so 1 d n). Then < a k >=< a d >, and o(a k ) = The subgroups of G are precisely: {H =< a n d >: d n}. n gcd(n, k) = n d = o(ad ). Specifically, for every d n, there is a unique subgroup, H =< a n d > = Z d of order d. We have < a k >= G (so a k generates G) gcd(n, k) = 1. It follows that #{generators of G} = #{elements of order n} = φ(n). Furthermore, if d n, we have Suppose that m, k n. Then we have #{elements of order d} = φ(d). < a m > < a k > k m. Recall that G = Z n in this case. If [m] Z n, we write < [m] >= mz n = {[m], [2m],..., [(n 1)m], [nm] = [0]}. The facts above translate as follows: If k Z and d = gcd(n, k), then kz n = dz n, and o([k]) = The subgroups of Z n are precisely { H = n } d Z n : d n ; 4 n gcd(n, k) = n d = o([d]).
5 the subgroup n d Z n = Z d has order d. The class [k] generates Z n gcd(n, k) = 1, so the number of generators is φ(n). The number of elements of order d n in Z n is φ(d). Lastly, if d 1, d 2 n, then we have d 1 Z n d 2 Z n d 2 d 1. The above facts imply that the number of subgroups of a cyclic group of order n is given by the number of divisors of n = p a 1 1 p a j j, ν(n) = (a 1 + 1) (a j + 1). Furthermore, we find that if n = p a 1 1 p a j is the prime factorization of n, then Z n = Zp a 1 1 Z a p j. j 5
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