301 Abstract Algebra 2:00 - 2:50 MWF Hentzel 432 Carver hentzel@iastate.edu Web Page http://www.math.iastate.edu/hentzel/class.301 Text: A First Course in Abstract Algebra, sixth Edition by John B. Fraleigh Course: Chapter 1,2,3,4,5 Friday, October 5. Main idea: Counting Arguments. Key words: Class Equation. Goal: Fill in a lot of loose ends which things people should know. Previous Assignment: 1. Let G be a group of order 56. (a) How many 2-sylow and 7-sylow subgroups are there. 1 2 4 7 8 14 28 56 2-sylow x x 7-sylow x x (b) If there are 8 7-sylow subgroups, how many elements of order 7 are there. In any 7-sylow subgroup with 7 elements, all the elements except the identity are of order 7. Therefore there are 6 elements of order 7 in each 7-sylow subgroup. Since 7 is a prime, none of these elements of order 7 can be in more than one 7-sylow subgroup. Thus the number of elements of order 7 is 8x7 = 48. (c) If there are 6 7-sylow subgroups, how many elements are there which are not of order 7. Is there enough to make more than one 2-sylow subgroup? There are 56-48 = 8 elements which are not of order 7. This is only enough for one 2-sylow subgroup. 2. Let H be a subgroup of the group G. Show that N(H) = {g in G | gHg' = H} is a subgroup of G. We have to show closure and inverses. closure: Let g1 and g2 elements in N(H). (g1g2)H(g1g2)' = (g1g2)H(g2'g1') = g1(g2Hg2')g1' = g1Hg1' = H. inverse: Let g be an element of N(H). g'H(g')' = g'Hg = g'(gHg')g = (g'g)H(g'g) = H. r 3. Show that there are no simple groups of order p m where p is prime and m < p. I will show that the p-sylow subgroup is normal. The number of p sylow subgroups divides the order of the group so the number of them is of the form p^s m. 0 <= s <= r and m divides n. The number of p sylow subgroups is also equal to one mod p. Therefore s = 0. Since 1 <= n <= m < p, the only possible number for n is 1. Therefore there is only one p-sylow subgroup and it is normal. Remark: If H and K are subgroups of a group G, then HnK is a subgroup of G. Proof: We have to show closure and inverses. closure: Given g1, g2 in HnK then g1,g2 are in H and g1, g2 are in K. Thus g1g2 is in H and g1g2 is in K since H and K are closed. Thus g1g2 is in HnK. inverse: Given g in HnK, then g is in H and g is in K so g' is in H and g' is in K because H and K have inverses. Thus g' is in HnK. Page 227 Lemma 4.3.8 If H and K are finite subgroups of a group G, then |H||K| |HK| = ------- |HnK| Given h in H and k in K, then hk = ht t'k for every t in HnK. Therefore the product hk can arise at least |HnK| times for different h's and different k's. Now suppose that h1k1 = hk. then h'h1 = kk1' = t and t is in HnK. h1 = ht, k1 = t'k This shows that duplication can arise only by this process involving elements of HnK. This means in HK, each element is duplicated |HnK| times and the number of distinct elements is |H||K|/|HnK|. Page 226 Theorem 4.3.7 If p and q are distinct primes with p < q, then every group G of order pq has a single subgroup of order q and this subgroup is normal in G. Hence G is not simple. If q is not congruent to 1 modulo p, then G is abelian and cyclic. Proof: 1 p q pq p-sylow x x q-sylow x There is only one q-sylow subgroup. So there is only one subgroup of order q and it is normal. If q =/= 1 mod p then the p-sylow group is also normal. H = p-sylow K = q-sylow HnK = {e} since any common elements would have to be of order p and also q which is impossible. |HK| = pq = G so every element of the Group is in HK. hkh'k' is in HnK so hkh'k' = e and hk = kh. So G = HK so given two elements in G, then can be written as h1k1 and h2k2. h1k1 h2k2 = h1h2 k1k2 since elements from H and K commute. = h2h1 k2k1 since H and K are cyclic and therefore commutative. = h2k2 h1k1 since elements from H and K commute. Since G is commutative, then G = ZpZq = Zpq and G is cyclic. The center of a group is the set of all elements which commute with everything else. Z[G] = {z in G | zg=gz for g in G}. The identity is always in the center of G. Page 225 Theorem 4.3.4: A group of order p^e where p is a prime and e >=1 always has a nontrivial center. Proof: Let G act on itself using conjugation. The orbits will which are singletons come from elements in the center. Thus G = Z[G] + orbits with more than one element. The size of the orbits divides |G| = p^e. Thus p divides the number of elements which are not in the center. Since p divides |G|, we must have p divides |Z[G]|. Since |Z[G]| >1, Z[G] must have at least p elements in it and so Z[G] is not a trivial subgroup. Page 226: Theorem 4.3.6. For a prime number p, every group G of order p^2 is abelian. Proof: Z[G] has order p or p^2. If the center has order p^2 then Z[G] is abelian. If the center has order p, pick g some element not in Z[G]. has order p so G = Z[G]. But if zg^i and wg^j are element of G with z,w in the center then zg^i wg^j = zw g^(i+j) = wg^j zg^i Assignment: Suppose that H and K are subgroups of a group G. 1. Show that HK need not be a group. You can find a counter example in S3. 2. Show that if H or K is normal, then HK is a subgroup. 3. Show that if H and K are normal, then HK is a normal subgroup.