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 Main idea: Subgroups of size p^e. Key words: p-sylow subgroup, conjugate, normal subgroup. Simple Goal: Prove the parts (b) and (c) of the sylow theorems. e Theorem: If | G | = p m, ( p is a prime and p does not divide m). e (a) Then G has a subgroup of order p . e (b) Any two subgroups of order p are conjugate. e (c) The number of subgroups of order p divides |G| and is equal to one mod p. Previous Assignment: 1. What are the 2-sylow subgroups of S4. ans: They are the symmetries of the square: {(1234), (1432), (13), (24), (12)(34), (13)(24), (14)(23), I} {(1243), (1342), (14), (23), (12)(34), (13)(24), (14)(23), I} {(1324), (1423), (12), (34), (12)(34), (13)(24), (14)(23), I} 2. What are the 3-sylow subgroups of S4. ans: They are the subgroups generated by 3-cycles. {(123),(132),I} {(124),(142),I} {(134),(143),I} {(234),(243),I} 3. Show that every group of order 20 has a normal subgroup. Divisors of 20 1 2 4 5 10 20 2-Sylow x x 5-Sylow x The 5-Sylow subgroup with 5 elements is normal 4. Show that every group of order 75 has two normal subgroups. Divisors of 75 1 3 5 15 25 75 3-Sylow x x 5-Sylow x The 5-sylow subgroup of order 25 is always normal. If the 3-Sylow subgroup is not normal, then there must be 25 of the 3-sylow subgroups. We count the number of elements of each order. Order 3 is 50 Order 5 or 25 is 24 Identity 1 ------- 75 I cannot prove with the knowledge you already have that the 3-sylow subgroup is normal. Proof of parts (b) and (c) done simultaneously. Yesterday, we showed that we have at least one subgroup of size p^e. We take one such subgroup. We will call it H. We look at all the conjugates of H. These conjugates are the orbit of H under conjugation. ---------------------------------------------------------- | _________________________ | | | H | | | |___________________ | | | orbit of H | | | | under conjugation | | | | | | | | | |_________________ | | | | | | |_____________________| | | | ---------------------------------------------------------- The orbit of H under conjugation is given by orbit of H = { gHg' | g is in G.} We will need the following lemma. Lemma: If H and K are two p-sylow subgroups and hKh' = K for all h in H, then H=K. We will give the proof later. We first notice that since this is an orbit with respect to the action by the group G, then the size of the orbit divides |G|. This says that the number of conjugates of H is a divisor of G. We now decompose the orbit of H in two ways. (a) The first way is to use the group H to act on the orbit. Since the orbit contains all conjugates by all of G, it will contain all conjugates by elements of H. However, there will be lots of elements which are conjugate under the action by G which are no longer conjugate using just the elements of H. Therefore this one big orbit becomes partitioned into a series of smaller orbits. The size of each of these smaller orbits is a divisor of |H| = p^e. In these smaller orbits, one of them, {H} is a singleton orbit. And p divides the sizes of all the other smaller orbits. This says that the number of conjugates of H is equal to 1 mod p. (b) We will now argue that all of the p-sylow subgroups are conjugate to H. Suppose that there were a p-sylow subgroup K which was not conjugate to H. If we decompose the orbit of H into smaller orbits using action by K, then none of these smaller orbits will be singletons. Using part(a) and part(b) We know that the size of the orbit of H is both a multiple of p and not a multiple of p. The source of this contradiction was the assumption that there is some K outside the orbit of H. This means that all of the p-sylow subgroups are in the orbit of H and consequently, all of the p-sylow subgroups are conjugate. We finish the proof by proving the two necessary lemmas. Lemma: If p is a prime and p does not divide m, then Combinations[m p^e, p^e] is not divisible by p, Proof: Combinations[m p^e, p^e] = (m p^e)(m p^e-1) (m p^e-2) (m p^e-3) .....(m p^e-i).... (m p^e-(p^e-1)) ----------------------------------------------------------------------- ( p^e)( p^e-1) ( p^e-2) ( p^e-3)......( p^e-i).....( p^e-(p^e-1) ) The argument is that all powers of p have to cancel in the above fraction. If there is a power of p which is introduced by mp^e-i, that power of p must also divide i. But then that exact same power of p is a factor of p^e-i and so cancels out. Lemma: If H and K are p-sylow subgroups and hKh' = K for all h in H, then H=K. If we look at the cosets generated by choosing the elements form h, they look like ------------------------------------- | | ------------------------------------- | | ------------------------------------- | | ------------------------------------- | | ------------------------------------- | | ------------------------------------- | | ------------------------------------- h2 | h2, h2k1, h2k2, h2k3, ... | ------------------------------------- h1 | h1, h1k1, h1k2, h1k3, ... | ------------------------------------- | e, k1, k2, k3, ... | ------------------------------------- The total number of elements in this table is |K|*(number of distinct cosets). The order of H is p^e and the number of cosets is also a power of p because these cosets are the orbit of K under the group action by H and the size of the orbit is a divisor of |H|. The above construction is always true, whenever H and K are p-groups. But when H and K are p-sylow subgroups there is a contradiction because the above elements form a group with more than p^e elements. We will show HK is a group. h1k1 h2k2 = h1 (h2 h2') k1 h2 k2 = (h1 h2) (h2' k1 h2) k2 <------> <-------------> H K (hk)' = k' h' = (h'h)(k'h' = h' (hk'h') H K Problems: 1. Let G be a group of order 56. (a) How many 2-sylow and 7-sylow subgroups are there. (b) If there are 8 7-sylow subgroups, how many elements of order 7 are there. (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? 2. Let H be a subgroup of the group G. Show that N(H) = {g in G | gHg' = H} is a subgroup of G. r 3. Show that there are no simple groups of order p m where p is prime and m < p.