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 Previous assignment; 1. Pi = 3.141592654 approximately. Find the best rational approximations to Pi using numerators and denominators no larger than 1000. 3.141592654 -------------- 1.000000000 3 1000000000 3141592654 3000000000 ---------- 7 141592654 1000000000 991148578 ---------- 15 8851422 141592654 132771330 --------- 1 8821324 8851422 8821324 ------- 30098 293 30098 8821324 8818714 ----- 11 2610 30098 28710 ----- 1 1388 2610 1388 ---- 1 1222 1388 1222 ---- 7 166 1222 1162 ---- 2 60 166 120 --- 1 46 60 46 -- 3 14 46 42 -- 3 4 14 12 -- 2 2 4 4 --- 0 3 7 15 1 293 11 1 1 7 2 1, 3, 22, 333, 355, 104348, 1148183, 1252531, 2400714, 18057529, 0, 1, 7, 106, 113, 33215, 365478, 398693, 764171, 5747890, 1 3 3 2 38515772, 56573301, 208235675, 681280326, 1570796327 12259951, 18007841, 66283474, 216858263, 500000000 Notice that we did not get back the same numbers we started with. This is because the GCD is 2. If we multiply the last two numbers by the GCD we get the numbers we started with. The best approximation for Pi with numerator and denominator less than 1000 is 355/113. The error is -7 Pi - 355/113 = -2.67 10 = 0.000000267 355/113 = 3.141592920 Pi = 3.141592654 | error in the 7th place 2. There are three kinds of years. Calendar Year is 365 days or 366 days during leap year. Tropical year is 365 days, 5 hours, 48 minutes, 46 seconds ========================================= It is the time spend by the sun in its apparent passage from vernal equinox to vernal equinox. Sidereal year 365 days, 6 hours, 9 minutes, 9 seconds ======================================== It is the time spent by the sun making is apparent passage from a fixed star and back to the same position again. The difference in time from the Tropical year to the sidereal year is due the the precession of the equinoxes. Use the Tropical year against the day to get the best approximating fractions. Tropical year is ((365*24 + 5)*60 + 48 )*60 + 46 seconds long. One day is 24*60*60 seconds long. (a) Best approximate Tropical year/day. Here is the start of the magic tableau. Finish filling it out. 365, 4, 7, 1, 3, 5, 64, 1, 365, 1461, 10592, 12053, 46751, 245808, 15778463 <--- days 0, 1, 4, 29, 33, 128, 673, 43200 <--- years 1, 0, 1, 7, 8, 31, 163, 10463 days to insert 0, 0, 1, 7, 8, 31, 163, 10476 Gregorian Rule (b) The numbers in the 6 th column are 46751 and 128. This says that in 128 years there are 46751 days. If we compute 46751-365*128 = 31. This means that in a period of 128 years we have to insert 31 days. How many days have to be inserted for the other columns? (c) The Gregorian calendar rule is this. Put in a February 29 if the year is divisible by 4 unless it is a century. Centuries are leap years if they are divisible by 400. In 128 years the number of extra days added by this rule is Floor[128/4] - Floor[128/100] + Floor[128/400] = 32 - 1 + 0 = 31. Thus the Gregorian rule works for 128 years. Find how many of the extra days are accounted for by the Gregorian rules? (d) Approximately how often to we have to add an additional day in addition to the Gregorian rules? So in 43200 years we have to subtract 13 days because the Gregorian calendar puts in too many. That is, we have to drop a day every 3323 years from here on out. This material is found in Section 4.2 on pages 217 through 223 starting with "The Sylow Theorems." We give the proofs differently. Main idea: Subgroups of size p^e. Key words: p-sylow subgroup, conjugate, normal subgroup. Simple Goal: Prove the three 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. Remarks: Two subgroups H and K of a group G are called conjugate if there exists g in G such that g'Hg = K. A subgroup H of a group G is called normal if g'Hg = H for all g in G. That is, it has no conjugates. If there is only one Sylow Subgroup, it must be normal. A group G is called simple if there are {e} and G are the only normal subgroups. Proof of (a). Consider the set S of all the subSets ==== e e e of G of order p . There are C(p m,p ) of them and this number is NOT a multiple of p. ------------------------------------------------- | T1 T2 T3 T4 ............. | | | | | | | | | | | | | = S | | | | | | | | | ................Tn | ------------------------------------------------- One can let G act on these subsets by left multiplication. For Ti in S g*Ti = { gt | t is in Ti }. And g*Ti is again a subSet of size p^e and so must also be in S. S is partitioned into orbits. It cannot be that p divides the size of each of the orbits, since the total number of these subsets = |S| is not a multiple of p. There must be some Ti whose orbit size is not a multiple of p. Remember: | G | = | Stabilizer of Ti |*| size of the orbit of Ti | Let Hi = the stabilizer of such a Ti. We are going to show that Hi is the subgroup we are looking for because it must have exactly p^e elements. First, since all the factors of p must be in |Hi| because non can be left over to be in the orbit of Ti. This means | Hi | >= p^e. Second, picking any t in Ti, we must have Ht c Ti so |Hi t| <= |Ti| = p^e. Since | Hi | = | Hi t| we get |Hi| = p^e. We are not going to prove parts (b) and (c) today. Let us now give an application. Prove No group of order 15 is simple. Proof; 15 = 3*5 1 3 5 15 3-sylow x 5-sylow x Both the 3-sylow and the 5-sylow subgroups are normal. Thus any group of order 15 has two normal subgroups and is not simple. For a group of order 60, How many sylow subgroups are possible? 1 2 3 4 5 6 10 12 15 20 30 60 2-sylow x x x x 3-sylow x x x 5-sylow x x Assignment: 1. What are the 2-sylow subgroups of S4. 2. What are the 3-sylow subgroups of S4. 3. Show that every group of order 20 has a normal subgroup. 4. Show that every group of order 75 has two normal subgroups.