TEST Abstract Algebra Math 301 FRIDAY October 19 Hentzel 1. How would you measure 12 cups using a 61 cup measure and a 29 cup measure? 2 29 61 58 -- 9 3 29 27 -- 1 2 3 2 -- 2 1 2 2 -- 0 2 9 1 2 1 2 19 21 61 0 1 9 10 29 -21 29 + 10 61 = 1 -252 29 + 120 61 = 12 61t 29 - 29t 61 = 0 small = 61t-252 big = 120-29t 61 t -252 should be about 0 t approximately 4.13 120-29t should be about 0 t approximately 4.1379 f[t_] := {61 t -252,120-29t} take 4 of the big scoops and remove 8 of the small scoops. 4 61 - 8 29 = 12 2. In the RSA code, n = 17*19 and r = 7. What is the decoding exponent. (17-1)(19-1) = 288 41 7 288 287 --- 7 1 7 7 --- 0 41 7 1 41 288 0 1 7 -41 7 +1 288 = 1 The private key is -41 which is 288-41 = 247 Do[Print[i," ",Mod[(i^7)^247,17 19]],{i,1,17 19}] 3. (a) State the first Sylow Theorem. Let G be a group of order p^e m where p is a prime and p does not divide m. Then G has a subgroup of order p^e. (b) State the second and third Sylow Theorem. The number of p-sylow subgroups divides the order of G and is equal to 1 mod p. (c) Now prove that the number of conjugates of a p-sylow subroup is equal to 1 mod p. You may use the first Sylow subgroup as well as the lemma that says if P and Q are both p-sylow subgroups and pQp' = Q for all p in P, then P=Q. _________________________________________ | | | | | P * * * * * * * * * * * * * * * * | | | |_______________________________________| We know that there exists at least one p-sylow subgroup. Call it P. The asterisks are the orbit of P under conjugation by all elements of G. We Now let only P itself act on this orbit. This will decompose the orbit into several smaller orbits. _________________________________________ | | | | || P|* * * *|* * * * *|* *|* * * * *| | | | |_______________________________________| By the lemma only one of these smaller orbits is a singleton and that smaller orbit is {P}. The rest of the smaller orbits have more than one element and the number of these elements in the smaller orbits is always a divisor of p^e. All of the smaller orbits except one have a multiple of p elements. The total number of element in the original orbit is one more than a multiple of p. 4. Give the possible numbers of p-sylow subgroups for a group of size 36. 1 2 3 4 6 9 12 18 36 2 x x x 3 x x 5. For the following puzzle peg board, if it could be reduced to one peg by horizontal and vertical jumps, where must the last peg be located? (o) = peg, ( ) = hole (o) (o) (o) (o) (o) (o) (o) (o) (o) (o) (o) (o) ( ) (o) (o) (o) (o) ( ) (o) (o) (o) (o) (o) (o) (o) (o) (o) (o) (o) (o) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (X) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (X) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 6. Find a permutation pi such that -1 pi (12)(34) pi = (13)(24) ans (23) 7. Suppose that G is a group and H is a normal subgroup of G. (a) What is the definition of a normal subgroup. H is a normal subgroup of G if gHg' = H for all g in G. (b) Prove that Hx*Hy c H(xy) for all x,y in G. If h1 and h2 are elements of H, then h1 x*h2 y = h1 (x h2) y = h1 (x h2 x') x y = h1 h3 x y since x h2 x' is in H. So (h1 x)*(h2 y) is in H(xy) 8. Suppose that f:G-->H is a homomorphism of group G into Group H with kernel K. We define a map g(Ka) = f(a). (a) Explain why one might suspect that g is not really defining a function. The domain of G is G/K. The cosets of G/K have lots of different names. Thus Ka = Kb = Kc ... where a,b,c are not equal. From the definition it seems that perhaps one can have Ka = Kb but f(a) =/= f(b). (b) Prove that the map g is well defined. If k is in the kernel of f, then f(k) = eH. If Ka = Kb then there exists k1 and k2 in K such that k1 a = k2 b. f(k1 a) = f(k1) f(a) = f(a). f(k2 b) = f(k2) f(b) = f(b). Therefore, f(a) = f(b) whenever Ka = Kb.