Second Course in Abstract Algebra
Second Course in Abstract Algebra MATH 114
Popular in Course
Popular in Mathematics (M)
This 11 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 114 at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 55 views. For similar materials see /class/226606/math-114-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.
Reviews for Second Course in Abstract Algebra
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 10/22/15
REVIEW MATH 114 What do you have to know for the rst midterm Groups De nitions of a group a subgroup a normal subgroups and a quo tient Lagrange7s theorem the order of a subgroup divides the order of the group lsomorphism theorems Action of G on a set X is the map G gtlt X a X such that 1x z and gh z 90m A G orbit O of z E X is the set 9x for all g E G The stabilizer of z E X is the subgroup Gm g E G l 9x There is the identity lOHGzl Sylow theorems Let lGl pnq p be prime and p q 1 A Sylow subgroup is a subgroup of order p in G There exists at least one Sylow p subgroup The number of Sylow p subgroups is 1 modulo p All Sylow p subgroups are conjugate A group G is solvable if there exist a chain of subgroups GG1G23Gn1 such that Gi1 is normal in G for all 239 S n 7 1 and GiGHL is abelian If N is a solvable normal subgroup of G and GN is solvable then G is solvable A subgroup and a quotient group of a solvable subgroup is solvable A group of order p is solvable for any prime p and n 2 1 Fundamental theorem of abelian groups Any nitely generated abelian group is a direct product of cyclic subgroups Polynomials lrreducible polynomials division algorithm factorization theorem theorem 11 in Artin Eisenstein criterion and other ways to check if a polynomial is irreducible Field theory Field extensions F C E The degree is the dimension of E over F If F C E C B then Algebraic element mini mal polynomial splitting elds automorphism group AutE and Autp E Theorems 78910 and 13 in Artin Date March 1 2006 2 REVIEW MATH 114 Review exercises 1 Let G be a group of order 312 Prove that G is solvable 2 Let n 2 5 Prove that a proper non trivial normal subgroup of Sn coincides with An 3 Peter makes toys by coloring the faces of wooden cubes in such way that no faces have the same color How many different toys can he make if he has 9 colors 4 Let p be a prime number and x2 1 is reducible over Z17 Prove that p E 1 mod 4 5 Find the degree of the splitting eld of the polynomial 5 7 7 over Q 6 Find the degree of the splitting eld of the polynomial 13 1 over Q 7 Let F be a eld Prove that the number of irreducible polynomials in F is in nite 8 Check that the number 04 cos 20 is algebraic and nd the minimal polynomial of 04 over Q REVIEW MATH 114 3 Solutions 1 Use 312 3 gtlt 13 gtlt 8 The number of 13 subgroups divides 24 and is congruent to 1 modulo 13 Therefore there is only one 13 subgroup Denote it by N Then N is cyclic therefore solvable GN has 24 elements and therefore solvable by homework problem Hence G is solvable 2 Let N be a proper non trivial subgroup of S Then N An is a normal subgroup in An But An is simple hence N An 1 or N An An In the latter case N A since N is proper and the index of An in Sn is 2 If N An 1 then N can contain at most one odd permutation Indeed if it contains two odd permutations s and t then 52 st be even permutations in N hence 52 st 1 Thus N 15 On the other hand one can nd a permutation u such that 115114 74 5 But 115114 E N Contradiction 3 Let G be the group of rotations of a cube We proved in class that G is isomorphic to S4 in particular G1 24 Enumerate faces in some way There are 9 gtlt 8 gtlt 7 gtlt 6 gtlt 5 gtlt 4 ways to assign a color to a number The group G acts on the set of assignments Each orbit has 24 elements since the stabilizer of each color assignment is trivial Therefore the number of orbits is 9X8X7X6X5X4 24 4 If x2 1 is reducible then it has a root a E Z Then 042 71 and 044 1 Therefore a has order 4 in the multiplicative group Z By Lagrange7s theorem 4 divides lZz p71 5 The polynomial x5 7 7 is irreducible by Eisenstein criterion and has one real root Denote it by a All other roots are aw DMZ 04103 and 04104 where w is the fth root of 1 Therefore the splitting eld is Q wa Note that Q w Q 4 because the minimal polynomial of an over Q is x4 3 2 z 1 irreducible as proved in homework and Q a Q 5 since the minimal polynomial of a over Q is 5 7 7 Thus 4 and 5 divide Q aw On the other hand Q CW1 Q Q a Q Q aw Q M S 20 Therefore Q aw Q 20 6 Note that the roots of x13 1 are 71 76 782 7 812 where 8 is the 13 th root of 1 Therefore Q 8 is the splitting eld of 13 1 The degree Q 8 Q 12 because the minimal polynomial for 8 is 2520 1211H139 7 Assume that the number of irreducible polynomials is nite Let p1 pn be all irreducible polynomial Then 12 p1xpn1 must have an irreducible divisor but p do not divide q Contradiction 4 REVIEW MATH 114 8 Note that cos 60 Use the formula cos3ltp cos3 p 7 3cos psin2ltp cos3 p 7 3cos ltplt17 cos2 p 4cos3 p 7 3cosltp Therefore 1 4M 7 3a 7 E 0 We claim that the minimal polynomial for 04 is 8x3 7 6x 7 1 We need to check that it is irreducible Make the substitution m lt suf ces to show that y3 7 3y 7 1 is irreducible Possible rational roots of y3 7 3y 7 1 are 1 and 71 but they are not roots by direct checking REVIEW EXERCISES MATH 114 1 Let G be a transitive subgroup of Sn 1 Prove that if n is prime7 then G contains an n cycle b Show that a is not true if n is not prime 2 Let F be a eld such that the multiplicative group F is cyclic Prove that F is nite 3 Let G be a transitive subgroup of 56 which contains a 5 cycle Prove that G is not solvable 4 Let F be a eld and charF 31 27 0476 E F Prove that F F if and only of 046 is a square in F 5 Find the minimal polynomial for 1 3V BxZl over Q 6 Prove that any algebraically closed eld is in nite 7 ls x3 z 1 irreducible over F256 8 Which of the following extensions are normal QCQ1 7 QCQWEMBT QCQWEME Qmo 10 Let Q C F be a nite normal extension such that for any two sub elds E and K of F either K C E or E C K Then the Galois group of F over Q is cyclic of order p for some prime number p 11 Let F C B C E be a chain of extensions such that F C B is normal and B C E is normal ls it always true that F C E is normal 12 Find the Galois group of 2 7 3 2 1 3 7 6 over Q 13 Find the Galois group of 4 3x 5 over Q 14 Let p be a prime number Prove that n p is constructible if and only if n 2k for some k 9 Determine if Date May 67 2006 2 REVIEW EXERCISES MATH 114 15 Prove that any sub eld of Q coincides with Q dxE for some divisor d of n 16 Prove that there exists a polynomial of degree 7 whose Galois group over Q is Z7 17 Let f E Q be an irreducible polynomial of odd prime degree p solvable in radicals Prove that the number of real roots of f equals p or 1 18 Let fz 6 F2 be an irreducible polynomial Prove that fz divides x256 7 x if and only if the degree of f is 12 74 or 8 19 Suppose that the Galois group over Q of a polynomial f E Q has odd order Prove that all roots of f are real 20 Find the Galois group of 6 7 8 over Q SOLUTIONS FOR REVIEW EXERCISES MATH 114 1 Let G be a transitive subgroup of S a Prove that if n is prime then G contains an n cycle b Show that a is not true if n is not prime Solution The number of elements in an orbit divides the order of G Since G is transitive n divides If n is prime then by Sylow theorems G contains an element of order n which is an n cycle If n is not prime the statement is false For example let n 4 G be the Klein subgroup of S4 2 Let F be a eld such that the multiplicative group F is cyclic Prove that F is nite Solution Let u be a generator of F Assume rst that charF 31 2 Then 71 u for some n hence uh 1 and therefore F Zgn is nite Let now charF 2 Then 1 u u for some 71 Hence F Z2 is a nite extension of Z2 and therefore F is nite 3 Let G be a transitive subgroup of 56 which contains a 5 cycle Prove that G is not solvable Solution Observe rst that G divides 6 Hence a cyclic 5 subgroup is a Sylow subgroup of G Assume that G is solvable We have a chain GG03G133Gk1 such that Gi1 is normal in G and GiGHL is cyclic of prime order Among such chains of subgroups choose one such that Z5 2 GiGHL appears for maximal 239 We claim that then Z5 G154 lndeed GHlGHZ g Zp for some p lt 5 Hence GiGHZ E Z5 gtlt Z1 and one can nd G24r1 normal in G such that GQHGHZ E Z5 GiGQJr1 g Zp Hence we moved Z5 to the right Now we claim that Z5 G571 is normal in G To prove proceed by induction Assume that G5L Z5 is normal in G then it is the unique Sylow subgroup in G Hence for any 9 E Gi1 gGk1g 1 G154 and therefore G5L is normal in Gi1 Finally let Z5 be generated by a cycle 5 12345 G is transitive therefore there is a permutation t E G such that t1 6 Then clearly tst l 31 5 Hence Z5 is not normal Contradiction 4 Let F be a eld and charF 31 2 046 E F Prove that F F if and only of 046 is a square in F Date May 13 2006 2 SOLUTIONS FOR REVIEW EXERCISES MATH 114 Solution Assume that F F E The Galois group of E over F is Z2 Let s 31 1 be the element of the Galois group Then 5 7 7w 5W 7 Write xB a bxE for some a7b E F Then a7bxE77a7bE implies xB lnE Then xB ba and we obtain 046 bzaz is a square Conversely7 if 046 02 then xB Therefore F 5 Find the minimal polynomial for 1 WE 3xZl over Q Solution Let u Bxi Solve the equation ablt1uu2 0lt1uu22 1uu23 0 for 17b707d7 using the relation u3 2 1uu22 1u2u42u2u22u3 1u22u2u2u24 5Zlu3u27 1 u u23 5 4U 3u2 1 u u2 54u3u25u4u23u35u24u33u4 5 9u 1211 ms 31 19 151 12112 The solution a 717b c 73 The minimal polynomial is x3 7 3x2 7 3x 7 1 6 Prove that any algebraically closed eld is in nite Solution Let F be a nite eld and have q elements Choose 71 relatively prime to q 7 1 and q Then x 1 implies z 1 by Lagrange7s theorem Therefore x 7 1 does not split in F7 and F is not algebraically closed 7 ls 3 m 1 irreducible over F256 Solution The polynomial is irreducible over F2 because it does not have roots in F2 The degree F256F2 87 therefore F256 does not contain a eld of degree 3 Thus7 F256 does not contain a root of the polynomial Hence x3 z 1 is irreducible over F256 8 Which of the following extensions are normal QCQ 17W QCQltW M gt7 QCQWEME SOLUTIONS FOR REVIEW EXERCISES MATH 114 3 Solution The minimal polynomial of V 1 7 is x4 7 2x2 7 1 the Galois group of this polynomial is D4 Hence the splitting eld has degree 8 But Q V 1 7 Qgt 4 Hence Q V1 7 is not normal The extension Q C Q BxE is not normal because it contains a real root of x3 7 2 but does not contain two complex roots since Q Bx3 is a sub led of R The extension Q C Q BxE V73 is normal because it is a splitting eld of 37 2 9 Determine if Q 17WQltj1 Solution No The rst eld is not a normal extension of Q the second one is normal 10 Let Q C F be a nite normal extension such that for any two sub elds E and K of F either K C E or E C K Then the Galois group of F over Q is cyclic of order p for some prime number p Solution Let G denote the Galois group Then for any two subgroups H and H either H C H or H C H First we prove that G is cyclic lndeed consider an element 9 E G of maximal order For any h E G lt h gtClt g gt hence G is generated by 9 Now let us prove that G p Assume the contrary then G has two distinct prime divisors p and q Then G has Sylow p subgroup and Sylow q subgroup which have trivial intersection Contradiction 11 Let F C B C E be a chain of extensions such that F C B is normal and B C E is normal ls it always true that F C E is normal Solution False Counterexample QCQlt CQ17 12 Find the Galois group of 2 7 3 2 1 3 7 6 over Q Solution The splitting eld of 3 7 6 contains Therefore the splitting eld of 2 7 3 3 7 6 contains the roots of x2 1 Let E be the splitting eld of 2 7 3 2 1 3 7 6 Then E FB where B is a splitting eld of x3 7 6 whose Galois group is 3 and F is a splitting eld of x2 7 3 whose Galois group is Z2 Let us prove that F H B Q If not then F C B Since Sg has only one subgroup of index 2 then F Q but B is real Contradiction By Corollary of the natural irrationalities theorem Ath E AthB gtlt AthF SQ gtlt Z2 13 Find the Galois group of x4 3x 5 over Q Solution The polynomial is irreducible over Z2 Hence the Galois group contains a 4 cycle The resolvent cubic is x3 7 20x 9 which is irreducible over Q So the Galois group is S4 or A4 and contains a 4 cycle Hence the Galois group is S4 4 SOLUTIONS FOR REVIEW EXERCISES MATH 114 14 Let p be a prime number Prove that n 2 is constructible if and only if n 2k for some k Solution The minimal polynomial is x 7p irreducible by Eisenstein criterion If n is not a power of 2 a root is not constructible since the order of the Galois group is not a power of 2 If n is a power of 2 then is constructible because it can be obtained by taking square root several times 15 Prove that any sub eld of Q coincides with Q ix2 for some divisor d of 71 Solution Since x 7 2 is irreducible over Q the degree of Q over Q is nLet F be a sub eld of Q Consider the minimal polynomial f for 32 over F Let k denote the degree of f Since k Q F k divides n and FQ d All roots of f are roots of x 7 2 which are x2w5 where w is a primitive n th root of 1 Let 10 be the free coef cient of Then 10 equals plusminus the product of roots of fx a0 i x2 Since 10 E F C R if i1 Thus iao dx2 E F But Q d has degree d over Q because 7 2 is irreducible over Q Therefore F Q ix2 16 Prove that there exists a polynomial of degree 7 whose Galois group over Q is Z7 Solution For example consider the splitting eld E for 29 7 1 The Galois group of E over Q is Z28 which contains a subgroup Z4 Let F EZ4 Then the Galois group of F over Q is Z7 Pick up an element 04 in Fa Q The minimal polynomial of 04 has the Galois group Z7 17 Let f E Q be an irreducible polynomial of odd prime degree p solvable in radicals Prove that the number of real roots of f equals p or 1 Solution Let G be the Galois group of f Then G is a subgroup of FTP Let 039 be the complex conjugation After suitable enumeration of roots by elements of Zp we have 039 t at b for some a E Zb E Z The number of real roots is the number of t xed by 039 But the number of solutions for the equation at b t is 01 or p Since any polynomial of odd degree has at least one real root the number of real roots is either 1 or p 18 Let fz 6 F2 be an irreducible polynomial Prove that fz divides x256 7 x if and only if the degree of f is 12 4 or 8 Solution Let f divide 256 7 x The elements of F256 are the roots of 256 7 x therefore fz splits in F256 Conversely if fz splits in F256 then fz divides 256 7 x The irreducible polynomial splits in F256 if and only if its degree divides the degree of F256 which is 8 19 Suppose that the Galois group over Q of a polynomial f E Q has odd order Prove that all roots of f are real Solution Complex conjugation is an element of order 2 unless all roots are real 20 Find the Galois group of 6 7 8 over Q SOLUTIONS FOR REVIEW EXERCISES MATH 114 5 Solution The polynomial factors 678 272 4224 The Galois group of x4 2x2 4 is Z2 gtlt Z2 Let a and B g be two roots of x4 2x2 47 then a 2a2622a 7242 Hence xi is in the splitting eld of x4 2x 47 Thus7 the Galois group is Z2 gtlt Z2
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'