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# Second Course in Abstract Algebra MATH 114

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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.

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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

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