189-570A: Higher Algebra I

Take-Home Final: Correction

1. a) The eigenvalues of r(x), where r is a representation of G, are nth roots of unity. Hence the characters evaluated on x are sums of nth roots of unity.

b) Since chi(x) is a sum of roots of unity, it is an algebraic integer. Hence it is rational if and only if it is integral. If s is the automorphism sending the nth roots of unity to their jth powers, then note that s(chi(x)) = chi(x^j). In particular, if x is conjugate to x^j for all j, then chi(x) is fixed by all automorphisms s, so that chi(x) is rational. Conversely, if chi(x) is rational, then chi(x)=chi(x^j) for all characters chi, which can only happen if x is conjugate to x^j, since the characters span the complex vector space of class functions on G.

c) The field generated by the chi(x) is the subfield of the field of nth roots of unity which is fixed by H, where H is the subgroup of (Z/nZ)^* containing all j such that x is conjugate to x^j.

d) Let r be a non-trivial representation. Its kernel consists only of 1, since G has no non-trivial normal subgroup, so that M=r(x) is a matrix of order p. The fact that x is conjugate to x^j for all j prime to n forces the characteristic polynomial f of M to be in Q[x]. Since x^p -1 factors as (x-1) (x^p-1 + ... + 1), and not all the eigenvalues of M are equal to 1, the second factor divides f. Hence the degree of f is at least p-1 and the result follows.

e) If r is such a representation, it must map an element x of order p to a matrix M of order p, since any normal subgroup containing x contains SL₂(F_p). Furthermore, x is conjugate to x^j, so that the characteristic polynomial f of M belongs to Q[x]. The result now follows by the same argument as for d).

2. Most of you had no difficulties with a) and the first part of 2), but no one was able to do the second part. The idea here was to let G be a finite group and A and B be two subgroups with the property that the permutation represesentations on G/A and on G/B are isomorphic as linear representations, even though they are not isomorphic as permutation representations. (We saw some examples of this in the assignment, involving two types of parabolic subgroup of GL₃(F₂), of index 7.) Now let E be an extension of Q with Galois group G, and let alpha and beta be generators for the subextensions of E/Q of degree 7 associated to the groups A and B respectively by the Galois correspondence. They are not isomorphic, since A and B are not conjugate. However, any element in G induces permutations on G/A and on G/B of the same cycle shape...

3. People did fairly well on this question - this came as a pleasant surprise to me, since I thought it was the most difficult in the exam! The main difficulties seemed to occur with part c: most of the solutions were too computational, involving alot of calculations with 3 by 3 matrices. Here is a short conceptual solution. Using the formula of b, one can se that Sigma(C3,C7,C7') has cardinality 168, where C3 is the conjugacy class of elements of order 3 and C7 and C7' are the two conjugacy classes of elements of order 7. Let (x,y,z) be an element of Sigma(C3,C7,C7'). To see that the group H generated by x,y, and z is equal to G, note that 21 divides the order of H, and that this order cannot be 42, or 84, since G has no subgroup of index 4 or 2 (it can be seen by glancing at the character table of G, that G has no permutation representation of degree 4 or 2.) If H is of order 21, it has a unique Sylow 7-subgroup by the Sylow theorem, so that y and z must commute. But this is impossible, since yz is of order 3. Therefore, H must be equal to G. Now, it becomes immediate that G acts transitively on Sigma. For the stabilizer of (x,y,z) consists of elements which commute with x, y and z, and hence with all of G. But G has trivial center. It follows that the orbit of (x,y,z) has cardinality 168, and so is equal to all of Sigma(C3,C7,C7').

5. People seemed to have a surprising amount of difficultly with this question - surprising, because we covered something similar in the asigments. Here is a short solution. Let L be the splitting field of xⁿ-alpha over F, and let K be the intersection of E and L. The field K is normal over F, since it is the intersection of two normal extensions. The Galois group of K over F is a quotient of a solvable group, since gal(L/F) is solvable. But gal(E/F) has no non-trivial solvable quotients, and hence K=F. The result follows directly from this.

6. Many people found a counterexample here; it helped that something very close to this question was covered in an assigment. The standard counterexample was to take G-tilde = Z/4Z, H = 2Z/4Z, G= Z/2Z, and E=Q(i). A solution E-tilde to the embedding problem does not exist. If it did, the generator of gal(Q(i)/Q) would lift to an element of order 2 in gal(E-tilde/Q), induced by a complex conjugation in gal(E-tilde/Q) attached to a complex embedding of E-tilde. But the generator of G does not lift to an element of order 2 in G-tilde.