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(xj).
In particular, if x is conjugate to xj 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(xj) for all characters chi, which can only happen if x
is conjugate to xj, 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 xj.
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 xj for all j prime to n forces the
characteristic polynomial f of M to be in Q[x]. Since xp -1
factors as (x-1) (xp-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
Furthermore, x is conjugate to xj, 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
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.