MATH 511B, SOLUTIONS 4
1.
(a) One must check that the given operations are well-dened. For instance,
we verify that (a f ) is -linear: we have
(a f )(cv) = af (cv) = a(c)f (v) = (c)(af (v) = (c)(a f )(v)
as desired. The rest of the check is similarly stra
MATH 511A, SOLUTIONS TO HOMEWORK 5
1. The space V certainly has an orthogonal basis: simply take any basis and
apply the Gram-Schmidt process to it. Let v1 , . . . , vn be such a basis. By Lemma
0.3 from the handout, we either have [vi , vi ] > 0 for all
1. We see that [K : Q] = [K : Q( 8 2)][Q( 8 2) : Q] = 8 = 16. We deduce
2
j
that the automorphisms of K are determined by 8 2 8 8 2 with 0 j < 8 and
i i. Note that 8 = (1 + i)/ 2.
By
the fundamental theorem of Galois theory, the subelds Q(i), Q( 2), and
MATH 511A, SOLUTIONS 7
1. Since H K is a subgroup of H, we know from Lagranges Theorem that
#(H K) divides #H. Similarly #(H K) divides #K. Since the GCD of #H
and #K is 1, we have #(H K) = 1.
2. There are many ways to solve this problem. One way is to no
MATH 511B, SOLUTIONS 6
1. Since V is nite dimensional, to prove that L is invertible it suces to prove
that L is injective. If L w = 0 then
v, w = L(L1 v), w = L1 v, L w = L1 v, 0 = 0
for all v V ; since the pairing is perfect, it follows that w = 0, as d
MATH 511A, SOLUTIONS TO HOMEWORK 10
1. We can dene a map : NG (H) Aut(H) by sending g to the automorphism
g : H H obtained from conjugation by g. (Note that this may not be an inner
automorphism of H, since the conjugation is by an element of G, not of H.
MATH 511A, SOLUTIONS TO HOMEWORK 11
1.
(a) To check that an element lies in the center of G, it suces to check that
it commutes with all the generators of G: since the centralizer CG (x) is a
subgroup of G, if it contains a set of generators of G then it
MATH 511A, SOLUTIONS 8
1. The relations bd = db and bd = d2 b imply d = 1. Then c2 = dcd1 = c
implies c = 1. Now b2 = c1 ac = a, so a commutes with b. Then a2 = bab1 = a,
so a = 1, and b has order at most 2. Summary: a, c, d = 1 and b has order at most
2.
MATH 511A, SOLUTIONS 2
v
1. Let us follow Knapps notation and let cfw_vi denote the column vector
(c1 cn )t Fn such that v = c1 v1 + + cn vn . (Here the superscript t denotes
the transpose.) Then A is the matrix such that
A
v
cfw_vi
Tv
cfw_wi
=
for all
MATH 511A, SOLUTIONS TO HOMEWORK 12
1. For the rst equality, note that (1, k) CG (H) if and only if
(h, 1) = (1, k)(h, 1)(1, k 1 ) = (k)(h), 1)
for all h. This is the case if and only if (k) = 1, i.e., if and only if k ker().
For the second equality, note
MATH 511B, SOLUTIONS 3
1. Let v1 , . . . , vn be a basis for V (over C). We claim that v1 , . . . , vn , iv1 , . . . , ivn
are a basis for VR . Indeed, suppose v VR . Regarded as an element of V , we may
write
v = z1 v1 + + zn vn
with each zj C. Write zj
MATH 511B, SOLUTIONS 1
1.
(a) If T is invertible then it is invertible as a map of sets, and so it is one-to-one
and onto.
Conversely, if T is one-to-one and onto then there exists a map of sets
T 1 : W V such that T 1 T is the identity on V and T T 1 is