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 )
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 f
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
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. Ther
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 ; si
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,
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
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
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 i
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.
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
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 :