Download Sample Solutions for MATH 731 Final: Group Theory, Representations, Field Extensions and more Exams Linear Algebra in PDF only on Docsity!
MATH 731, FALL 2008
FINAL EXAM sample solutions
A. Let G be a group and let N, K be normal subgroups with G = N ⊕ K. (1) Prove that K is Abelian if and only if K ⊆ Z(G). Proof. If K ⊆ Z(G), then certainly K is Abelian. Suppose conversely that K is Abelian and let x ∈ K , g ∈ G. Then we can write g = nk for some n ∈ N , k ∈ K. Since G = N ⊕ K , elements of N and K commute. Thus gx = n(kx) = (kx)n = xkn = xnk = xg. This proves x ∈ Z(G).
(2) Prove that if [G, G] ⊆ N , then K ⊆ Z(G). Proof. Let x, y ∈ K. Then xyx−^1 y−^1 ∈ [G, G] ∩ K = {e}. Thus xy = yx, so K is Abelian. By (1), this implies K ⊆ Z(G).
B. Show that there is no square complex matrix A such that A∗^ = −A−^1.
Proof. Suppose A∗^ = −A−^1. We have AA∗^ = −I = A∗A, so A is normal. Thus there is a unitary U with D = U AU −^1 diagonal. Then D∗^ = (U −^1 )∗A∗U ∗^ = U (−A−^1 )U −^1 = −(U AU −^1 )−^1 = −D−^1. Let z be a diagonal entry of D. The corresponding diagonal entry of D−^1 is z−^1. Thus ¯z = −z−^1 , so |z|^2 = z z¯ = −1. This is impossible, so such a D cannot exist. Thus A cannot exist. Alternative proof: Note that if v ∈ Cn^ , then v∗v = ‖v‖^2. Suppose v 6 = 0. Then −‖v‖^2 = −v∗v = v∗(A∗A)v = (Av)∗(Av) = ‖Av‖^2. This is impossible, since the left hand term of the equality is negative and the right hand term is positive.
C. Let G be a finite group, let V be a finite dimensional complex representation of G, and let 〈 , 〉 be an arbitrary inner product on V. Define ( , ) : V × V → C by (v, w) =
|G|
g∈G〈g.v, g.w〉^ for^ v, w^ ∈^ V^. (1) Show that ( , ) is an inner product on V.
Proof. (a) (v + v′, w) =
|G|
g∈G〈g.(v^ +^ v
′), g.w〉 = 1 |G|
g∈G(〈g.v, g.w〉^ +^ 〈g.v
′, g.w〉) =
1 |G|
g∈G〈g.v, g.w〉^ +^
|G|
g∈G〈g.v
′, g.w〉 = (v, w) + (v′, w).
(cv, w) =
|G|
g∈G〈g.(cv), g.w〉^ =^
|G|
g∈G c〈g.v, g.w〉^ =^ c(v, w).
(b) (w, v) =
|G|
g∈G〈g.w, g.v〉^ =^
|G|
g∈G 〈g.v, g.w〉^ =^
|G|
g∈G〈g.v, g.w〉^ = (v, w).
(c) If v 6 = 0, then (v, v) =
|G|
g∈G〈g.v, g.v〉^ >^ 0, since each^ 〈g.v, g.v〉^ >^ 0.
(2) Show that (g.v, g.w) = (v, w) for all g ∈ G, v, w ∈ V.
Proof. (g.v, g.w) =
|G|
h∈G〈h.g.v, h.g.w〉^ =^
|G|
x∈G〈x.v, x.w〉^ = (v, w).^ The second-to-last equality follows from the change of variable x = hg , since as h runs over G, so does x. 1
2
D. Let E/F be an extension of fields and let α, β ∈ E. Suppose φ ∈ Gal(E/F ) (that is, φ is an F -automorphism of E ) and φ(α) = β. Prove that α and β have the same minimal polynomial over F. Proof.∑ Let f be the minimal polynomial of α, so f is monic irreducible. Write f = n i=0 aix i (^) and note φ(ai) = ai for each i. Then f (β) = f (φ(α)) = ∑n i=0 φ(ai)(φ(α)) i (^) = φ(
∑n i=0 aiα i) = φ(f (α)) = φ(0) = 0. Since f is monic irreducible and f (β) = 0, f must be the minimal polynomial of β. An alternative end to the proof: The same argument shows if g(β) = 0, then g(α) = g(φ−^1 (β)) = φ−^1 (g(β)) = 0. Thus α and β are roots of exactly the same polynomials in F [x], which means they have the same minimal polynomial.
E. Suppose K/E/F is a tower of fields and f ∈ F [x]. (1) If K is a splitting field for f over F , prove that K is a splitting field for f over E. Proof. Let β 1 ,... , βr be the roots of f in K – these are all of the roots of f. The defini- tion of splitting field tells us K = F [β 1 ,... , βr]. Clearly this implies K = E[β 1 ,... , βr], so K is a splitting field of f over E.
(2) If E = F [α 1 ,... , αk], where α 1 ,... , αk are roots of f , and K is a splitting field for f over E , prove that K is a splitting field for f over F. Proof. Let β 1 ,... , βr be the roots of f in K. The definition of splitting field tells us K = E[β 1 ,... , βr] = F [α 1 ,... , αk, β 1 ,... , βr] = F [β 1 ,... , βr], where the last equality holds since each αi is one of the βj. Thus K is a splitting field of f over F.
(3) Let F = Q, E = Q[
2], K = Q[
2 , i], f = x^2 + 1 ∈ F [x]. Show that K is a splitting field for f over E , but K is not a splitting field for f over F. Proof. K = E[i, −i] and i, −i are the roots of f , so K is a splitting field for f over E. However, K 6 = F [i, −i], so K is not a splitting field for f over F.
