MATH 451 First Mid-Term Exam, Exams of Algebra

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NAME: John Q. Public

Question Marks 1 12 2 25 3 28 4 25 5 10 1

Question 1. Let H be a nonempty subset of the group G. Prove that H is a subgroup of G iff a b−^1 ∈ H for all a, b ∈ H.

First suppose that H is a subgroup of G. Then H is closed under multiplication and taking inverses. Hence if a, b ∈ H, then b−^1 ∈ H and so ab−^1 ∈ H. Next suppose that ∅ 6 = H ⊆ G is such that ab−^1 ∈ H for all a, b ∈ H. Since H 6 = ∅, there exists an element a ∈ H and hence 1 = aa−^1 ∈ H. It follows that if a ∈ H, then a−^1 = 1 a−^1 ∈ H. Finally suppose that a, b ∈ H. Then b−^1 ∈ H and so ab = a(b−^1 )−^1 ∈ H. Thus H is a subgroup of G.

Question 3. Suppose that G is a finite group and that S is a G-set. For each s ∈ S, let Os denote the corresponding G-orbit.

(a) Prove that if s ∈ S, then [ G : Gs ] = |Os|. (b) Prove that if G is a finite p-group and p does not divide |S|, then there exists a fixed point for the action of G; i.e. an element s ∈ S such that gs = s for all g ∈ G. (Hint: Let s 1 , · · · st be representatives of the distinct G-orbits and consider the equation |S| = |Os 1 | + · · · + |Ost |.) (a) It is easily checked that if g, h ∈ G, then g s = h s iff gGs = hGs.

Hence we can define an injective map ϕ : G/Gs → Os by ϕ(gGs) = g s. To see that ϕ is also surjective, let r ∈ Os be arbitrary. Then there exists g ∈ G such that g s = r and hence ϕ(gGs) = g s = r. (b) Let s 1 , · · · st be representatives of the distinct G-orbits. Then |S| = |Os 1 | + · · · + |Ost |.

Since p does not divide |S|, there exists i such that p does not divide |Osi |. Since

|Osi | = [G : Gsi ] = |G|/|Gsi |

and G is a p-group, it follows that |Osi | = 1 and so si is a fixed point for the action of G.

Question 4. If G is a group and H 6 G is a subgroup, then the centralizer of H in G is defined to be

CG(H) = { g ∈ G | gh = hg for all h ∈ H }

and the normalizer of H in G is defined to be

NG(H) = { g ∈ G | gHg−^1 = H } (a) Prove that NG(H) is a subgroup of G. (b) Prove that CG(H) is a normal subgroup of NG(H). (Hint: This can either be proved directly or else by considering a suitable homomorphism ϕ : NG(H) → Aut(H)). (a) First note that 1 H 1 −^1 = H and so 1 ∈ NG(H). Next if g ∈ NG(H), then gHg−^1 = H and so H = g−^1 Hg. Thus g−^1 ∈ NG(H). Finally if g, h ∈ NG(H), then ghH(gh)−^1 = ghHh−^1 g−^1 = g(hHh−^1 )g−^1 = gHg−^1 = H

and so gh ∈ NG(H). Thus NG(H) is a subgroup of G. (b) First note that if g ∈ CG(H), then gHg−^1 = H and so CG(H) ⊆ NG(H). It follows that

CG(H) = {g ∈ NG(H) | ghg−^1 = h for all h ∈ H}.

Next note that if g ∈ NG(H), then gHg−^1 = H and so we can define an associated automorphism cg ∈ Aut(H) by cg (h) = ghg−^1. Consider the map

ϕ : NG(H) → Aut(H)

defined by ϕ(g) = cg. Then it is easily checked that ϕ is a homomorphism; and that ker ϕ = {g ∈ NG(H) | ghg−^1 = h for all h ∈ H} = CG(H).

Hence CG(H) is a normal subgroup of NG(H).