# Nonempty Subset - Abstract Algebra - Exam, Exams for Algebra. Acharya Nagarjuna University

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This is the Exam of Abstract Algebra which includes Weighted Equally, Group Theory, Vector Spaces, Linear Algebra, Unique Factoriztaion, Normal Subgroup etc. Key important points are: Nonempty Subset, Group, Subgroup, Mu...
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MATH 451 FIRST MID-TERM

NAME: John Q. Public

Question Marks

1 12

2 25

3 28

4 25

5 10

1

2 MATH 451 FIRST MID-TERM

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.

MATH 451 FIRST MID-TERM 3

Question 2. (a) Prove that the alternating group A5 does not have a sub-

group which is isomorphic to the symmetric group S4.

(b) Prove that the alternating group A5 does not have a subgroup of order 15.

(a) Suppose that H 6 A5 is a subgroup such that H ∼= S4. Then |H| = |S4| = 24.

Applying Lagrange’s Theorem, we must have that 24 = |H| divides |A5| = 60, which

(b) Suppose that H 6 A5 is a subgroup such that |H| = 15. Consider the

transitive action of A5 on the coset space S = A5/H and let

ϕ : A5 → Sym(S)

be the associated homomorphism. Let N = kerϕ. Since A5 acts transitively on S

and |S| > 1, we have that N 6= A5. Hence, since A5 is simple, it follows that N = 1.

But this means that ϕ is an injection of A5 into Sym(S), which is impossible since

|A5| = 60 and |Sym(S)| = 24.

4 MATH 451 FIRST MID-TERM

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 s1, · · · st be representatives of the distinct G-orbits and consider

the equation |S| = |Os1 |+ · · ·+ |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 s1, · · · st be representatives of the distinct G-orbits. Then

|S| = |Os1 |+ · · ·+ |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.

MATH 451 FIRST MID-TERM 5

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 1H 1−1 = H and so 1 ∈ NG(H). Next if g ∈ NG(H), then

gHg−1 = H and so H = g−1Hg. Thus g−1 ∈ NG(H). Finally if g, h ∈ NG(H),

then

ghH(gh)−1 = ghHh−1g−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

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

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

6 MATH 451 FIRST MID-TERM

Question 5. Let G be a finite group. Prove that if G r {1} is a single conjugacy

class, then |G| = 2.

Suppose that G r {1} is a single conjugacy class. If a ∈ G r {1}, then

|G| − 1 = |G r {1}| = |aG| = [G : CG(a)] = |G|/|CG(a)|.

Thus |G| − 1 divides |G| and this implies that |G| = 2.