MATH 413-01 Abstract AlgebraHomework Solutions: #8 – #11

1 Homework #8

1. [Berry, K.] Find all isomorphic finite groups in our Group Atlas. Justify your work.

There are 18 groups within the group atlas. We can eliminate isomorphisms between groups that donot have the same order. By grouping these groups based on their order we can see,Groups of order 3: A3

Groups of order 4: (Z4,+), U(10) , Color-4 , Rigid-body-motion-of-a-rectangle, and Klein-four

Groups of order 5: (Z5,+), Color-5

Groups of order 6: D3, (Z6,+), GL(2,Z2), S3 , Color-6 , and Hexaflexagon

Groups of order 8: D4, Quaternions

Groups of order 10: D5

Groups of order 12: A4

The groups in red and underlined are abelian. By properties of isomorphisms, we know that groupsof the same order that are abelian can be isomorphisms. By definition we know that to be isomorphicthe elements within the groups must also be of the same order. Using the group atlas we can seethat,

(Z4,+) ' U(10) ' Color-4

(Z4,+) (Z4,+) Color − 4

0 1 Blue1 7 Green2 9 Y ellow3 3 Red

0, 1, Blue 1, 7, Green 2, 9, Y ellow 3, 3, Red

0, 1, Blue 0, 1, Blue 1, 7, Green 2, 9, Y ellow 3, 3, Red1, 7, Green 1, 7, Green 2, 9, Y ellow 3, 3, Red 0, 1, Blue2, 9, Y ellow 2, 9, Y ellow 3, 3, Red 0, 1, Blue 1, 7, Green

3, 3, Red 3, 3, Red 0, 1, Blue 1, 7, Green 2, 9, Y ellow

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The mapping of Rigid-body motions of a rectangle ' Klein-four is as follows,

Rectangle Klein− 4

R0 aR180 bF| c

F− d

R0, a R180, b F|, c F−, d

R0, a R0, a R180, b F|, c F−, d

R180, b R180, b R0, a F−, d F|, c

F|, c F|, c F−, d R0, a R180

F−, c F−, c F|, c R180 R0, a

The mapping of (Z5,+) ' Color-5 is as follows,

(Z5,+) Color − 5

0 Clear1 Orange2 Oxblood3 Red4 Blue

0, Clear 1, Orange 2, Oxblood 3, Red 4, Blue

0, Clear 0, Clear 1, Orange 2, Oxblood 3, Red 4, Blue1, Orange 1, Orange 2, Oxblood 3, Red 4, Blue 0, Clear2, Oxblood 2, Oxblood 3, Red 4, Blue 0, Clear 1, Orange

3, Red 3, Red 4, Blue 0, Clear 1, Orange 2, Oxblood4, Blue 4, Blue 0, Clear 1, Orange 2, Oxblood 3, Red

The mapping of (Z6,+) ' Color-6 is as follows,

(Z6,+) Color − 6

0 Y ellow1 Pumpkin2 Lavender3 Red4 Blue5 Green

0, Y ellow 1, Pumpkin 2, Lavender 3, Red 4, Blue 5, Green

0, Y ellow 0, Y ellow Pumpkin 2, Lavender 3, Red 4, Blue 5, Green1, Pumpkin 1, Pumpkin 2, Lavender 3, Red 4, Blue 5, Green 0, Y ellow2, Lavender 2, Lavender 3, Red 4, Blue 5, Green 0, Y ellow 1, Pumpkin

3, Red 3, Red 4, Blue 5, Green 0, Y ellow 1, Pumpkin 2, Lavender4, Blue 4, Blue 5, Green 0, Y ellow 1, Pumpkin 2, Lavender 3, Red

5, Green 5, Green 0, Y ellow 1, Pumpkin 2, Lavender 3, Red 4, Blue

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Exploring the remaining groups we find that,D3 ' S3 ' Hexaflexagon ' GL(2,Z2) . The mapping follows,

D3 S3 Hexaflexagon GL(2,Z2)

R0 ε n IR120 (123) ↑ AR240 (321) ↓ BF1 (23) f CF2 (13) ↑f DF3 (12) ↓f E

R0, n, ε, I R120, (123), ↑, A R240, (321), ↓, B F1, (23), f, C F2, (13), ↑f , D F3, (12), ↓f , E

R0, n, ε, I R0, n, ε, I R120, (123), ↑, A R240, (321), ↓, B Fl, (23), f, C F2, (13), ↑f , D F3, (12), ↓f , ER120, (123), ↑, A R120, (123), ↑, A R240, (321), ↓, B R0, n, ε, I F3, (12), ↓f , E F1, (23), f, C F2, (13), ↑f , DR240, (321), ↓, B R240, (321), ↓, B R0, n, ε, I R120, (123), ↑, A F2, (13), ↑f , D F3, (12), ↓f , E F1, (23), f, CF1, (23), f, C F1, (23), f, C F2, (13), ↑f , D F3, (12), ↓f , E R0, n, ε, I R120, (123), ↑, A R240, (321), ↓, BF2, (13), ↑f , D F2, (13), ↑f , D F3, (12), ↓f , E F1, (23), f R240, (321), ↓, B R0, n, ε, I R120, (123), ↑, AF3, (12), ↓f , E F3, (12), ↓f , E F1, (23), f, C F2, (13), ↑f , D R120, (123), ↑, A R240, (321), ↓, B R0, n, ε, I

2. [Brubaker, N.] Find a subgroup of S4 isomorphic to D4. Justify your work.

Let φ be a bijective function that maps elements from D4 to a subgroup of S4, and call this subgroupP (D4).

φ : D4 → P (D4) =

[R0 R90 R180 R270 F14 F12 F13 F24

ε (1234) (13)(24) (1432) (14)(23) (12)(34) (13)(2)(4) (1)(3)(24)

]To show that φ is an isomorphism we need to show that it is bijective, and that it preserves theoperation from D4. By definition, φ is bijective. To show that φ preserves the operation in D4, weneed the Cayley tables to be identical in structure, which can be seen below.

D4 R0 R90 R180 R270 F| F− F� F�

R0 R0 R90 R180 R270 F| F− F� F�R90 R90 R180 R270 R0 F� F� F| F−R180 R180 R270 R0 R90 F− F| F� F�R270 R270 R0 R90 R180 F� F� F− F|F| F| F� F− F� R0 R180 R90 R270

F− F− F� F| F� R180 R0 R270 R90

F� F� F− F� F| R270 R90 R0 R180

F� F� F| F� F− R90 R270 R180 R0

P (D4) ε (1234) (13)(24) (1432) (12)(34) (14)(23) (24) (13)

ε ε (1234) (13)(24) (1432) (12)(34) (14)(23) (24) (13)(1234) (1234) (13)(24) (1432) ε (13) (24) (12)(34) (14)(23)

(13)(24) (13)(24) (1432) ε (1234) (14)(23) (12)(34) (13) (24)(1432) (1432) ε (1234) (13)(24) (24) (13) (14)(23) (12)(34)

(12)(34) (12)(34) (24) (14)(23) (13) ε (13)(24) (1234) (1432)(14)(23) (14)(23) (13) (12)(34) (24) (13)(24) ε (1432) (1432)

(24) (24) (14)(23) (13) (12)(34) (1432) (1234) ε (13)(24)(13) (13) (12)(34) (24) (14)(23) (1234) (1432) (13)(24) ε

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3. [Conger, A.]

(a) Show that U(8) is not isomorphic to U(10).

U(8) = {1, 3, 5, 7}1 3 5 7

1 1 3 5 73 3 1 7 55 5 7 1 37 7 5 3 1

U(10) = {1, 3, 7, 9}1 3 7 9

1 1 3 7 93 3 9 1 77 7 1 9 39 9 7 3 1

These groups are not isomorphic because in U(8) : |1| = |3| = |5| = |7| = 2. In U(10) : |1| =|9| = 2 and |3| = |7| = 4. Since U(8) has no elements of order 4, we cannot map U(8) to U(10).Thus they are not isomorphic.

(b) Show that U(8) is isomorphic to U(12).

U(8) = {1, 3, 5, 7}1 3 5 7

1 1 3 5 73 3 1 7 55 5 7 1 37 7 5 3 1

U(12) = {1, 5, 7, 11}1 5 7 11

1 1 5 7 115 5 1 11 77 7 11 1 511 11 7 5 1

These groups are isomorphic since they share a Cayley table and are bijective if mapped as

1→ 13→ 55→ 77→ 11

(c) Prove or disprove that U(20) and U(24) are isomorphic.

U(20) = {1, 3, 7, 9, 11, 13, 17, 19}U(24) = {1, 5, 7, 11, 13, 17, 19, 23}In U(24), the element 1 has order 1 and all other elements have order 2. In U(20), the element3 has order 4. Since U(20) has an element of order 4 but U(24) does not, the two groups cannotbe isomorphic because isomorphic groups preserve order.

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4. [Gruenig, S.] Find a homomorphism φ from U(30) to U(30) with kernel {1, 11} and φ(7) = 7.

We know that φ(1) = 1, φ(11) = 1, and φ(7) = 7. Working in U(30) we know that 7 ◦ 11 = 17. Then

φ(7) ◦ φ(11) = φ(17)

7 ◦ 1 = φ(17)

7 = φ(17).

So φ(17) = 7. Similarly, we can find the remaining four elements:

φ(29):φ(7) ◦ φ(17) = φ(29)

7 ◦ 7 = φ(29)

19 = φ(29).

φ(19):φ(19) ◦ φ(11) = φ(29)

φ(19) ◦ 1 = 19

φ(19) = 19.

φ(13):φ(7) ◦ φ(19) = φ(23)

7 ◦ 19 = φ(13)

13 = φ(13).

φ(23):φ(13) ◦ φ(11) = φ(23)

13 ◦ 1 = φ(23)

13 = φ(23).

Thus, the homomorphism φ is defined as follows:

φ(1) = 1 φ(11) = 1φ(7) = 7 φ(17) = 7φ(13) = 13 φ(23) = 13φ(19) = 19 φ(29) = 19

∗ 1 7 13 19

1 1 7 13 197 7 19 1 1313 13 1 19 719 19 13 7 1

Note that the above Cayley table is also valid for the factor group U(30)/ ker(φ), where the symbolsof the Cayley table are representatives of the distinct left (or right) cosets of ker(φ) in U(30).

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5. [Havelka, R.] If φ is a homomorphism from G to H and σ is a homomorphism from H to K, showthat σ ◦ φ is a homomorphism from G to K. How are ker (φ) and ker (σ ◦ φ) related?

Proof. (a) We have φ : G→ H and σ : H → K.Observe that since φ is a homomorphism, then φ(G) is a subset of H, likewise σ(H) is a subsetof K.

Now we have φ(g) ∈ H, then it follows that σ(φ(g)) ∈ K.We can also see that ∀a, b ∈ G, then σ(φ(a ◦ b)) = σ(φ(a) ◦ φ(b)) = σ(φ(a)) ◦ σ(φ(b)). So σ ◦ φis a homomorphism from G to K.

(b) Claim: ker(φ) ≤ ker(σ ◦ φ).The definition of the kernel states that ker(φ) ≤ G and ker(σ ◦ φ) ≤ G.Also observe that ∀p ∈ ker(φ), then φ(p) = e2 and σ(e2) = e3, so p ∈ ker(σ ◦ φ). Thusker(φ) ⊆ ker(σ ◦ φ). By the two step test, we need to find that ker(φ) to be nonempty, hasclosure, and it has inverses. By the above satement, it can be seen that e1 ∈ ker(φ), so ker(φ)is nonempty.Consider a, b ∈ ker(φ). Then φ(a) = e2 and φ(b) = e2, so φ(a ◦ b) = φ(a) ◦ φ(b) = e2 ◦ e2 = e2.Thus a ◦ b ∈ ker(φ).Now for inverses, since ker(φ) is closed, then ∀a ∈ ker(φ), there exists a−1 such that

a ◦ a−1 = e1φ(a ◦ a−1) = φ(e1)φ(a) ◦ φ(a−1) = e2e2 ◦ φ(a−1) = e2φ(a−1) = e2

Thus a−1 ∈ ker(φ). Therefore ker(φ) ≤ ker(σ ◦ φ).

6. [Hoffman, L.] If φ and γ are isomorphisms from the cyclic group 〈a〉 to some group, and φ(a) = γ(a),then prove that φ = γ.

Proof. We need to show that ∀b ∈ 〈a〉, φ(b) = γ(b). We know the following facts.

(a) For b ∈ 〈a〉, ∃n ∈ Z such that b = an.

(b) For n ∈ Z, φ(an) = (φ(a))n and γ(an) = (γ(a))n.

Then, for b ∈ 〈a〉,φ(b) = φ(an) = (φ(a))n = (γ(a))n = γ(an) = γ(b).

Thus φ = γ.

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7. [Leonard, R.] Show that Z has infinitely many subgroups isomorphic to Z.

Proof. Consider φ : Z→ nZ. We must prove the following:

(a) nZ ⊆ Z(b) φ is a homomorphism

(c) φ is bijective

(a) nZ = {nz : z ∈ Z and n ∈ Z+}, so nZ consists of multiples of n. Because there is no commondivisor other than 1 for all integers, it must be that nZ ⊆ Z. We already know nZ is a group,so it must be that nZ ≤ Z. This is true for all n ∈ Z. Since there are an infinite number ofpossibilities for n, there are an infinite number of subgroups.

(b) We must show the group operation is preserved for φ

φ(a ◦ b) = n(a ◦ b) = na ◦ nb = φ(a) ◦ φ(b)

So the group operation is preserved, and φ is a homomorphism.

(c) It must be shown that φ is both injective and surjective. Consider φ(a) = φ(b) If φ(a) = naand φ(b) = nb, it must be that a = b by cancellation. This indicates that φ is injective. Takeφ(a) ∈ nZ. The only input that will result in na is a, so φ is surjective.

Thus φ : Z→ nZ is an isomorphism, and because there are an infinite number of sets nZ, there arean infinate number of subgroups of Z isomorphic to Z.

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2 Homework #9

1. [Stotz, D.] Show that the mapping a→ log10 a is an isomorphism from R+ under multiplication toR under addition.

Proof. It must be shown that:

(a) φ is a homomorphism.

(b) φ is injective.

(c) φ is surjective.

We now prove the following:

(a) Consider a, b ∈ R+. By the properties of logarithms,

φ(ab) = log10(ab) = log10 a+ log10 b = φ(a) + φ(b).

Therefore, the group operations are preserved and φ is a homomorphism from {R+, ∗} to {R,+}.(b) Consider x, y ∈ R+ such that φ(x) = φ(y). By our definition of φ,

φ(x) = φ(y)log10(x) = log10(y)

10log10(x) = 10log10(y)

x = y.

Since x = y, φ is injective, as each output has one unique input.

(c) Consider a ∈ R. Let x = 10a, and note that even if a < 0, 10a ∈ R+. By our definition of φ,

φ(x) = log10(x) = log10(10a) = a.

Thus φ is surjective.

So, φ is a bijective homomorphism, and therefore an isomorphism.

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2. The following problems deal with finding the left cosets of subgroups.

(a) [Smith, C.] Let n be a positive integer. Let H = {0,±n,±2n,±3n, . . .}. Find all left cosets ofH in Z. How many are there?

When the operation for a group is addition, cosets are usually designated by a + H instead ofaH.

Possible left cosets are 0 +H, 1 +H, 2 +H, 3 +H, . . . , (n− 1) +H,n+H, . . ..

Then

0 +H = H,

1 +H = {1, 1± n, 1± 2n, 1± 3n, . . .},2 +H = {2, 2± n, 2± 2n, 2± 3n, . . .},3 +H = {3, 3± n, 3± 2n, 3± 3n, . . .},...

(n− 1) +H = {(n− 1), (n− 1)± n, (n− 1)± 2n, (n− 1)± 3n, . . .},n+H = {n, n± n, n± 2n, n± 3n, . . .} = {0,±n,±2n,±3n, . . .} = H.

All the left cosets of H in Z are 0 +H, 1 +H, 2 +H, 3 +H, . . . , (n− 1) +H, since n+H is reallyjust H. Therefore there are n cosets of H in Z.

(b) [Hjelmfelt, C.] Find all the left cosets of {1, 11} in U(30).

If H = {1, 11} and G = U(30) = {1, 7, 11, 13, 17, 19, 23, 29}, we know by Lagrange’s theoremthat the number of distinct cosets is: |G|/|H| = 8/2 = 4.1 ◦H = {1, 11} 11 ◦H = 1 ◦H7 ◦H = {7, 17} 17 ◦H = 7 ◦H13 ◦H = {13, 23} 23 ◦H = 13 ◦H19 ◦H = {19, 29} 29 ◦H = 19 ◦HThe unique left cosets in U(30) are 1H, 7H, 13H, 19H.

(c) [Keene, L.] Suppose that a has order 15. Find all the left cosets of 〈a5〉 in 〈a〉.We know that |a| = 15, and so

a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a ◦ a = e.

Or, a5 ◦ a5 ◦ a5 = e

Thus |a5| = 3, and 〈a5〉 = {e, a5, a10} ≤ 〈a〉. Then the cosets are

e〈a5〉 = {e, a5, a10},

a〈a5〉 = {a, a6, a11},

a2〈a5〉 = {a2, a7, a12},

a3〈a5〉 = {a3, a8, a13},

a4〈a5〉 = {a4, a9, a14},

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a5〈a5〉 = {a5, a10, a15} = {e, a5, a10} = e〈a5〉.

At which point the cycle repeats, giving.

e〈a5〉 = a5〈a5〉 = a10〈a5〉,

a1〈a5〉 = a6〈a5〉 = a11〈a5〉,

a2〈a5〉 = a7〈a5〉 = a12〈a5〉,

a3〈a5〉 = a8〈a5〉 = a13〈a5〉,

a4〈a5〉 = a9〈a5〉 = a14〈a5〉.

And there are 5 unique left cosets of 〈a5〉 in 〈a〉.

3. [Spurlock, J.] If H and K are subgroups of G and g ∈ G, show that g(H ∩K) = gH ∩ gK.

Proof. (⊆) Consider some element a ∈ g(H ∩K). This means there is some element m ∈ (H ∩K)such that a = g ◦m. Since m ∈ (H ∩K), it must be that m ∈ H and m ∈ K.

This means that g ◦m ∈ gH and g ◦m ∈ gK, or a ∈ gH and a ∈ gK, so a ∈ gH ∩ gK. Since a isarbitrary, g(H ∩K) ⊆ gH ∩ gK.

(⊇) Consider some element a ∈ gH ∩ gK. This means there is some element h ∈ H such thata = g ◦ h, and there is some element k ∈ K such that a = g ◦ k. Since they are both equal to a, wecan say that g ◦ h = g ◦ k. By cancellation, this means h = k.

Therefore, h ∈ H and h ∈ K, so h ∈ H ∩ K. This means that g ◦ h = a ∈ g(H ∩ K). Since a isarbitrary, g(H ∩K) ⊇ gH ∩ gK.

Having proven containment in both directions, we can say g(H ∩K) = gH ∩ gK.

4. [Taggart, C.] Suppose that H and K are subgroups of G and that there are elements a, b ∈ G suchthat aH ⊆ bK. Prove that H ⊆ K.

Proof. (Taggart, C.) Note that b ∈ bK. Also a ∈ aH. Since a ∈ aH that means ∃k ∈ K such thata = b ◦ k so a ∈ bK. Since a ∈ bK that means aK = bK and a−1 ◦ b ∈ K.

Consider x ∈ H. That means ∃c ∈ K such that a ◦ x = b ◦ c. By performing a left operation witha−1, we get x = a−1 ◦ b ◦ c. We know that c ∈ K and that a−1 ◦ b ∈ K. This means that theircomposition is in K, which means x ∈ K. Therefore H ⊆ K.

Proof. (Conger, A.) Suppose aH ⊆ bK. Since H and K are groups, there exist e ∈ H and f ∈ Ksuch that a ◦ e = a = b ◦ f since aH ⊆ bK. Applying b−1 to the left we find b−1 ◦ a = f . SinceaH ⊆ bK,

H = a−1(aH) ⊆ a−1(bK) = (a−1 ◦ b)K = (a−1 ◦ b)(fK) = ((a−1 ◦ b) ◦ (b−1 ◦ a))K = K.

Thus H ⊆ K.

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Proof. (Cosand, K.) We know that b ∈ bK and that a ∈ aH. Since aH ⊆ bK, we know there existsa k ∈ K such that a = b ◦ k, and therefore a ∈ bK. Since we know that a ∈ bK, aK = bK. We alsonow know that a−1 ◦ b ∈ K.

If we take x ∈ H, then we know there must exists a c ∈ K such that a◦x = b◦ c. If we perform a leftoperation with a−1, we then have a−1 ◦ a ◦ x = a−1 ◦ b ◦ c, which can be simplified to x = a−1 ◦ b ◦ c.We now know that a−1 ◦ b ◦ c ∈ K, since K has closure. Thus x = a−1 ◦ b ◦ c ∈ K. Therefore H ⊆ K.

Proof. (Dyke, M.) Note that because H is a subgroup of G, a ∈ aH by Coset Lemma #1. Thena ∈ bK, which means ∃k ∈ K such that a = b ◦ k and a−1 = (b ◦ k)−1 = k−1 ◦ b−1.Now let x ∈ H, which means that a ◦ x ∈ bK and ∃y ∈ K such that a ◦ x = b ◦ y. Then

a−1 ◦ a ◦ x = a−1 ◦ b ◦ yx = a−1 ◦ b ◦ yx = (k−1 ◦ b−1) ◦ b ◦ yx = k−1 ◦ (b−1 ◦ b) ◦ yx = k−1 ◦ y

Because K is a subroup of G, k−1 ∈ K. Since k−1, y ∈ K, and K is closed, k−1 ◦ y ∈ K. Therefore,x ∈ K, and H ⊆ K.

Proof. (Franck, D.) Since we know that H is a subgroup of G, the identity must be in H (e ∈ H).Therefore, a = a ◦ e ∈ aH ⊆ bK. Then ∃c ∈ K such that a = b ◦ c.Note that a−1 = (b ◦ c)−1 = c−1 ◦ b−1, so c−1 ∈ K. If we take x ∈ H, we need to show x ∈ K tosupport the fact that H ⊆ K. Using our knowledge of groups, we know a ◦ x ∈ a ◦ H. Therefore,a ◦x ∈ bK, which suggests that there exists y ∈ K with the property that a ◦x = b ◦ y. Let’s lay thisout:

x = (a−1 ◦ a) ◦ x = a−1 ◦ (a ◦ x) = a−1 ◦ (b ◦ y) = (c−1 ◦ b−1) ◦ b ◦ y = c−1 ◦ y.

As c, y ∈ K, and K is a subgroup, then c−1 ◦ y ∈ K. Thus, x ∈ K, and H ⊆ K.

Proof. (Gruenig, S.) By (1) of our coset lemma, we know that a ∈ aH. Since aH ⊆ bK, a is also anelement of bK. Then by (4) of our coset lemma, since a ∈ bK, we know that bK = aK, so replacingbK with aK in our relationship gives aH ⊆ aK. By definition, this means that for all h ∈ H thereexists some k ∈ K such that ah = ak, and by cancellation, h = k. Since this is true ∀h ∈ H, H ⊆ K.

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Proof. (Havelka, R.) Observe that the definitions of aH and bK are aH = {a ◦ h : h ∈ H} andbK = {b ◦ k : k ∈ K}. Since H ⊆ G and K ⊆ G, then h, k ∈ G.

We have aH ⊆ bK, so ∀a, h ∈ aH, there exists a b ◦ k ∈ bK where a ◦ h = b ◦ k. Note, it can be seenthat h = (a−1 ◦ b) ◦ k and (b−1 ◦ a) ◦h = k. So h ∈ (a−1 ◦ b)K and k ∈ (b−1 ◦ a)H. Now H is a group,so e ∈ H. This means that (b−1 ◦ a) ◦ e = (b−1 ◦ a) = k, thus b−1 ◦ a ∈ K. Since K is a group, thenthe inverse of b−1 ◦ a is also in K. So (b−1 ◦ a)−1 = a−1 ◦ b ∈ K.

Let a−1 ◦ b = k2, and observe that k2 ∈ K. Now h = (a−1 ◦ b) ◦ k becomes h = k2 ◦ k, and it it canbe seen that since k2, k ∈ K and k2 ◦ k ∈ K. Thus, from the fact that h = k2 ◦ k, then h ∈ K.

Thus all elements h ∈ H are also elements in K, therefore H ⊆ K.

Proof. (Leonard, R.) We know aH ⊆ bK, so there exists a (b ◦ k) ∈ bK such that b ◦ k = a ◦ h forevery (a ◦ h) ∈ aH. Therefore

h = a−1 ◦ b ◦ k.

We can see from the above that

a−1 ◦ b ◦ k ∈ H.

We want to show a−1 ◦ b ∈ K. If this is in K, then we know h ∈ K, because K is closed anda−1 ◦ b ◦ k = h. We know by the coset lemma that a ∈ aH and b ∈ bK, and so there exists k ∈ Ksuch that a = b ◦ k and a ∈ bK. By the lemma, we then see that aK = bK, and we can write

a ◦ k1 = b ◦ k2,

ork1 = a−1 ◦ b ◦ k2.

Therefore a−1 ◦ b ∈ K, because k1 ∈ K. Therefore h ∈ K. Since h ∈ K, we see that H ⊆ K .

Proof. (Norstedt, Z.) Let c ∈ aH. We know that ∃h ∈ H such that c = a ◦ h. We also know thatc ∈ bK, so c = b ◦ k, for some k ∈ K.

Then c = a ◦ h = b ◦ k, and a = b ◦ k ◦ h−1. Note that a ∈ aH, so a = b ◦ (k ◦ h−1) ∈ bK, and therebyk ◦ h−1 ∈ K. Because k was taken out of K, k−1 ∈ K, since K is a subgroup. Thus,

k−1 ◦ (k ◦ h−1) = (k−1 ◦ k) ◦ h−1 = h−1 ∈ K,

by closure. Therefore, (h−1)−1 = h ∈ K, and H ⊆ K.

Moreover, as H and K are both groups, H ≤ K.

12

Proof. (Smith, C.) We know H is subgroup of G, and by the coset Lemma (1) a ∈ aH. AlsoaH ⊆ bK, and since aH is a subset of bK, we know a ∈ bK. Because bK is a coset and K is asubgroup of G, a = b ◦ k for some k ∈ K.

Consider h ∈ H. We know a ◦ h ∈ aH and using substitution

a ◦ h = (b ◦ k1) ◦ h = b ◦ (k1 ◦ h) ∈ aH.

Because aH ⊆ bK and b ◦ (k1 ◦ h) ∈ bK, we now know b ◦ k1 ◦ h = b ◦ k2, for some k1, k2 ∈ K. Usingcancellation we get

k1 ◦ h = k2k−11 ◦ k1 ◦ h = k−11 ◦ k2

e ◦ h = k−11 ◦ k2h = k−11 ◦ k2.

Because k−11 and k2 are both in K and h = k−11 ◦ k2, then h ∈ K, and h ∈ H. Therefore H ⊆ K.

Proof. (Spurlock, J.) Since H is a subgroup of G, we know that a ∈ aH by part 1 of the cosetlemma. Since aH ⊆ bK, this means there exists some k ∈ K such that a = b ◦ k. Applying b−1, wefind that k = b−1 ◦ a. By part 6 of the coset lemma, this means bK = aK, and therefore aH ⊆ aK.

This means that, for any h ∈ H, there exists a k ∈ K such that a ◦ h = a ◦ k. By cancellation, thismeans h = k, and thus H ⊆ K.

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3 Homework #10

1. [Cochran-Bjerke, L.] Let H and K be subgroups of a finite group G with H ⊆ K ⊆ G. Prove that|G : H| = |G : K| · |K : H|.

Proof. Note that H ⊆ K, and since H is a subgroup of G, then by the Finite Subgroup Test H is asubgroup of K. Since H ≤ K, K ≤ G, and H ≤ G, then using Lagrange’s Theroem we have

|G : H| = |G||H|

, |G : K| = |G||K|

, |K : H| = |K||H|

.

Consider

|G : K| · |K : H| = |G||K|· |K||H|

=|G||H|

= |G : H|.

Thus|G : H| = |G : K| · |K : H|.

2. [Havelka, R.] Let |G| = 15. If G has only one subgroup of order 3 and only one of order 5, provethat G is cyclic. Generalize to |G| = pq, where p and q are prime.

Proof. By Lagrange’s Theorem, for any g ∈ G then the divisors of |G| can be the order of g, or|g| = 1, p, q, pq. Let H,K ≤ G such that |H| = p and |K| = q. Then the following conditions arepossible for g ∈ G.

(a) If |g| = 1, then g = e, and g is in all the subgroups of G.

(b) If |g| = p, then g ∈ H, and H = 〈g〉.(c) If |g| = q, then g ∈ K and K = 〈g〉.(d) If |g| = pq, then g ∈ G and no proper subgroups of G.

Then consider g ∈ G such that |g| = pq. It is necessary to verify that such an element exists in G.Since |H| = p, |K| = q, and H ∩ K = {e}, then the number of elements that are not exclusivelyin only G is p + q − 1. Observe that pq > p + q − 1, for all primes p and q, or in other words,|G| > p+ q − 1. Thus ∃g1 ∈ G, neither in H nor K, such that |g1| = pq.

Claim: 〈g1〉 = G, for this g1 ∈ G.

(⊆) It follows by definition that 〈g〉 ⊆ G.

(⊇) Since |g1| = |〈g1〉| = pg, then G ⊆ 〈g1〉, since every element of G must be created by 〈g1〉.Therefore 〈g1〉 = G, and G is cyclic.

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3. [Franck, D.] Let G be a group of order 25. Prove that G is cyclic or g5 = e for all g ∈ G. Generalizeto any group of order p2, where p is prime. Does your proof work for this generalization?

Proof. If we let g ∈ G and p be prime, then we know that |G| = p2, so the order of g divides theorder of p2. Then |g| could be 1, p, or p2.

If we assume |g| = 1, this means that g = e, so gp = e.

If we have |g| = p, then gp = e.

Lastly, if we assume |g| = p2, then by definition, we know that |〈g〉| = p2. Clearly, 〈g〉 ⊆ G. Also,since |〈g〉| = p2 and |G| = p2, then G ⊆ 〈g〉. Thus G = 〈g〉, and G is cyclic.

Thus, G is either cyclic or gp = e, ∀g ∈ G.

4. [Kittler, S.] Prove that if G is a finite group, the index of Z(G) cannot be prime.

Proof. (Kittler, S.) By contradiction, assume |G : Z(G)| = p, where p is prime, and |Z(G)| = k, forsome positive integer k. Thus |G| = pk. Take g ∈ G \ Z(G). Consider C(g), where |G : C(g)| = x,|C(g) : Z(G)| = y, and xy = p. The only way for this to be true is if either (1) x = 1 and y = p, or(2) x = p and y = 1.

For case 1: |G : C(g)| = 1 and |C(g) : Z(G)| = p. Thus G = C(g), which means g ◦x = x◦g, ∀x ∈ G,which is the definition of the center and g ∈ Z(G), and this is a contradiction.

For case 2: |G : C(g)| = p and |C(g) : Z(G)| = 1. Thus C(g) = Z(G), and g ∈ C(g) = Z(G). Thusg ∈ Z(G), and this is a contradiction.

Thus |G : Z(G)| cannot be prime.

Proof. (Dyke, M.) There are two cases to consider.

Case 1:

Assume G is an abelian group. Then Z(G) = G. The index of Z(G) is |G||Z(G)| , which equals 1 in

this case. Therefore, the index of Z(G) is not prime, because 1 is not a prime number.

Case 2:

Assume G is non-abelian. By the contrapositive of Theorem 9.3, G/Z(G) is not cyclic. Thenby the contrapositive of our second corollary of Lagrange’s Theorm, G/Z(G) cannot have primeorder. Because G/Z(G) is the collection of left cosets of Z(G) in G, in this case there is anon-prime number of left cosets of Z(G) in G. Therefore, by definition of the index, the indexof Z(G) is not prime.

Since the index of Z(G) is not prime in both cases, it can never be prime.

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Proof. (Taggart, C.) Assume to the contrary that |G : Z(G)| = p, where p is prime. Since Z(G)/G,there is a factor group G/Z(G), which has p elements. Since |G/Z(G)| is prime, it is cyclic, and thus∃a ∈ G such that G/Z(G) = {Z(G), aZ(G), a2Z(G), ..., ap−1Z(G)}. Since this group contains all theleft cosets of Z(G), all the elements of G are contained within one of these cosets.

Consider b ∈ G such that b = ak ◦ z1, ∃k ∈ Z and ∃z1 ∈ Z(G). Also consider c ∈ G such thatc = aj ◦ z2, ∃j ∈ Z and ∃z2 ∈ Z(G). Then

b ◦ c = ak ◦ z1 ◦ aj ◦ z2 = ak ◦ aj ◦ z1 ◦ z2 = ak+j ◦ z2 ◦ z1 = aj ◦ ak ◦ z2 ◦ z1 = aj ◦ z2 ◦ ak ◦ z1 = c ◦ b.

Since b, c are arbitrary elements in G, then b◦c = c◦b, ∀b, c ∈ G. So |Z(G)| = |G|, and |G : Z(G)| = 1.This is a contradiction. Thus |G : Z(G)| cannot be prime.

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5. [Karatekeli, K.] You are going to rewrite pages 151–153 of the book. Starting with the title AnApplication of Cosets to Permutation Groups through the title The Rotation Group ofa Cube and a Soccer Ball, you are going to typeset a better version of what the book does.Everything the book doesn’t prove, you are going to prove. You are going to provide every detail thebook skims over. You are going to come up with your own Examples 7 and 8, as well.

4 An Application of Cosets to Permutation Groups

Lagrange’s Theorem and its corollaries dramatically demonstrate the fruitfulness of the coset concept. Wenext consider an application of cosets to permutation groups.

Definition 1 (Stabilizer of a Point). Let G be a group of permutations of a set S. For each i in S, letstabG(i) = {φ ∈ G : φ(i) = i}. We call stabG(i) the stabilizer of i in G.

Proposition 1. For all i ∈ S, the stabilizer, stabG(i), is a subgroup of G.

Proof. Using the two-step subgroup test, we must show:

1. stabG(a) is non-empty.

2. αβ ∈ stabG(a).

3. α−1 ∈ stabG(a).

Then

1. Let a ∈ S. Note that because ε(a) = a, we know ε ∈ stabG(a). Therefore, stabG(a) is non-empty.

2. Let α, β ∈ stabG(a). Since we know β ∈ stabG(a), then β(a) = a. Note (αβ)(a) = α(β(a)) = α(a) =a, so αβ ∈ stabG(a).

3. Consider α(a) = a. By performing a left hand operation using α−1, we can show α−1 ∈ stabG(a).Note that

α(a) = a

α−1(α(a)) = α−1(a)

(α−1α)(a) = α−1(a)

ε(a) = α−1(a)

a = α−1(a).

Since α ∈ stabG(a), we can also conclude that α−1 ∈ stabG(a). Thus, stabG(a) ≤ G, by the two-stepsubgroup test.

Definition 2 (Orbit of a Point). Let G be a group of permutations of a set S. For each s in S, letorbG(s) = {φ(s) : φ ∈ G}. The set orbG(s) is a subset of S called the orbit of s under G. We use |orbG(s)|to denote the number of elements in orbG(s).

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Example 1. Let G = {(1)(2)(3)(4), (12)(3)(4), (1)(2)(34), (12)(34)}, which is a group. Then

orbG(1) = {1, 2}, stabG(1) = {(1)(2)(3)(4), (1)(2)(34)},orbG(2) = {2, 1}, stabG(2) = {(1)(2)(3)(4), (1)(2)(34)},orbG(3) = {3, 4}, stabG(3) = {(1)(2)(3)(4), (12)(3)(4)},orbG(4) = {4, 3}, stabG(4) = {(1)(2)(3)(4), (12)(3)(4)}.

Example 2. Let G = {(1)(2)(3), (1)(23), (2)(13), (3)(12), (123), (132)}, which is a group. Then

orbG(1) = {1, 3, 2}, stabG(1) = {(1), (1)(23)},orbG(2) = {2, 3, 1}, stabG(2) = {(2), (2)(13)},orbG(3) = {3, 2, 1}, stabG(3) = {(3), (3)(12)}.

Theorem 1 (Orbit-Stabilizer Theorem). Let G be a finite group of permutations of a set S. Then, forany i from S, |G| = |orbG(i)||stabG(i)|.

Proof. By Lagrange’s Theorem, |G|/|stabG(i)| is the number of distinct left cosets of stabG(i) in G. Toprove this, we must show

1. The relation between the left cosets of stabG(i) and the elements in the orbit of i is well-defined.

2. There is an injective(one-to-one) relation between the left cosets of stabG(i) and the elements in theorbit of i.

3. There is a surjective(onto) relation between the left cosets of stabG(i) and the elements in the orbitof i.

Then

1. To show that T is a well-defined function, we must show that αstabG(i) = βstabG(i) implies α(i) =β(i). But αstabG(i) = βstabG(i) implies α−1β ∈ stabG(i), so that (α−1β)(i) = i, and therefore,β(i) = α(i).

2. To show the mapping is injective, we define a correspondence T by mapping the coset αstabG(i)to α(i) under T . Assume that β(i) = α(i). Then (α−1β)(i) = i. But this implies that there is aα−1β ∈ stabG(i). Therefore, αstabG(i) = βstabG(i), which shows it is injective.

3. To show T is onto orbG(i), let j ∈ orbG(i). Then α(i) = j for some α ∈ G. Clearly T (αstabG(i)) =α(i) = j, therefore T is onto.

Definition 3 (Partition). A partition of a set S is a collection of non-empty disjoint subsets of S whoseunion is S.

Example 3 (Quaternions). Since a partition is made of non-empty disjoint subsets, we can take elementsfrom an existing group to create disjoint subsets. Using the Quaternions, a partition would consist of

{1, j,−1}, {−1}, {k,−k}, {−j, i}.

Note that every set is disjoint and the collection of the elements makes up the Quaternion group.

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Proposition 2. The orbits of the members of S constitute a partition of S.

Proof. We need to show

1. Each orbit is non-empty and every element is contained in an orbit.

2. The cycles are either equal, orbG(i) = orbG(j), or disjoint, orbG(i) ∩ orbG(j) = ∅.

Then

1. We know that ∀a ∈ S will have an orbG(a) such that a ∈ orbG(a).

2. If the cycles are disjoint, we are done. If the intersection of the orbits is equal to the empty set,we need to show the orbits are equal to zero. Assume orbG(i) ∩ orbG(j) 6= ∅, where i 6= j. Letk ∈ orbG(i) ∩ orbG(j). So, for some α, β ∈ G, α(i) = k = β(j). Then we can show

β−1(k) = (β−1β)(j)

β−1(α(i)) = ε(j)

(β−1α)(i) = j.

Similarly, we can use the same process to show that i = (α−1β)(j). Also, since β−1α ∈ G andα−1β ∈ G, then j = (β−1α)(i) ∈ orbG(i), and i ∈ orbG(j).

Take x ∈ orbG(i). We need to show x ∈ orbG(j), by finding γ ∈ G, such that γ(j) = x. There existsa µ ∈ G, such that µ(i) = x. Then

µ(α−1(β(j))) = µ(i) = x

(µα−1β)(j) = x.

Let γ = µα−1β ∈ G. Then γ(j) = x, and x ∈ orbG(j).

Thus orbG(i) ⊆ orbG(j). Similarly, orbG(j) ⊆ orbG(i). Hence orbG(i) = orbG(j).

Therefore, the orbits of the members of S constitute a partition of S.

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5 Homework #11

1. [Berger, E.] Let φ be a homomorphism from G1 to G2. Prove that ker (φ) is a normal subgroup ofG1.

Proof. Recall the Normal Subgroup Test: A subgroup H of G is normal in G if and only if xHx−1 ⊆H, for all x in G. Assume k ∈ ker(φ) and g ∈ G1. We are trying to prove that gkg−1 ∈ ker(φ). So

φ(g ◦ k ◦ g−1) = φ(g) ◦ φ(k) ◦ φ(g−1) = φ(g) ◦ e2 ◦ (φ(g))−1 = φ(g) ◦ (φ(g))−1 = e2.

As such, gkg−1 ∈ ker(φ), so ker(φ) is a normal subgroup of G1.

2. [Kranstz, S.] In D4, let K = {R0, F�}, and let L = {R0, F�, F�, R180}. Show that K / L / D4 butthat K is not normal in D4. (Normality is not transitive.) Hint: a problem in the extra exerciseswill help you avoid the endless calculations needed to show normality.

Proof. In order to show K / L / D4 more easily, use Problem 7 from below.

Note that

|D4 : L| = |D4||L|

=8

4= 2.

Therefore L / D4. Also

|L : K| = |L||K|

=4

2= 2.

Thus K / L.

However, K is not normal in D4. A counterexample will be provided in order to prove that ∃x ∈ D4

such that xK 6= Kx. Let x = F−. So

xK = {F−, R270},

andKx = {F−, R90}.

Since xK does not equal Kx, then K is not normal in D4.

3. [Henson, T.] Consider the following factor groups using subgroups of Z. (I used Excel to helporganize and compute the cosets.)

(a) Viewing 〈3〉 and 〈12〉 as subgroups of Z, prove that 〈3〉/〈12〉 is isomorphic to Z4.

Note that for this group 〈a〉 = {an : n ∈ Z}. Thus

〈3〉 = {. . . ,−3, 0, 3, 6, 9, 12, . . .},

〈12〉 = {. . . ,−12, 0, 12, 24, 36, 48, . . .},

and〈3〉/〈12〉 = {〈12〉, a+ 〈12〉, b+ 〈12〉, . . .},

∀a, b ∈ 〈3〉.

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The resulting cosets are

0 + 〈12〉 = {. . . , 0, 12, 24, 36, 48, 60, . . .},

3 + 〈12〉 = {. . . , 3, 15, 27, 39, 51, 63, . . .},

6 + 〈12〉 = {. . . , 6, 18, 30, 42, 54, 66, . . .},

and9 + 〈12〉 = {. . . , 9, 21, 33, 45, 57, 69, . . .}.

We know that these are the only cosets as any other multiple of 3 will just be a shifted versionof one of these four cosets. Now, let’s put these cosets into a Cayley table to compare them to(Z4,+).

+ 0 + 〈12〉 3 + 〈12〉 6 + 〈12〉 9 + 〈12〉0 + 〈12〉 0 + 〈12〉 3 + 〈12〉 6 + 〈12〉 9 + 〈12〉3 + 〈12〉 3 + 〈12〉 6 + 〈12〉 9 + 〈12〉 0 + 〈12〉6 + 〈12〉 6 + 〈12〉 9 + 〈12〉 0 + 〈12〉 3 + 〈12〉9 + 〈12〉 9 + 〈12〉 0 + 〈12〉 3 + 〈12〉 6 + 〈12〉

Compare this with the Cayley Table for (Z4,+).

+ 0 1 2 3

0 0 1 2 31 1 2 3 02 2 3 0 13 3 0 1 2

Clearly the Cayley tables match and the elements map as follows:

0 + 〈12〉 → 0,

3 + 〈12〉 → 1,

6 + 〈12〉 → 2,

9 + 〈12〉 → 3.

As a result, 〈3〉/〈12〉 is isomorphic to Z4.

(b) Similarly, prove 〈8〉/〈48〉 is isomorphic to Z6.

Here〈8〉 = {. . . ,−8, 0, 8, 16, 24, 32, . . .},

〈48〉 = {. . . ,−48, 0, 48, 96, 144, 192, . . .},

and〈8〉/〈48〉 = {〈48〉, a+ 〈48〉, b+ 〈48〉, . . .},

∀a, b ∈ 〈8〉.The resulting cosets are

0 + 〈48〉 = {. . . , 0, 48, 96, 144, 192, 240, . . .},

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8 + 〈48〉 = {. . . , 8, 56, 104, 152, 200, 248, . . .},

16 + 〈48〉 = {. . . , 16, 64, 112, 160, 208, 256, . . .},

24 + 〈48〉 = {. . . , 24, 72, 120, 168, 216, 264, . . .},

32 + 〈48〉 = {. . . , 32, 80, 128, 176, 224, 272, . . .},

40 + 〈48〉 = {. . . , 40, 88, 136, 184, 232, 280, . . .}.

We know that these are the only cosets as any other multiple of 8 will simply be a shifted versionon these six cosets. Now we make a Cayley table of the cosets and compare them to (Z6,+).

+ 0 + 〈48〉 8 + 〈48〉 16 + 〈48〉 24 + 〈48〉 32 + 〈48〉 40 + 〈48〉0 + 〈48〉 0 + 〈48 8 + 〈48〉 16 + 〈48〉 24 + 〈48〉 32 + 〈48〉 40 + 〈48〉8 + 〈48〉 8 + 〈48〉 16 + 〈48〉 24 + 〈48〉 32 + 〈48〉 40 + 〈48〉 0 + 〈4816 + 〈48〉 16 + 〈48〉 24 + 〈48〉 32 + 〈48〉 40 + 〈48〉 0 + 〈48 8 + 〈48〉24 + 〈48〉 24 + 〈48〉 32 + 〈48〉 40 + 〈48〉 0 + 〈48 8 + 〈48〉 16 + 〈48〉32 + 〈48〉 32 + 〈48〉 40 + 〈48〉 0 + 〈48 8 + 〈48〉 16 + 〈48〉 24 + 〈48〉40 + 〈48〉 40 + 〈48〉 0 + 〈48 8 + 〈48〉 16 + 〈48〉 24 + 〈48〉 32 + 〈48〉

Compare this with the Cayley table for (Z6,+).

+ 0 1 2 3 4 5

0 0 1 2 3 4 51 1 2 3 4 5 02 2 3 4 5 0 13 3 4 5 0 1 24 4 5 0 1 2 35 5 0 1 2 3 4

Clearly, the Cayley tables match and the elements map as follows:

0 + 〈48〉 → 0,

8 + 〈48〉 → 1,

16 + 〈48〉 → 2,

24 + 〈48〉 → 3,

32 + 〈48〉 → 4,

40 + 〈48〉 → 5.

As a result, 〈8〉/〈48〉 is isomorphic to Z6.

(c) Generalize to arbitrary integers k and n. You don’t need to provide a proof, just a conjecture.

Let k, n ∈ Z. If k|n, then 〈k〉/〈n〉 is isomorphic to Zn/k.

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4. [Norstedt, Z.] If H is a normal subgroup of a group G, prove that

C(H) = {g ∈ G : g ◦ h = h ◦ g,∀h ∈ H},

the centralizer of H in G, is a normal subgroup of G.

Proof. Begin by the Normal Subgroup Test. First, this test requires that C(H) be a subgroup of G.Then, if gC(H)g−1 ⊆ C(H),∀g ∈ G, we can conclude that C(H) is a normal subgroup of G.

As the identity commutes with all elements of H, e ∈ C(H), and C(H) is non-empty. Continue bythe One-Step Subgroup Test.

Let a, b ∈ C(H) and consider h1 ◦(a◦b−1), where h1 ∈ H. Because anything in C(H) commutes witheverything in H, h1 ◦ (a ◦ b−1) = a ◦ h1 ◦ b−1. We know that b commutes with h1, so b ◦ h1 = h1 ◦ b.Applying b−1 on both the left and the right results in

b−1 ◦ h1 ◦ b ◦ b−1 = b−1 ◦ b ◦ h1 ◦ b−1b−1 ◦ h1 ◦ e = e ◦ h1 ◦ b−1b−1 ◦ h1 = h1 ◦ b−1,

and b−1 also commutes with h1. Hence, a◦(h1◦b−1) = a◦b−1◦h1, and h1◦(a◦b−1) = (a◦b−1)◦h1, ∀h1 ∈H. Thus, a ◦ b−1 ∈ C(H), and C(H) ≤ G by the One-Step Subgroup Test.

Next, consider h1 ◦ g. Because H is a normal subgroup of G, we know that h1 ◦ g = g ◦ h2 for someh2 ∈ H. Applying g−1 on both the left and right yields

g−1 ◦ h1 ◦ g ◦ g−1 = g−1 ◦ g ◦ h2 ◦ g−1,g−1 ◦ h1 ◦ e = e ◦ h2 ◦ g−1,

g−1 ◦ h1 = h2 ◦ g−1.

Thus, if h1 ◦ g = g ◦ h2, then g−1 ◦ h1 = h2 ◦ g−1.Finally, let k ∈ C(H) and consider h1 ◦ (g ◦ k ◦ g−1). Since H CG, h1 ◦ g = g ◦ h2, and

(h1 ◦ g) ◦ k ◦ g−1 = g ◦ h2 ◦ k ◦ g−1.

By the definition of C(H), we know that k commutes with all elements of H, so

g ◦ (h2 ◦ k) ◦ g−1 = g ◦ k ◦ h2 ◦ g−1.

By what was shown above, h2 ◦ g−1 = g−1 ◦ h1, and hence

g ◦ k ◦ (h2 ◦ g−1) = g ◦ k ◦ g−1 ◦ h1.

Thus, h1 ◦ (g ◦ k ◦ g−1) = (g ◦ k ◦ g−1) ◦ h1, and g ◦ k ◦ g−1 commutes with all h1 ∈ H for all k ∈ K.Therefore, by definition g◦k◦g−1 ∈ C(H), ∀g ∈ G, or equivalently gC(H)g−1 ⊆ C(H), and C(H)CGby the Normal Subgroup Test.

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5. [Kizzier, A.] Prove the following properties of rings. Let a, b, and c be elements of a ring R. Then

3. (−a)(−b) = ab.

0 = 0(−b)0 = (−a+ a)(−b)0 = (−a)(−b) + a(−b)0 = (−a)(−b)− (ab)

ab = (−a)(−b).

4. a(b− c) = ab− ac and (b− c)a = ba− ca.

a(b− c) = a(b+ (−c))= ab+ a(−c)= ab+ (−(ac))

= ab− ac.

Furthermore, if R has unity 1, then

5. (−1)a = −a.

Using rule number 2, (−1)a = −(1a) = −a.

6. (−1)(−1) = 1.

Using rule number 3, (−1)(−1) = 1 · 1 = 1.

7. The unity element is unique.

Let x, y ∈ R such that for each r ∈ R, we have rx = xr = r and ry = yr = r. This means that xand y are both unity elements. If we choose r = y and use our first equation, then xy = yx = y.If we choose r = x and use our second equation, then xy = yx = x. Then x = y, and the unityelement is unique.

8. Multiplicative inverses are unique.

Let r ∈ R. Suppose that x, y ∈ R such that rx = xr = 1 and ry = yr = 1. This means that xand y are both multiplicative inverses of r. Then xr = yr. Since x is a multiplicative inverse ofr, then xrx = yrx. So

x = x · 1= x(rx)

= y(rx)

= y · 1= y.

Since x = y, then multiplicative inverses are unique.

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6. [Rotert, A.] Suppose that R is a ring and that a2 = a for all a ∈ R. Show that R is commutative.(A ring in which a2 = a for all a is called a Boolean ring, in honor of the English mathematicianGeorge Boole (1815-1864).) Hint: consider for a, b ∈ R the quantity (a+ b)2.

Proof. Consider a, b ∈ R. Then

(a+ b)2 = a2 + ab+ ba+ b2

(a+ b) = a+ ab+ ba+ b

(a+ b) = ab+ ba+ (a+ b)

0 = ab+ ba (By cancellation)

ab = −ba.

This second step works because R is a ring, and is thus closed, so (a + b) = c, ∃c R, and c2 = c.Using this, then, we can say

ab = (ab)2

= (−ba)2

= (−ba)(−ba)

= (ba)(ba)

= (ba)2

= ba.

Therefore, we can conclude that a Boolean ring is commutative.

7. [Franck, D.] Prove that if H has index 2 in G, then H is normal in G. Hint: use the definition of anormal subgroup, not the Normal Subgroup Test.

Proof. Let H be a subgroup of index 2 in group G. Choose g ∈ G. We need to show that gH = Hg.

By definition, we know that the index suggests that there will be two distinct left cosets of H in G.This leaves two cases to consider.

If g ∈ H, then gH = Hg, because we know that gH = H and Hg = H, by a previous proof. ThusgH = Hg in this case.

Otherwise, if we have g ∈ G\H, we know that gH 6= H, since g 6∈ H. This suggests that gH = G\Hand Hg = G\H. Therefore, gH = Hg in this case, as well.

Thus, by the definition of a normal subgroup, the subgroup H is normal in G.

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8. [Cosand, K.] Let G be a finite group, and let H be a normal subgroup of G. Prove that the orderof the element gH in G/H must divide the order of g in G.

Proof. (Cosand, K.) Let |g| = n, then we know gn = e. Let |gH| = k. If we consider (gH)n, thenwe know (gH)n = (gn)H = eH = H. Note that H is the identity for the factor group G/H. ByCorollary 2 on Page 79 in the book, we know for a group G and a ∈ G with |a| = k, that if an = e,then k divides n. If we apply this idea to our current problem, we can conclude that k|n.

Proof. (Gates, S.) Consider the mapping ψ : G → G/H given by ψ(g) = gH. If φ preservesthe group operation, ψ is a homomorphism. By definition of the quotient group, the operation isaH ◦ bH = (a ◦ b)H. Consider, then, for a, b ∈ G:

ψ(a) ◦ ψ(b) = aH ◦ bH = (a ◦ b)H = ψ(a ◦ b).

Thus the group operation is preserved, and ψ is a homomorphism.

Also consider |g| = k, and recall that H is the identity of G/H. Then for gH ∈ G/H

(gH)k = ψ(g)k = ψ(gk) = ψ(e) = H.

Thus (gH)k = H, or the order of gH in G/H divides k, the order of g in G.

9. [Leonard, R.] Let φ be an isomorphism from a group G1 onto a group G2. Prove that if H is anormal subgroup of G1, then φ(H) is a normal subgroup of G2.

Proof. The first step of this proof is to prove that φ(x)φ(H) = φ(xH). To do this, we will use acontainment argument.

(⊇) Take a ∈ φ(x)φ(H). Then a = φ(x)φ(h), ∃h ∈ H. Moreover

a = φ(x)φ(h)

= φ(x ◦ h).

Thus a ∈ φ(xH).

(⊆) Take a ∈ φ(xH). Then a = φ(x ◦ h), ∃h ∈ H. Moreover

a = φ(x ◦ h)

= φ(x) ◦ φ(h).

Thus a ∈ φ(x)φ(H).

Since a is in both φ(x)φ(H) and φ(xH), it must be that φ(xH) = φ(x)φ(H). Similarly, we can showφ(Hx) = φ(H)φ(x).

Since H / G1, we know that xH = Hx, ∀x ∈ G1, and that φ(H) ≤ G2.

Consider y ∈ G2. Since φ is surjective, we know ∃x ∈ G1 such that φ(x) = y. Thus

yφ(H) = φ(x)φ(H) = φ(xH) = φ(Hx) = φ(H)φ(x) = φ(H)y,

and we see that φ(H) / G2, because yφ(H) = φ(H)y,∀y ∈ G2.

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10. [Franck, D.] Show, by example, that for fixed nonzero elements a and b in a ring, the equation ax = bcan have more than one solution. How does this compare with groups?

If we use the ring Z12, and choose a = 2 and b = 6, then the equation ax = b will have two solutions:x = 3 and x = 9.

Let’s prove these. If x = 3, then 2 · 3 ≡ 6 (mod 12). Likewise, if x = 9, then 2 · 9 ≡ 18 ≡ 6 (mod 12).Thus there are two solutions for the equation ax = b.

Alternatively, with groups, if a ◦ x = b, then x = a−1 ◦ b, so there is always one unique solution.

11. [Gruenig, S.] Find an integer n that shows that the rings Zn need not have the following propertiesthat the ring of integers has.

(a) a2 = a implies a = 0 or a = 1.

(b) ab = 0 implies a = 0 or b = 0.

(c) ab = ac and a 6= 0 imply b = c.

Consider the ring Z6.

(a) Let a = 3. Then 32 = 9 ≡ 3 (mod 6). So a2 = a, and a2 = a does not imply a = 0 or a = 1 forZ6.

(b) Let a = 2 and b = 3. Then 2 · 3 = 6 ≡ 0 (mod 6). So ab = 0 does not imply a = 0 or b = 0 forZ6.

(c) Let a = 3, b = 3, and c = 1. Consider ab and ac. Then ab = 3 · 3 = 9 ≡ 3 (mod 6), andbc = 3 · 1 = 3. So ab = ac, but b 6= c. Thus, ab = ac does not imply b = c for Z6.

Therefore, the integer 6 shows that the rings Zn need not have these properties.

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