Solutions to Abstract Algebra (Dummit and Foote 3e)

Chapter 2 : Subgroups

Jason Rosendalejason.rosendale@gmail.com

February 11, 2012

This work was done as an undergraduate student: if you really don’t understand something in one of theseproofs, it is very possible that it doesn’t make sense because it’s wrong. Any questions or corrections can bedirected to jason.rosendale@gmail.com.

Exercise 2.1.1a

0 + 0i is the identity element of G, and a+ ai, b+ bi ∈ G implies −b+ (−b)i ∈ G with (a+ ai) + (−b+ (−b)i) =(a− b) + (a− b)i ∈ G.

Exercise 2.1.1b

1 is the identity element of G, and a, b ∈ G implies |ab−1| = |a||b−1| = |1||1| = 1 so that ab ∈ G.

Exercise 2.1.1c01 is the identity element of G, and a/b, p/q ∈ G implies b|n and q|n so that b = n/x and q = n/y for somex, y ∈ Z. Thus:

a

b− p

q=ax

n− py

n=ax− py

n

which, when reduced, becomes

=(ax− py)/ gcd(ax− py, n)

n/ gcd(ax− py, n)

The denominator of this fraction clearly divides n, and thus the fraction is ∈ G.

Exercise 2.1.1d01 is the identity element of G, and a/b, p/q ∈ G implies

a

b− p

q=aq − bpbq

And if b, q are relatively prime to n then bq is also relatively prime to n so the fraction is ∈ G.

Exercise 2.1.1e

1 is the identity element of G, and a, b ∈ G implies

(ab)2 = a2b2

which is a product of elements of Q and is thus an element of Q.

1

Exercise 2.1.2a

(1 2)(2 3) = (1 2 3), so the set is not closed if n ≥ 3.

Exercise 2.1.2b

(sr−1)(sr) = r2, so the set is not closed (unless r2 = 1, thus the requirement n ≥ 3).

Exercise 2.1.2c

Suppose n is composite. Then we can choose a, b such that ab = n with 1 < a < n. We’re told that there isan element x such that o(x) = n. If {x ∈ G| |x| = n} were a subgroup then the requirement of closure wouldrequire that all elements in {1, x, x2, . . . , xn−1} be elements of the subgroup. But o(xb) = a, so xb is not in theset and thus it is not closed under the operation.

Exercise 2.1.2d

The operation isn’t closed under addition (1+3 is not odd), and not closed under inverses for multiplication(3−1 = 1/3 6∈ Z).

Exercise 2.1.2e

The operation isn’t closed. (√

2 +√

3)2 = 5 + 2√

6.

Exercise 2.1.3a

The identity exists, and each element is its own inverse.

Exercise 2.1.3b

The identity exists, r2 is its own inverse, and (sr)−1 = sr3.

Exercise 2.1.4

Let G = Q−{0} under the operation of multiplication and let H = Z. H is closed under its operation but doesnot contain inverses.

Exercise 2.1.5

By Langrange’s theorem, |H| = n− 1 must divide n which is not possible when n > 2.

Exercise 2.1.6

Let T represent the torsion subgroup of G. Clearly 1 ∈ T . Suppose now that a, b ∈ T . There must be finite i, jsuch that ai = bj = 1. From this, we have:

7→ ai = bj = 1 assumed

→ (ai)j = (bj)i = 1 1i = 1j = 1

→ aij = bij = 1 algebraic substitution

→ aijbij = 1

→ (ab)ij = 1 from abelianism

To prove that abelianism is necessary, consider the group of bijective functions Z → Z. The torsion group Tcontains every element of finite order: in particular, it contains the following two bijective functions of order 2:

f(x) = x+ 1 if x is even , f(x) = x− 1 if x is odd

g(x) = x− 1 if x is even , f(x) = x+ 1 if x is odd

2

But the composite of these two functions is:

(f ◦ g)(x) = x− 2 if x is even , (f ◦ g)(x) = x+ 2 if x is odd

so that f ◦ g is of infinite order, so that f ◦ g 6∈ T . Thus T is not a group.

Exercise 2.1.7

Note that the operation on Z× (Z/nZ) must be addition in order for it to be a group. Every element of Z/nZis of finite order, and every nonzero element of Z is of infinite order. So the torsion group must be

T = {1} × {Z/nZ}

To show that the set of elements of infinite order together with the identity do not form a subgroup of this directproduct, we simply take the two elements 3× 0 and −2× 1. Their sum, 1× 1, is a nonidentity element of finiteorder.

Exercise 2.1.8

Let H ∪K be a subgroup. Proof by contradiction: Suppose neither H ⊆ K nor K ⊆ H. Then we could findsome a ∈ H −K and some b ∈ K −H. But ab 6∈ H, for this would imply a−1ab = b ∈ H; and ab 6∈ K, for thiswould imply abb−1 = a ∈ K. Thus ab 6∈ H ∪K, which contradicts our assumption that H ∪K was a group.

Exercise 2.1.9

To show that it’s a subgroup of GL, then we can just appeal to basic linear algebra and rely on the facts thatdet(I)=1 and det(AB)=det(A)det(B).

Exercise 2.1.10a

Both H and K contain the identity, so 1 ∈ H ∩ K. If a, b ∈ H ∩ K then a, b−1 ∈ H and a, b−1 ∈ K; thusab−1 ∈ H, ab−1 ∈ K; therefore ab−1 ∈ H ∩K.

Exercise 2.1.10b

Let Aα be an arbitrary, possibly non-countable collection of subgroups of G. All subgroups contain the identity,so 1 ∈

⋂αAα. If a, b ∈

⋂αAα then a, b−1 ∈ Aα for all α; thus ab−1 ∈ Aα for all α; therefore ab−1 ∈

⋂αAα.

Exercise 2.1.11a

This set contains the identity (1,1); and if the set contains (a1, 1) and (a2, 1) then it contains (a1 − a2, 1).

Exercise 2.1.11b

This set contains the identity (1,1); and if the set contains (1, b1) and (1, b2) then it contains (1, b1 − b2).

Exercise 2.1.11c

This set contains the identity (1,1); and if the set contains (a1, a1) and (a2, a2) then it contains (a1−a2, a1−a2).

Exercise 2.1.12a

This set contains the identity 1n = 1, and if it contains an, bn then it contains (b−1)n (from closure underinverses) (ab−1)n (from abelianism).

Exercise 2.1.12b

This set contains the identity 1 since 1n = 1, and if it contains a, b then an = 1, bn = 1 and (bn)−1 = 1, therefore(ab−1)n = 1 (from abelianism).

3

Exercise 2.1.13

Note: the weird requirement on H just guarantees us that every element of H has a multiplicative inverse eventhough H is only a group under addition. We prove that if H 6= {0} then H = Q: Suppose we have somenonzero element a

b ∈ H. From closure under addition we have bab = a ∈ H. From the existence of multi-plicative inverses we then have 1

a ∈ H, and therefore a 1a = 1 ∈ H. Now that we have 1 ∈ H, closure requires

n ∈ H for all n ∈ Z and the existence of multiplicative inverses requires 1n ∈ H for all n ∈ Q. And every rational

number pq is some combination of these elements, so Q ⊆ H. And since we were told that H ⊆ Q, we have H = Q.

Exercise 2.1.14

The set contains s and sr, but it does not contain ssr = r when n ≥ 3.

Exercise 2.1.15

For ease of notation, let this infinite union be represented by H. Since 1 ∈ H1, we know that 1 ∈ H. Nowsuppose a, b ∈ H. There must be some m,n such that a ∈ Hm and b ∈ Hn. So that for N = max(m,n) we havea, b, b−1 ∈ HN and thus ab−1 ∈ HN . This gives us ab−1 ∈ H.

Exercise 2.1.16

The identity matrix fulfills the criteria for the given set (let’s call it UTn). Proof by induction that UTn is asubgroup for all n ∈ N: UT1, the set of 1 × 1 matrices over F , is trivially a subgroup of GL1(F ). Assumethat UTk is a subgroup of GLk(F ): we prove that UTk+1 is also a subgroup. Let A,B be arbitrary elements ofUTk+1. The inverses are also in UTk+1:

A =

[a11 UTk0 ann

], A−1 =

[(a11)−1 −UTk0 (ann)−1

]And the product AB−1 is in UTk+1:

AB−1 =

[a11 UT1k0 ann

] [(b11)−1 −UT2k0 (bnn)−1

]=

[a11(b11)−1 UT1k − UT2k0 ann(bnn)−1

]so that UTk+1 is a group. By induction, UTn is a subgroup of GLn(F ) for all n.

Exercise 2.1.17

This proof is almost identical to that of the preceeding exercise. The identity matrix fulfills the criteria for thegiven set (let’s call it ULn). Proof by induction that ULn is a subgroup for all n ∈ N: UL1, 1× 1 matrice overF , is trivially a subgroup of GL1(F ). Assume that ULk is a subgroup of GLk(F ): we prove that ULk+1 is alsoa subgroup. Let A,B be arbitrary elements of ULk+1. The inverses are also in ULk+1:

A =

[1 ULk0 1

], A−1 =

[1 −ULk0 1

]And the product AB−1 is in ULk+1:

AB−1 =

[1 UL1k0 1

] [1 −UL2k0 1

]=

[1 UL1k − UL2k0 1

]so that ULk+1 is a group. By induction, ULn is a subgroup of GLn(F ) for all n.

Exercise 2.2.1

CG(A) is a group, so it contains g iff it contains g−1, so the two sets contain the same elements. More formally:choose g ∈ CG(A). The following steps are all bidirectional (iff):

4

7→ g ∈ CG(A) assumed

↔ g−1 ∈ CG(A) the group CG(A) contains inverses

↔ (∀a ∈ A)g−1ag = a definition of membership in CG(A)

↔ g ∈ {g ∈ G|g−1ag = a for all a ∈ A}

Exercise 2.2.2

Members of Z(G) commute with every g ∈ G, and CG(Z) is the set of elements that commute with elements ofZ(G): that is, G itself. More formally:

7→ g ∈ CG(Z(G)) assumed

↔ gag−1 = a for all a ∈ Z(G) def. of CG

↔ gag−1 = a for all a such that aga−1 = g for all g ∈ G def. of Z

↔ ga = ag for all a such that ag = ga for all g ∈ G right-multiplication

↔ g ∈ G last statement is true for all g ∈ G

Exercise 2.2.3

We know that centralizers are groups, so we need only prove that CG(B) ⊆ CG(A).

7→ g ∈ CG(B) assumed

↔ gbg−1 = b for all b ∈ B def. of CG

↔ gbg−1 = b for all b ∈ A A ⊆ B↔ g ∈ CG(A) def. of CG

Exercise 2.2.6a

Choose h ∈ H. We will prove that h ∈ NG(H) by proving that hHh−1 = H. From the closure of H, we know thathHh−1 ⊆ H. To prove that H ⊆ hHh−1 ⊆ H, we choose h2 ∈ H and note that h(h−1h2h)h−1 = h2 ∈ hHh−1.Thus hHh−1 = H and so h ∈ NG(H).

We’re also asked to show that (a) does not hold if H is not a subgroup. I find this a bit confusing, since theclaim is that H is a subgroup. I’ll assume that we’re really being asked to show that H is necessarily a subsetof NG(H) if H is not a subgroup. Consider the group D8 with H = {r, s}. Only the identity element commuteswith every element of H, so NG(H) = {1}.

Exercise 2.2.6b

This is true practically by definition. By definition, CG(H) is the set of elements that commute with everyelement of H. So to say H ≤ CG(H) is to say that every element of H commutes with every element of H.More formally:

7→ H is abelian assumed

↔ (∀h1, h2 ∈ H)h1h2 = h2h1 def. of abelian

↔ (∀h1, h2 ∈ H)h1 = h2h1h−12 right-multiplication

↔ (∀h1 ∈ H)h1 ∈ CG(H) def. of CG(H)

↔ H ≤ CG(H) def. of ≤Each step was bidirectional (iff), so we have proven that H is abelian iff H ≤ CG(H).

5

Exercise 2.2.7

Part (a) is a direct consequence of exercise 1.2.5. Part (b) is a direct consequence of exercise 1.2.4.

Exercise 2.2.8

The text of this chapter contains a proof that stabilizers of G are subgroups of G. We could also proceed bydefining S = {i} and noting that the kernel of the action of G on S is precisely Gi; and the kernel of an actionis a subgroup.

Exercise 2.2.9

To show NH(A) ⊆ NG(A) ∩H, choose an arbitrary element h ∈ H:

7→ h ∈ H ∧ h ∈ NH(A) assumed

↔ h ∈ H ∧ (∀a ∈ A)hAh−1 = A def. of NH(A)

↔ h ∈ H ∧ h ∈ NG(A) def. of NG(A)

↔ h ∈ H ∩ h ∈ NG(A)

Each step was bidirectional, so the two sets are equal.

Exercise 2.2.10

H must contain the identity element and one nonidentity element h: that is, H = {1, h}. We proceed with abidirectional proof:

7→ g ∈ NG(H) assumed

↔ (∀h ∈ H)gHg−1 = H def of NG(H)

↔ g1g−1 = 1 and ghg−1 = h see below for justification

↔ (∀h ∈ H)ghg−1 = h H = {1, h}↔ g ∈ CG(H) def. of CG(H)

The middle step is justified in the downward (’only if’) direction by noting that we clearly have g1g−1 = 1, sothat if we require gHg−1 = H then it’s necessary that ghg−1 = h. The step is justified in the upward (’if’)direction by noting that {1, h} is a complete list of the elements of H.

For the final deduction, we note that if CG(H) = G this means each h ∈ H commutes with every elementof G. And, by definition, Z(G) is the set of all elements that commute with every element of G. ThereforeH ⊆ Z(G).

Exercise 2.2.11

Choose some z ∈ Z(G). Then zg = gz for all g ∈ G. Therefore zgz−1 = g for all g ∈ G, which means zAz−1 = Afor any set A ⊆ G. By definition, this means z ∈ NG(A).

Exercise 2.2.12b

The identity action of S4 would be the identity permutation (1). It’s clear from the nature of permutations that(σ1 ◦ σ2) · p = σ1 · (σ2 · p).

Exercise 2.2.12c

An element of S4 stabilizes 4 iff its disjoint cycles contain the 1-cycle (4). Thus we can construct the ismorphism

ϕ : S3 7→ S4 ϕ(σ) = σ ◦ (4)

6

Exercise 2.2.12d

The permutations that stabilize x1 + x2 are those that either stabilize both 1 and 2 or swap the elements of 1and 2:

(1), (1 2), (1 2)(3 4), (3 4)

Showing that this is an abelian group is trivial.

Exercise 2.2.12e

The permutations that stabilize x1x2 +x3x4 are the above permutations, plus the permutations that swap both1,3 and 2,4 or 1,4 and 2,3:

(1 3)(2 4), (1 4)(2 3), (1 3 2 4), (2 3 1 4)

To prove isomorphism, let R = (1324) and S = (1 3)(2 4). We obvious have o(R) = 4, o(S) = 2. And to verifyRS = SR−1:

RS = (1 3 2 4)(1 3)(2 4) = (1 2) = (1 3)(2 4)(2 3 1 4) = SR−1

We need now only prove that ϕ(SiRiSjRj) = sirisjrj is a homomorphism from this set of permutations to D8.And this follows immediately from the associativity of D8:

ϕ(SiRiSjRj) = sirisjrj = (siri)(sjrj) = ϕ(SiRi)ϕ(SjRj)

Exercise 2.2.12f

The same logic applies: the permutations that stabilize (x1 + x2)(x3 + x4) are those permutations that swapelements in exactly the same way as in part (e).

Exercise 2.2.13

The proof is identical to that given for n = in the preceeding exercise.

Exercise 2.2.14

It’s not specified if we’re looking for the center of H(F ) in H(F ) or the center of H(F ) is GL3(F ). We’ll assumethe more general case for now and look for the center of H(F ) in GL3(F ).

H(F ) is just the set of 3 × 3 upper-triangular matrices with diagonal elements of 1. The center, Z(H(F )),consists of the elements A ∈ GL3(F ) such that AH = HA for all H ∈ H(F ). Calculating AH and HA:

AH =

a b cd e fg h i

1 x y0 1 z0 0 1

=

a ax+ b ay + bz + cd dx+ e dy + ez + fg gx+ h gy + hz + i

HA =

1 x y0 1 z0 0 1

a b cd e fg h i

=

a+ dx+ gy b+ ex+ hy c+ fx+ iyd+ gz e+ hz f + izg h i

We need to see when these two matrices are equal for all H, so we’re solving for A. We see immediately thatwe must have d = g = h = 0, so we simplify to

AH =

a ax+ b ay + bz + c0 e ez + f0 0 i

, HA =

a ex+ b iy + fx+ c0 e iz + f0 0 i

So that we must have a = e = i. Further simplifying:

AH =

a ax+ b ay + bz + c0 a az + f0 0 a

, HA =

a ax+ b ay + fx+ c0 a az + f0 0 a

7

So that our final requirement is that bz = fx. The only way for this to hold for all values of x, z is for b = f = 0.Therefore, the center of H(F ) is the set of matrices of the form

A =

a 0 c0 a 00 0 a

Now we see that the exercise must have been asking us to find the center of H(F ) in H(F ), for there is notnecessarily any isomorphism between F and the set of all such Z(H(F )) (consider F = Z2: |F | = 2 but|Z(H(F )))| = 4). Restricting our center to H(F ) gives us

A =

1 0 c0 1 00 0 1

which has an obvious isomorphism with F :

ϕ : F → H(F ), ϕ(x) =

1 0 x0 1 00 0 1

Exercise 2.3.2

If o(x) = n, then x0, x1, . . . , xn−1 are distinct by exercise 1.1.32. Thus | 〈x〉 | = |G| and 〈x〉 ⊆ G, and they’reboth finite sets so they must contain the same elements. To show that this does not hold for infinite groups, letG = Z and let x = 2.

Exercise 2.3.3

Any n such that gcd(n, 48) = 1 (i.e., numbers with no factors of 2 or 3).

Exercise 2.3.4

Any n such that gcd(n, 202) = 1 (i.e., numbers with no factors of 2 or 101).

Exercise 2.3.5

It’s tempting to resort to Euler’s Phi function and say the answer is ϕ(49000). To find this number explicitly,we first find how many integers have factors in common with 49000. Using the theorem from set theory that|A ∪B ∪ C| = |A|+ |B|+ |C| − |A ∩B| − |A ∩ C| − |B ∩ C|+ |A ∩B ∩ C|, we have

(factors of 2 + factors of 5 + factors of 7)− (factors of 2,5 + factors of 5,7 + factors of 2,7) + (factors of 2,5,70)

= (24500 + 9800 + 7000)− (4900 + 1400 + 3500) + (700) = 32200

So that ϕ(49000) = 4900− 32200 = 16800.

Exercise 2.3.8

Any a such that gcd(a, 48) = 1 (i.e., numbers with no factors of 2 or 3).

Exercise 2.3.9

We’ll use the additive notation instead of the multiplicative notation and find the integers a such that ψa : 1 7→ ax(i.e., ψa(k) = kax) extends to a well-defined homomorphism. We’ll see that ϕa is well-defined only iff 3|a. Weprove that well-definedness requires this. Suppose psia is a well-defined homomorphism. Choose i = j ∈ Z:

8

7→ i = j given: equality in equality in Z/48Z→ i = j + 48 equality in Z→ ψa(i) = ψa(j + 48) ψa is well-defined: equality in Z36

→ ψa(i) = ψa(j + 48) + 36n ψa is well-defined: equality in Z→ ψa(i)− ψa(j) = ψa(12) + ψa(36) + 36n) ψ is a homomorphism: equality in Z

the operations on the groups Z and Z/48Z are addition, so xa is repeated addition. Thusψ(kx) = kψ(x) :

→ ψa(i)− ψa(j) = 12ψa(1) + 36(ψa(1) + n) ψ is a homomorphism: equality in Z

Moving to Z36, the LHS of this equation is zero because ψ is well defined and 36k = 0 in Z36:

→ 12ψa(1) = 0 ψa is well-defined: equality in Z36

→ (∃n ∈ Z)12ψa(1) = 36n equality in Z→ (∃n ∈ Z)psia(1) = 3n equality in Z→ 3|ψa(1) def. of divisibility

→ 3|a def. of ψa

This really just shows that ψ is a well-defined homomorphism only if 3|a. To prove the if part, we assumea = 3k and define ψa : 1 7→ 3kx. We need only prove well-definedness, since homomorphism follows trivially.

7→ i = j assumed: equality in equality in Z/48Z→ i = j + 48n equality in Z→ ψa(i)− ψa(j + 48n) = 3kxi− 3kx(j + 48) def of ψa: equality in Z36

→ ψa(i)− ψa(j + 48n) = 3kxi− 3kx(j + 48) + 36m equality in Z36

→ ψa(i)− ψa(j + 48n) = 3kx(i− j) + 3(48)kx+ 36m algebraic rearrangment: equality in Z→ ψa(i)− ψa(j + 48n) = 3kx(i− j) + 36(4kx+m) algebraic rearrangment: equality in Z→ ψa(i)− ψa(j + 48n) = 3kx(i− j) equality in Z36

→ ψa(i)− ψa(j + 48n) = ψa(i− j) def. of ψa: equality in Z36

→ ψa(i)− ψa(j) = ψa(0) in Z/48Z we have j = j + 48 and i = j: equality inZ36

→ ψa(i)− ψa(j) = 0 def of ψa: equality in Z36

→ ψa(i) = ψa(j) algebraic rearrangement: equality in Z36

Exercise 2.3.10

The order islcm(30, 54)/30 = 54/ gcd(30, 54) = 9

Exercise 2.3.11

A subgroup that is not cyclic is {1, r2, s, sr2}. Each element has order 1 or 2.

Exercise 2.3.12a

We can prove this by enumeration. There are only four elements of Z2 × Z2, each of which has order 1 or 2.

Exercise 2.3.12b

Proof by contradiction. Suppose the element (a, b) generated Z2 × Z2. Then (a+ 1, b) cannot be an element of〈(a, b)〉 for this would imply that n(a, b) = (a+1, b) for some n which would mean that n = 1 and thus a = a+1,which is not true for any element of Z2.

9

Exercise 2.3.12c

Proof by contradiction. Suppose the element (a, b) generated Z × Z. Then (a + 1, b) cannot be an element of〈(a, b)〉 for this would imply that n(a, b) = (a+1, b) for some n which would mean that n = 1 and thus a = a+1,which is not true for any element of Z.

Exercise 2.3.13

For parts (a) and (b), the group on the left-hand side contains an element (0, 1) of order 2 while the right-handside contains no element of order 2.

Exercise 2.3.15

Proof by contradiction. Suppose the element (a, b) generated Q × Q. Then (a + 1, b) cannot be an element of〈(a, b)〉 for this would imply that n(a, b) = (a+1, b) for some n which would mean that n = 1 and thus a = a+1,which is not true for any element of Q.

Exercise 2.3.16

We’re told that o(x) = n, o(y) = m. Let A = lcm(m,n). From the relation mn = gcd(m,n)lcm(m,n) we have

(xy)A

= xAyA from commutativity

= (xm)A/m(yn)A/n A/m,A/n are both integers

= 1

Thus o(xy)|A. This does not necessarily hold if x, y do not commute (consider D8 with o(r) = 4, o(s) = 2, buto(sr) 6= 4). The easiest example of commuting elements such that o(xy) is not equal to lcm(x, y) is to take D8

with x = y = r. We have o(rr) = 2 while lcm(o(r), o(r)) = lcm(4, 4) = 4.

Exercise 2.3.17

〈x|xn = 1〉

Exercise 2.3.18

Choose x ∈ Zn such that 〈x〉 = Zn. Define ϕ : Zn → H as ϕ : xk → hk. To show that this is a homomorphismis trivial. To show uniqueness, assume we had another homomorphism f with f(x) = h. Then we have, fromhomomorphism:

f(xk) = f(x)k = hk = ϕ(xk)

so that f = ϕ.

Exercise 2.3.19

Define ϕ : Z→ H as ϕ(n) = hn. This is a homomorphism:

ϕ(a+ b) = ha+b = hahb = ϕ(a)ϕ(b)

And this function is unique: if f is any homomorphism with f(1) = h, we have (using multiplicative notationso that 1n = n):

f(n) = f(1n) = f(1)n = hn = ϕ(n)

Exercise 2.3.20

If xpn

= 1, then we know that o(x)|pn. But p is prime, so the only divisors of pn are p0, p1, p2, . . . , pn. Thuso(x) = pm for some m ≤ n.

10

Exercise 2.3.21

From the binomial theorem we have

(1 + p)pn−1

=

pn−1∑k=0

(n

k

)pk1n−k =

pn−1∑k=0

(n

k

)pk

We want to find the value of this term mod pn:

pn−1∑k=0

(pn−1

k

)pk mod pn

Any terms of pk with k ≥ n are going to become zero here, so this is equivalent to

=

n∑k=0

(pn−1

k

)pk mod pn

The coefficient(pn−1

k

)will be a multiple of pn−1 whenever pn−1 > k, which can (with a bit of algebra) be shown

to hold for all n, k whenever p > 2. Thus every term in this sum except the k = 0 term has a factor of pn andwe can simplify:

=

0∑k=0

(pn−1

k

)pk = 1 mod pn

It’s not enough to show that this doesn’t always hold for pn−2; if we want to make the deduction they’re askingus to make, it must *never* hold.

Exercise 2.3.24a

If g ∈ NG(〈x〉), then g 〈x〉 g−1 = 〈x〉, which means that gxg−1 ∈ 〈x〉, and therefore gxg−1 = xa for some a ∈ Z.

Exercise 2.3.24b

Note that(gxg−1)n = g(xg−1g)n−1xg−1 = gxng−1

So suppose gxg−1 = xa. Then we have gxng−1 = (gxg−1)n = (xa)n. This is sufficient to prove that g 〈x〉 g−1 ⊆〈x〉. Since G is a finite group, o(x) = n is finite. For i, j = 0, 1, . . . , n− 1 we have

⇐⇒ gxig−1 = gxjg−1 ⇐⇒ gxig−1gx−jg−1 = 1 ⇐⇒ xi−j = g−1g = 1 ⇐⇒ i = j

So that gxig−1 is distinct for each i = 0, 1, . . . , n − 1. Thus |g 〈x〉 g−1| = | 〈x〉 |. Since these are finite subsetswith g 〈x〉 g−1 ⊆ 〈x〉, this proves that g 〈x〉 g−1 = 〈x〉.

Exercise 2.3.25a

Suppose that G = 〈x〉 with |G| = o(x) = n. Let k be an integer relatively prime to n. We prove that o(xk) = n:

7→ o(xk) = i G is finite, so this must be true for some i > 0

→ (xk)i = 1

→ xki = 1

→ o(x)|ki→ n|ki o(x) = n

→ n|i k is relatively prime to n

The least i > 0 such that n|i is obviously i = n, so o(xk) = n. Thus the size of the range of this map is|⟨xk⟩| = n = |G|, which for finite groups is sufficient to prove

⟨xk⟩

= G.

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Exercise 2.3.25b

Let G be a finite group with |G| = n (not necessarily cyclic). Choose k to be relatively prime to n. Define themapping f : G→ G as f(g) = gk. We’re asked to show that this mapping is surjective.Choose an arbitrary element g ∈ G. Each element has finite order, so let o(g) = m. Since m|n (by Langragne’stheorem), we know that k is relatively prime to m. Thus, applying the previous exercise, f(〈g〉) = 〈g〉 and thusg is in the range of f . But g was arbitrary, so every g ∈ G is in the range of f . And thus f is surjective.

Exercise 2.3.26a

We proved this in exercise 25. We proved the ’if’ portion explicitly, and we were assured that the ’only if’portion is a direct consequence of Cauchy’s Theorem which will be presented in section 3.2.

Exercise 2.3.26b

σa(x) = σb(x) ⇐⇒ xa = xb ⇐⇒ xa−b = 1 ⇐⇒ (a− b)|n ⇐⇒ a = b(mod n)

Exercise 2.3.26c

Let Zn = 〈x〉. Suppose ϕ : Zn → Zn is an automorphism. Every element of Zn is expressible as xi for some i,so ϕ(x) = xi for some i. And this completely defines ϕ, since for each xj ∈ 〈x〉 we have ϕ(xj) = ϕ(x)j = (xi)xj .

Exercise 2.3.26d

σa ◦ σb(x) = σa(σb(x)) = σa(xb) = xab = σab(x)

Exercise 2.4.1

〈H〉 is defined to be the intersection ∩Ai of all subgroups Ai such that H ⊆ Ai ≤ G. Since H ∈ {Ai} we have∩Ai ⊆ H; since H ⊆ Ai for each Ai, we have H ⊆ ∩Ai. Thus H = ∩Ai = 〈H〉.

Exercise 2.4.2

If x ∈ 〈A〉 then x is the finite product of elements of A; since A ⊆ B, this means that x is the finite product ofelements of B; thus x ∈ 〈B〉. To see that we can have A ⊂ B with 〈A〉 = 〈B〉, consider the additive group Zwith A = {1} and B = {1, 2}.

Exercise 2.4.3

Each element of H commutes with every element of H (by abelianism) and every element of Z(G) (by definition).In turn, every element of Z(G) commutes with every element of H (since H ⊂ G) and every element of Z(G)(since Z(G) ⊂ G). This means that all the elements from H,Z(G) commute with one another. Thus 〈H,Z(G)〉is abelian.However, the elements of CG(H) do not necessarily commute with other elements of CG(H). Consider G = GLn,H = In. H is just the identity element, so it’s trivially abelian. Every element commutes with the identityelement, so we have CG(H) = G. Thus 〈H,CG(H)〉 = G. But we know that G is not abelian.

Exercise 2.4.4

First, suppose H = {1}. Then 〈H − {1}〉 = 〈∅〉. This represents the intersection of every subgroup A such that∅ ⊆ A ≤ G. But the empty set is a subset of every set, so 〈∅〉 is the intersection of every subgroup of G; theonly commonality is 1. Thus 〈H − {1}〉 = {1} = 〈H〉.Now suppose H contains some nonidentity element h. Then 1 = hh−1 ∈ 〈H − {1}〉, so that 〈H − {1}〉 = 〈H〉 =H.

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Exercise 2.4.5

The only elements of order 2 in S3 are the two 2-cycles (c.f. exercise 1.4.13). Without loss of generality, we canwrite these as (a b) and (b c). It’s trivial (but tedious) to construct six distinct elements from finite products ofthese two elements.

Exercise 2.4.6

This is equivalent to 〈(1 2), (3 4)〉. From exercise 1.4.13, we know that any product of commuting 2-cycles hasorder 2.

Exercise 2.4.7

We let R = AB = (1 3 2 4) and S = A = (1 2). Define the isomorphism ϕ : D8 → S4 as r 7→ R, s 7→ S. Weneed only show that the relations from the generator 〈o(r) = 4, o(s) = 2, sr = r−1s〉 hold in S4. And this isdone easily, since o(R) = 4, o(S) = 2, and SR = R−1S = B.

Exercise 2.4.8

Let A = (1 2 3 4), B = (1 2 4 3). Then 〈A,B〉 contains A−1 = (1 4 3 2) and A−1BA−1 = (1 2). On page 64,we’re assured that the two elements A, (1 2) generate all of S4 (although we’re told that the proof will come in alater chapter). If this proof isn’t convincing, we can always hunker down and generate the 24 distinct elementsof S4 from finite products of A,B.

Exercise 2.4.9

In a previous exercise, we’ve shown that |GL2(Fp)| = p4−p3−p2 +p. In this case, that gives us |GL2(F3)| = 48.Half of these have determinant 1, and half have determinant 2. Thus |SL2(F3)| = 24. We haven’t encounteredany groups that are isomorphic to this group, so I don’t see any way to prove that these two elements generateSL2(F3) other than actually generating all 24 distinct elements.

Exercise 2.4.10

In exercise 1.5.3, we developed a generator for Q8:

Q8 = 〈1, i, j, k|i2 = j2 = k2 = −1, ij = k〉

Let the two given elements of SL2(F3) be represented by I, J and define the following:

I =

[0 -11 0

], J =

[1 11 -1

], K = IJ

[-1 11 1

], 1 =

[1 00 1

]Remember that in F3 we have 2 = −1. It’s easy to verify that o(i) = o(j) = o(k) = 4, and that i2 = j2 = k2 = −1.From section 1.6, this mapping between the generators of Q8 and the generators of whatever we’re calling thissubset of SL2(F3) is sufficient to prove isomorphism.

Exercise 2.4.11

The subgroup described above has only one element of order 2, while S4 has more than one (e.g., (1 2) and(3 4)). Thus the two groups cannot be isomorphic by exercise 1.7.2.

Exercise 2.4.12

This isn’t true unless we restrict ourselves to upper-triangular matrices with 1s along the diagonal (cf exercise2.1.17). In this case, all such matrices are of the form 1 a b

0 1 c0 0 1

13

where a, b, c ∈ {0, 1}. There are clearly 23 = 8 such matrices. Now, define the matrices R,S to be

R =

1 1 00 1 10 0 1

, S =

1 0 00 1 10 0 1

then we have o(R) = 4, o(S) = 2. Furthermore, we have

R−1 = R3 =

1 1 10 1 10 0 1

, SR =

1 1 00 1 00 0 1

= R−1S

at which point we’ve reproduced the generator for D8.

Exercise 2.4.13

Let P = {1/p|p is prime }. Clearly 〈P 〉 ⊆ Q+. Now, choose r/q ∈ Q+. From the fundamental theorem ofalgebra, we know that q is the finite product of prime powers q =

∏pnii . Therefore we have

r

q= r

1∏pnii

= r∏ 1

pnii

∈ 〈P 〉

so that Q+ ⊆ 〈P 〉.

Exercise 2.4.14a

Every finite G is generated by itself: G = 〈G〉.

Exercise 2.4.14b

The additive group Z is generated by 〈1〉.

Exercise 2.4.14c

Let H be a finitely generated subgroup and let A be the finite set of generators for H such that 〈A〉 = H. Definek, 1/k as described in the exercise. It’s clear that 〈1/k〉 generates each element in A and therefore generates allof H. Thus H ⊆ 〈1/k〉.

Exercise 2.4.14d

If Q were finitely generated, then it would itself be a finitely generated subgroup of Q, and would therefore becyclic by part (c). But it’s not cyclic. By contrapositive, then, Q can not be finitely generated.

Exercise 2.4.15

Consider the additive group of the set of powers of 2:{2i|i ∈ Z

}This set cannot be generated by

⟨2k⟩, because the set generated by this element will never yield 2k−1 (remember:

operation is addition).

Exercise 2.4.16a

If |G| is finite, then there are only (|G|− |H|) elements in G−H, and therefore there are precisely (2|G|−|H|− 1)proper subsets of G containing H. If none of these finitely many subsets is a maximal group, then that meansthat H was its own maximal group in G.

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Exercise 2.4.16b

By Lagrange’s theorem, the order of any subgroup of D8 must divide 8. So the only subgroup of D8 with order> 4 must be D8 itself.

Exercise 2.4.16c

Suppose H ≤ G = 〈x〉 and |G| = o(x) = n. Each element of G, and thus H, is of the form xi for somei ∈ {0, . . . , n − 1}. To satisfy group closure, we must have H = 〈xa〉 where a = gcd({i|xi ∈ H}). We nowconsider three cases:

1. If (a, n) = 1 then H = G and H is not maximal (c.f. proposition 6 in section 2.3).

2. If a|n and a = m1m2 is composite we see that

H = 〈xa〉 ≤ 〈xm1〉 ≤ G

so that H is not maximal.

3. If a|n and a = p is prime, then 〈xp〉 is maximal. For suppose we have a group H ′ containing H and someadditional element xi so that H ′ =

⟨xp, xi

⟩. If p|i then xi is already in H and thus H = H ′. If p does not

divide i, then the primality of p means (p, i) = 1 and thus H ′ = G. In either case, no proper subgroup ofG contains H as a proper subgroup, so H is maximal.

We don’t need to consider the case where (a, n) > 1 but a 6 |n, since in this case we would have x(a,n) ∈ 〈xa〉 sothat 〈xa〉 =

⟨x(a, n)

⟩and clearly (a, n)|n. Therefore these three cases exhaust all possibilities and prove that H

is maximal iff H = 〈xp〉 for some prime p.

Exercise 2.4.17a

By the subgroup criterion, we need only show that for each a, b ∈ H we also have ab−1 ∈ H. Define H to bethe union of chain elements H1, H2, . . . so that H =

⋃Hi. Then we have:

a, b ∈ H → (∃i, j ∈ N)a ∈ Hi, b ∈ Hj → (∃j ∈ N)ab−1 ∈ Hj → ab−1 ∈ H

Exercise 2.4.17b

Let {g1, g2, . . . , gn} represent the finite generators of G. From our definition of S, each element in chain C mustbe a proper subgroup of G. So if H = G then for each gi we could find ki ∈ N such that gi ∈ Hki . Then, sincethis set is finite, we could find the maximum of {ki}: call it k. Thus each gi ∈ Hk, and so 〈g1, g2, . . . , gn〉 ≤ Hk,thus G = Hk. And this contradicts our claim that each Hi is a proper subgroup of G.

Exercise 2.4.17c

We’re pretty much done. We have a bunch of chains in S, all of which have an upper bound (by exercise2.4.16(a)); thus by Zorn’s Lemma S contains a maximal element.

Exercise 2.4.18a

Fix p prime, and let Z = {z ∈ C|zpn = 1 for some n ∈ Z+}. Define Hi = {z ∈ Z|zpi = 1}. We have abidirectional proof showing that Hk ≤ Hm ⇐⇒ m ≥ k.

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7→ Hk ≤ Hm assumed

↔ z ∈ Hk → z ∈ Hm

↔ zpk

= 1→ zpm

= 1

↔ (zpk

)a = zpm

, a > 0 (note: a ∈ N

To justify the claim that a > 0, we note that for zpk

= 1 to always imply zpm

= 1 we must have

zpm

be a positive integer power of zpk

.

↔ apk = pm, a > 0

To justify this last step in the → direction, note that we must have a = b if we want za = zb for allz ∈ Z.

↔ a > 0, a = pm−k

↔ m ≥ k

Exercise 2.4.18b

From the given assumption, we can immediately conclude that⟨e2πim/p

n⟩

= Hk

Exercise 2.4.18c

Let Y < Z. For each y ∈ Y , we have some smallest n such that ypn

= 1 so that y = exp(2πim/pn) for somem. Our choice of the smallest such n allows us to know that (m, p) = 1 (for otherwise (m, p) = pk and we have

y = exp(2πi(mp−k)/pn−k) so that ypn−k

= 1). From part (b), this means that 〈y〉 contains n distinct elementsincluding exp(2πi/pn). The closure of Y then assures us that exp(2πi/pn) ∈ Y .

Now now consider the set of all such exponentials generated by elements y ∈ Y , and further consider the setof powers of p in this set: {

n ∈ N |exp(

2πi

pn

)∈ Y

}For every n in this set, we have Hn ⊆ Y . Thus, if this set has no upper bound, then Y =

⋃∞n=1Hk = Z and

Y is not a proper subgroup. If this set does have an upper bound, then it has a least upper bound N and thusY =

⋃n=1N − 1Hk = HN−1. So we have proven that Y is a proper subgroup iff Y = Hk for some k ∈ Z+.

Exercise 2.4.18d

Let gi1 , . . . , gin be an arbitrary finite set of elements of Z. We prove that this cannot be a generator of Z. Usingthe logic of the previous exercise, for each gi we can find a subgroup Hj such that gi ∈ Hj . By exercise 2.4.2this implies 〈gi1〉 ≤ 〈Hj1〉. Thus by part (a) we have

〈gi1 , . . . , gin〉 ≤ 〈Hj1 , . . . ,Hjn〉 = Hj , j = max(j1, . . . , jn)

And clearly Hj 6= Z, so Z was not generated by this finite set.

Exercise 2.4.19a

For each p/q ∈ Q and each integer k we have

p

qk∈ Q, k

(p

qk

)=p

q

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Exercise 2.4.19b

Let A be a nontrivial finite abelian group. Every element in a finite group has a finite order. So define k to bethe finite product

k =∏ai∈A

o(ai)

Since A is nontrivial, we can find some nonidentity element b ∈ A. But this element has no k-th root: by ourdefinition of k we see that o(a)|k for all a ∈ A and thus ak = 1 6= b for all a ∈ A.

Exercise 2.4.20

First, assume that A × B is divisible. Then all of the elements of the form (1A, b) and (a, 1B) are divisible,which can only happen if A and B are both divisible. Next, assume that A and B are divisible: let (a, b) be anarbitrary element of A×B and let k be an arbitrary element of Z+. By divisibility we have a1/k ∈ A, b1/k ∈ Band thus

(a1/k, b1/k) = (a, b)1/k ∈ A×B

Exercise 2.5.1 through 2.5.20

These exercises would require way too much time to typeset and I’m kind of eager to move on to quotient groups.You’re on your own.

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