Exercises (§1.9, p.62)

Basic Algebra (Solutions)

by Huah Chu

Exercises (§1.9, p.62)

1. Let G = (Q, +, O), K = Z. Show that G/K ' the group of complex numbers of the form e^2πiθ, θ ∈Q, under multiplication.

Proof. Define a homomorphism φ : G → {e^2πiθ|θ ∈ Q} by θ → e^2πiθ. Then ker φ = K

and φ is surjective. ¤

2. Show that a → a⁻¹ is an automorphism of a group G if and only if G is abelian, and if G is abelian, then a → a^k is an endomorphism for every kZ.

Proof. (1) φ : a → a⁻¹ is an automorphism

⇔ For all a, b ∈ G, (ab)⁻¹ = φ(ab) = φ(a)φ(b) = a⁻¹b⁻¹.

⇔ For all a, b ∈ G, ab = ba, that is, G is abelian.

(2) G is abelian. ¿From (ab)^k = a^kb^k, we have a → a^k is an endomorphism. ¤

3. Determine Aut G for (i) G an infinite cyclic group, (ii) a cyclic group of order six, (iii) for any finite cyclic group.

Sol. (i) Let G = hai be an infinite cyclic group. The generators of G are a and a⁻¹. Hence, for φ ∈ Aut G, φ(a) = a or a⁻¹. Hence Aut G = {1_G, φ : a → a⁻¹} 'Z/2Z.

(ii) Let G = ha|a⁶ = 1i. The generators of G are a and a⁵ by exercise 4, §1.5. hence Aut G = {1G, φ : a → a⁵} 'Z/2Z.

(iii) Let G = hai by any finite cyclic group with |G| = n. Then all generators of G are a^k, (k, n) = 1. Then Aut G is the set of all homomorphisms defined by φ : a → a^k, (k, n) = 1.

Remark. In the case of (iii), Aut G is isomorphic to the group of units of the multi- plicative monoid (Z/nZ, ·). Its structure will be determined in Chap. 4, §11. (Thm.

4.19, 4.20).

4. Determine Aut S3.

Sol. We shall show that Aut S₃ ' S₃.

Step 1. The elements of S3 are 1, a = (123), a² = (132), b = (12), ab = (13), (a²b = (23). Then we have the relation ba = a²b. Using this relation, the reader can verify that

(a^mbⁿ)(a^pb^q) = a^m+(n+1)pb^n+q, m, p = 0, 1, 2; n, q = 0, 1 (∗) easily. Since an automorphism preserves the order of an element, hence, for φ ∈ Aut G, φ(a) = aⁱ and φ(b) = a^jb for some i = 1, 2, j = 0, 1, 2.

Step 2. Define the map φ_ij : G → G by φ_ij :

½ a → aⁱ

b → a^jb , i = 1, 2, j = 0, 1, 2. Then φ_ij ∈ Aut G:

We have φij(a^m) = a^im, φij(a^mb) = a^im+jb by the definition of φij. Using these, it is easy to see that φ_ij is bijective. Then we check that φ_ij is a homomorphism in the following four cases:

(i) x = a^mb, y = aⁿb. Then φ((a^mb)(aⁿb)) = φ(a^m+2n) (by (∗)) = a^i(m+2n). On the other hand, φ(a^mb)φ(aⁿb) = a^im+jba^in+jb = aim+j+2(in+j) = a^im+2in+3j = a^i(m+2n). The other three cases: (ii) x = a^mb, y = aⁿ, (iii) x = a^m, y = aⁿb, and (iv) x = a^m, y = aⁿ are left to the reader.

Step 3. It is easy to see that φ₁₀ = 1, φ₁₁ and φ₁₂ have order 3. φ₂₀, φ₂₁ and φ₂₂ have order 2. We define the mapping Φ : S₃ → Aut S₃ by aⁱ 7→ φ_1i, aⁱb 7→ φ_2i. The reader can verify that it is an isomorphism.

Remark. (1) Since S₃ = ha, b|a³ = b² = 1, ba = a²bi (See §1.11), to prove that φ_ij is a homomorphism, it is enough to check that (φ_ij(a))³ = (φ_ij(b))² = 1, φ_ij(b)φ_ij(a) = (φ_ij(a))²(φ_ij(b)).

(2) In fact, Aut S_n ' S_n for all n 6= 6, and Aut S₆/S₆ ' Z/2Z, (c.f. I. J. Rotman:

The theory of groups, p.132, or B. Huppert Endlich Gruppen I, p.173–177).

(3) For other remark, see the remark after exercise 5.

5. Let a ∈ G, a group, and define the inner automorphism (or conjugation) Ia to be the map x → axa⁻¹ in G. Verify that I_a is an automorphism. Show that a → I_a is a homomorphism of G into Aut G with kernel the center C of G. Hence conclude that Inn G ≡ {Ia|a ∈ G} is a subgroup of Aut G with Inn G ' G/C. Verify that Inn G is a normal subgroup of Aut G. Aut G/ Inn G is called the group of outer automorphisms.

Proof. The last statement follows from φI_aφ⁻¹(b) = I_φ(a)(b). We leave all the verifica-

tions to the reader. ¤

Remark. A group G is complete in case C(G⁰) = 1 and Aut G ' G. Exercise 2 in

§1.4 and the remark in the above exercise show that S_n is complete for n 6= 2, 6.

It can be shown that if G is simple of composite order, then Aut(G) is complete.

6. Let G be a group, GL the set of left translations aL, a ∈ G. Show that GLAut G is a group of transformations of the set G and that this contains G_R. G_LAut G is called the holomorph of G and is denoted as Hol G. Show that if G is finite, then

| Hol G| = |G|| Aut G|.

Proof. (1) If g_L∈ G_L, φ ∈ Aut G, then φg_Lφ⁻¹= φ(g)_L. ¿From this fact, we can prove that G_LAut G is a group.

(2) Since g_L⁻¹gR(x) = g⁻¹xg = I_g⁻¹ ∈ Aut G, hence gR = gLI_g⁻¹. And GR ⊂ G_LAut G.

(3) To prove | Hol G| = |G|| Aut G|, it suffices to show that G_L∩Aut G = {1}. Since gL(1) = g 6= φ(1) for φ ∈ Aut G, g 6= 1, the result follows. ¤

7. Let G be a group such that Aut G = 1. Show that G is abelian and that every element of G satisfies the equation x² = 1. Show that if G is finite then |G| = 1 or 2.

Proof. (1) let G be a group with Aut G = 1. Then G/C ' Inn G = 1 where C is the center of G (by exercise 5). Hence G is abelian. If G is abelian, a → a⁻¹ is an automorphism (by exercise 2). The assumption Aut G = 1 implies that a = a⁻¹ for all a, that is, a² = 1.

(2) Suppose |G| is finite and G 6= 1.

Step 1. We prove that G contains elements a₁, . . . , a_r such that every element of G can be written in a unique way in the form a − 1^k1· · · a^k_r^r, k_i = 0, 1:

For this purpose, we show that, for all i, there exists a normal subgroup H = ha₁, . . . , a_ii of G such that every element of H can be written as a^k₁¹· · · a^k_r^r, k_i = 0, 1, uniquely. We prove this statement by induction on i. Note that any subgroup of G is normal since G is abelian.

Take any 1 6= a₁ ∈ G, then ha₁i is normal in G. Suppose we have H = ha₁i × · · · × ha_ii. Take any a_i+1 ∈ G − H. Then H ∩ ha_i+1i = 1 since |ha_i+1i| = 2. Because G is abelian any element of hH, a_i+1i can be written in the form hb with h ∈ H, b ∈ ha_i+1i.

Moreover, the expression is unique: If h₁b₁ = h₂b₂, then h⁻¹₂ h₁ = b₂b⁻¹₁ ∈ H ∩ha_i+1i = 1 and h₁ = h₂, b₁ = b₂. Hence the statement.

Step 2. Suppose n ≥ 2. Define the mapping α : G → G by a^k₁¹a^k₂²· · · a^k_nⁿ = a^k₁²a^k₂¹a^k₃³· · · a^k_nⁿ. Obviously, α is a nontrivial automorphism. This contradicts to the hypothesis Aut G = 1. Thus n = 1 and |G| = 2.

Remarks. (1) We reprove Step 2 in the language of vector space. In Step 1, we have shown that G is abelian and x² = 1 for all x. Regard G as an additive group, then G is a vector space over finite fieldZ/2Z (§4.13) and an automorphism is just a nonsingular linear transformation. Let {a₁, . . . , a_n} be a basis of G. Suppose dim G ≥ 2, then G

has a nontrivial nonsingular linear transformation a1 7→ a2, a2 7→ a1, and ai → ai, i > 2. A contradiction.

(2) When G is an infinite abelian group with x² = 1 for all x, we can still regard G as a vector space over Z/2Z. In this case, using Zorn’s lemma, we can find a base for G. Hence it is not difficult to construct a nontrivial nonsingular linear transformation on G.

8. Let α be the automorphism of a group G which fixes only the unit of G(α(a) = a ⇒ a = 1). Show that a → α(a)a⁻¹ is injective. Hence show that if G is finite, then every element of G has the form α(a)a⁻¹.

Proof. Let α be a fixed point free automorphism (α(a) = a ⇒ a = 1). Suppose α(a)a⁻¹ = α(b)b⁻¹. Then α(b⁻¹a) = b⁻¹a. Hence b⁻¹a is fixed by α and b⁻¹a = 1.

Thus a → α(a)a⁻¹ is injective.

If |G| < ∞, by the pigeon hole principle, the mapping is surjective. ¤ 9. Let G and α be as in 8, G finite, and assume α² = 1. Show that G is abelian of odd order.

Proof. (1) For any element g of G, g has the form α(a)a⁻¹. α(g) = α(α(a)a⁻¹) = α²(a)α(a⁻¹) = aα(a)⁻¹ = g⁻¹. Thus G is abelian by exercise 2.

(2) Next we show that |G| is odd. Suppose to the contrary, there is a ∈ G with order 2 (exercise 13, §1.2). Then α(a) = a⁻¹ = a, contradicts to the hypothesis about

α. ¤

Remark. An automorphism α of G is said to be fixed point free if it leaves only the unit fixed. This exercise shows that: if G admits a fixed point free automorphism of order 2, then G is abelian. Some further results are:

Suppose that G admits a fixed point free automorphism α of order n. (1) If n = 3, then G is nilpotent (for the definition, see Basic Algebra, I, p.243, exercise 6) and x commutes with α(x) for all x. (2) If n is a prime, then G is nilpotent (John G.

Thompson). (3) G is solvable in general (for the definition, see Basic Algebra, I, p.237).

For more details, we refer to D. Gorenstein: Finite groups, chap. 10, pp.333–357 and D. Gorenstein. Finite simple groups.

10. Let G be a finite group, α an automorphism of G, and set I = {g ∈ G|α(g) = g⁻¹}.

Suppose |I| > ³|G|. Show that G is abelian. If |I| = ³|G|, show that G has an abelian

Proof. (1) Let I = {g ∈ G|α(g) = g⁻¹} and |I| > ³₄|G|. For any h ∈ I, claim:

I ∩ h⁻¹I ⊂ C(h). In fact, if x ∈ I ∩ h⁻¹I, then x − h⁻¹g with g, x ∈ I. Now α(h⁻¹g) = (h⁻¹g)⁻¹ = g⁻¹h; on the other hand α(h⁻¹g) = α(h)⁻¹α(g) = hg⁻¹. Thus g⁻¹ ∈ C(h). It follows that g ∈ C(h) and x = h⁻¹g ∈ C(h) also.

Since |I| = |h⁻¹I| > ³₄|G|, so |I ∩ h⁻¹I| > ¹₂|G|. Thus C(h) is a subgroup of order

> ¹₂|G|. Then C(h) = G and h ∈ C(G), the center of G. Because this holds for any h ∈ I, so |C(G)| ≥ ³₄|G| and G = C(G), G is a abelian.

(2) Suppose |I| = ³₄|G|. Then G can not be abelian, otherwise, I is a subgroup of G.

Hence there exists h ∈ I − C(G). Let K = I ∩ h⁻¹I, then K = C(h) and |K| = ¹₂|G|, by the proof of (1). Since [G : K] = 2, K is normal. The only property remains to prove is that K is abelian.

For any k = h⁻¹g ∈ K = C(h), then g ∈ C(h). Thus for k₁ = h⁻¹g₁, k₂ = h⁻¹g₂ ∈ K, g1g2 ∈ C(h) ⊂ I. Then (g1g2)⁻¹ = φ(g1g2) = φ(g1)φ(g2) = g₁⁻¹g₂⁻¹and g1commutes

with g₂. So k₁ commutes with k₂. ¤

Remark. The reader is urged to find a finite non-abelian group G and its automor- phism α such that |I| = ³₄|G|. In fact let G = {±1, ±i, ±j, ±k} be the quaternion group and α the inner automorphism determined by i. Then |{g ∈ G : α(g) = g⁻¹}| = 6.