Exercises (§1.2, pp.35–36)

Basic Algebra (Solutions)

by Huah Chu

Exercises (§1.2, pp.35–36)

1. Determine αβ, βα and α⁻¹ in S₅ if α =

µ 1 2 3 4 5 2 3 1 5 4

¶

, β =

µ 1 2 3 4 5 1 3 4 5 2

¶

Ans. αβ =

µ 1 2 3 4 5 2 1 5 4 3

¶

, βα =

µ 1 2 3 4 5 3 4 1 2 5

¶

, α⁻¹ =

µ 1 2 3 4 5 3 1 2 5 4

¶ .

2. Verify that the permutations 1 =

µ 1 2 3 1 2 3

¶ ,

µ 1 2 3 2 3 1

¶ ,

µ 1 2 3 3 1 2

¶

form a subgroup of S₃.

Proof. Omitted. ¤

3. Determine a multiplication table for S₃. Ans. Let 1 =

µ 1 2 3 1 2 3

¶ , a =

µ 1 2 3 1 3 2

¶ , b =

µ 1 2 3 2 1 3

¶ , c =

µ 1 2 3 2 3 1

¶ , e =

µ 1 2 3 3 2 1

¶

. The multiplication table is

1 a b c d e 1 1 a b c d e a a 1 d e b c b b c 1 a e d c c b e d 1 a d d e a 1 c b e e d c b a 1

4. Let G be the set of pairs of real numbers (a, b) with a 6= 0 and define: (a, b)(c, d) = (ac, ad + b), 1 = (1, 0). Verify that this defines a group.

Sol. (1,0) is the identity element and ¡₁

a,_a^b¢

is the inverse of (a, b).

5. Let G be the set of transformations of the real linearR defined by x → x⁰ = ax + b where a and b are real numbers and a 6= 0. Verify that G is a transformation group of R.

Sol. Let φ_(a,b):R → R is defined by x 7→ ax + b. Then φ(a,b)· φ_(c,d) = φ_(ac,ad+b). By the above exercise, we find that G = {φ_(a,b)|a, b ∈ R, a 6= 0} is a group.

6. Verify that the set of translations x → x⁰ = x + b is a subgroup of the group defined in exercise 5.

Sol. Omitted.

7. Show that if an element a of a monid has a right inverse b, that is, ab = 1; and a left inverse c, that is, ca = 1; then b = c, and a is invertible with a⁻¹ = b. Show that a is invertible with b as inverse if and only if aba = a and ab²a = 1.

Proof. (1) ab = ca = 1 ⇒ b = c and a⁻¹ = b.

Consider cab. cab = c(ab) = c · 1 = c; cab = (ca)b = 1 · b = b. Hence b = c. It follows that ab = ba = 1.

(2) aba = a and ab²a = 1 ⇒ a is invertible and a⁻¹ = b.

¿From ab²a = 1, we have a(b²a) = (ab²)a = 1. By part (1), we find that a is invertible.

¿From aba = a, we get (aba)a⁻¹ = a · a⁻¹, since a⁻¹ exists. Hence ab = 1. Now it

is clear that b = a⁻¹. ¤

8. Let α be a rotation about the origin in the plane and let ρ be the reflection in the x-axis. Show that ραρ⁻¹ = α⁻¹.

Let α : (x, y) → (x⁰, y⁰) where x⁰ = x cos θ − y sin θ, y⁰ = x sin θ + y cos θ and ρ : (x, y) → (x⁰, y⁰) where x⁰ = x, y⁰ = −y. You can check ραρ⁻¹ = α⁻¹ by direct computation.

9. Let G be a non-vacuous subset of a monoid M. Show that G is a subgroup if and only if every g ∈ G is invertible in M and g₁⁻¹g₂ ∈ G for any g₁, g₂ ∈ G.

Proof. We just need to prove the “if part” of the statement.

(1) Take any g ∈ G, then 1 = g⁻¹· g ∈ G. Thus G contains the unit.

(2) For any g ∈ G, g⁻¹ = g⁻¹· 1 ∈ G. Hence g is invertible in G.

(3) For any g1, g2 ∈ G, g1g2 = (g⁻¹₁ )⁻¹ · g2 ∈ G, since g₁⁻¹ ∈ G, therefore by the

definition of subgroup, G is a subgroup. ¤

10. Let G be a semigroup having the following properties: (a) G contains a right unit 1_r, that is, an element satisfying a1_r = a, a ∈ G, (b) every element a ∈ G has a right inverse relative to 1_r (ab = 1_r). Show that G is a group.

Proof. It suffices to show that, (1) 1r · a = a for all a ∈ G, (2) for all a ∈ G, ab = 1_r ⇒ ba = 1_r.

We first prove (2): Let c ∈ G such that bc = 1_r. Then ba = (ba)1_r = (ba)(bc) = b(ab) · c = (b · 1r)c = bc = 1r. Hence (2) follows.

Since 1_r· a = (ab)a = a(ba) = a · 1_r= a. Hence 1_r is the unit and (1) follows. ¤

11. Show that in a group, the equations ax = b and ya = b are solvable for any a, b ∈ G. Conversely, show that any semigroup having this property contains a unit and is a group.

Proof. In a group G, the equation ax = b has solution a⁻¹b and ya = b has solution ba⁻¹. Conversely, in a semigroup, ax = b and ya = b are solvable. In particular, ax = a is solvable for some a. Denote the solution by 1r. For any b ∈ G, there exists y such that ya = b by hypothesis. Then b1_r = ya1_r = ya = b, and 1_ris a right unit. Moreover, for any a ∈ G, ax = 1_r are solvable. Thus a has a right inverse relative to 1_r. Now

apply the result of Exercise 10. ¤

12. Show that both cancellation laws hold in a group, that is, ax = ay ⇒ x = y and xa = ya ⇒ x = y. Show that any finite semigroup in which both cancellation laws hold is a group.

Proof. For the first statement, multiply by a⁻¹ on both sides of ax = ay or xa = ya.

For the second, let S be a finite semigroup in which cancellation laws hold. We shall use exercise 11 to prove that G is a group. It suffices to show that the equations ax = b and ya = b are solvable for a, b ∈ G.

Consider the map m_a : S → S defined by m_a(s) = as. The cancellation law ax = ay ⇒ x = y implies that m_a is injective. Since |S| is finite, m_a is surjective by

the Pigeon-hole principle. Thus ax = b is solvable. ¤

13. Show that any finite group of even order contains an element a 6= 1 such that a² = 1.

Proof. Let S = {a|a ∈ G and a 6= a⁻¹}. |S| is even. Since |G| is even and 1 ∈ G − S, hence G − S 6= ∅ and |G − S| is even. Let 1 6= b ∈ G − S, then b = b⁻¹. Thus b² = 1,

b 6= 1. ¤

14. Show that a group G cannot be a union of two proper subgroups.

Proof. Suppose that H1, H2 are subgroups of G with Hi ( G. We shall show that G ) H₁∪ H₂.

Suppose that G = H₁∪ H₂. Choose a ∈ G − H₁. b ∈ G − H₂. Since G = H₁∪ H₂, we find that a ∈ H2, a /∈ H1, b ∈ H1, b /∈ H2. ¤

Claim: ab /∈ G, which will lead to a contradiction.

Suppose ab ∈ G = H1∪ H2. It is impossible that ab ∈ H1 because ab ∈ H1 implies a = (ab) · b⁻¹ ∈ H₁. Similarly ab /∈ H₂.

15. Let G be a finite set with a binary composition and unit. Show that G is a group if and only if the multiplication table (constructed as for monoids) has the following properties:

(i) every row and every column contains every element of G,

(ii) for every pair of elements x 6= 1, y 6= 1 of G, let R be any rectangle in the body of the table having 1 as one of its vertices, x a vertex in the same row as 1, y a vertex in the same column as 1, then the fourth vertex of the rectangle depends only on the pair (x, y), and not on the position of 1.

Proof. Let G be a group. Then exercise 11 implies (i). As for (ii), observe the table (1).

c d b 1 y a x z

(1)

We have bc = 1, ac = x, bd = y and b = c⁻¹. Then z = ad = (ac)(bd) = xy, which depends only on the pair (x, y).

Conversely, suppose G satisfies (i) and (ii). The condition (i) means that both cancellation laws hold. In virtue of exercise 12, we just need to check the associativity of G.

We give two different arguments to prove this law.

(I) Step 1. For all a ∈ G, there exists a unique a⁻¹ such that aa⁻¹ = 1 by (i).

Step 2. For all a ∈ G, a⁻¹a = 1: Comparing right-bottom vertices of the following two rectangles, we have a⁻¹a = 1 by (ii).

1 a

1 1 a

a⁻¹ a⁻¹ a⁻¹a

a⁻¹ 1

a 1 a

1 a⁻¹ 1

Step 3. For all a, b ∈ G, a⁻¹(ab) = b: ¿From the following rectangles.

1 ab

1 1 ab

a⁻¹ a⁻¹ a⁻¹(ab)

b bc b⁻¹ 1 c

a ab a(bc)

Owing to Step 3, the right-top vertex of the second rectangle is equal to c. The assertion is followed from these rectangles.

1 x

1 1 — x (A) y y — yx

1 — x (B) y — z

1 (∗) 1 1 — y — v

(∗∗) 1 — z

x — xy — w

(2) (3)

(II) We first observe the table (2). ¿From the rectangle (A) we see that the right- bottom vertex must be yx.

(α) Suppose one of x, y and z is 1, then (xy)z = x(yz) clearly. Hence we assume that none of x, y and z equal to 1.

(β) In table (3), we find an entry 1 in column (∗) (note that 1 may be equal to xy).

In the row (∗∗) which 1 lies we find an entry z. In the rectangle

y — v

1 — 1

, we have

v = yz by the above observation. ¿From the rectangle

1 — z xy — w

, we have w = (xy)z.

¿From the rectangle

1 — v x — w

, we have w = xv = x(yz). Thus (xy)z = w = x(yz), the

associative law holds. ¤

Remark. We can prove that G is a group under the following weaker conditions:





G is any set (not necessarily finite).

G has a binary composition with an identity elt. ∀x ∈ G, ∃y ∈ G, such that xy = 1.

condition (ii).