Exercises (§1.3, p.39)

Basic Algebra (Solutions)

by Huah Chu

Exercises (§1.3, p.39)

1. Use a multiplication table for S₃ (exercise 3, §1.2) and the isomorphism a → a_L(a_L the left translation defined by a) to obtain a subgroup of S₆ isomorphic to S₃.

Sol. We rewrite the elements of S3 as 1 =

µ 1 2 3 1 2 3

¶ , 2 =

µ 1 2 3 1 3 2

¶ , 3 = µ 1 2 3

2 1 3

¶ , 4 =

µ 1 2 3 2 3 1

¶ , 5 =

µ 1 2 3 3 1 2

¶ , 6 =

µ 1 2 3 3 2 1

¶

. Then the iso- morphism a → aL are:

1 →

µ 1 2 3 4 5 6 1 2 3 4 5 6

¶

= 1, 2 →

µ 1 2 3 4 5 6 2 1 5 6 3 4

¶

= (1 2)(3 5)(4 6), 3 →

µ 1 2 3 4 5 6 3 4 1 2 6 5

¶

= (1 3)(2 4)(5 6), 4 →

µ 1 2 3 4 5 6 4 3 6 5 1 2

¶

= (1 4 5)(2 3 6), 5 →

µ 1 2 3 4 5 6 5 6 2 1 4 3

¶

= (1 5 4)(2 6 3), 6 →

µ 1 2 3 4 5 6 6 5 4 3 2 1

¶

= (1 6)(2 5)(3 4).

2. Show that the two groups in examples 11 and 13 on pages 33 and 34 of “Basic Algebra, vol. 1” are isomorphic. Obtain a subgroup of S_n isomorphic to these groups.

Ans. For the rotation through angle ^2π_n in Rn, we map it into the complex number z₁ = cos^2π_n + i sin^2π_n in U_n. This mapping is an isomorphism. The subgroup of S_n generated by

µ 1 2 · · · n − 1 n 2 3 · · · n 1

¶

isomorphic to these groups.

3. Let G be a group. Define the right translation a_R for a ∈ G as the map x → xa in G. Show that G_R = {a_R} is a transformation group of the set G and a → a⁻¹_R is an isomorphism of G with GR.

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Proof. The proof is similar to that of Cayley’s Theorem. We leave it to the reader. ¤

4. Is the additive group of integers isomorphic to the additive group of rationals?

Ans. No. Let φ : (Z, +) → (Q, +) be any homomorphism. Let φ(1) = a. Then φ(n) = φ(1 + · · · + 1) = na for n ∈N, and φ(−n) = −na. Write a = ^r_s with r, s ∈ Z, s 6= 0, (r, s) = 1. If p is any prime number and p - s, then ¹_p ∈ φ(/ Z). Hence φ is not onto.

5. Is the additive group of rationals isomorphic to the multiplicative group of non-zero rationals?

Ans. No. Suppose η : (Q^∗, ·) → (Q, +) is an isomorphism. Then η(−1) + η(−1) = η((−1)(−1)) = η(1) = 0, η(−1) = 0.

A contradiction.

Remarks: (1) In fact, (Q^∗, ·) has an element of order 2, that is, −1. But none of the elements in (Q, +) has finite order.

(2) Furthermore, the additive group of rationals is never isomorphic to the multi- plicative group of positive rationals since nX = a is solvable for any n ∈ N, for any a ∈Q and Yⁿ = b is not always solvable for any n ∈N, for and b ∈ Q⁺.

6. In Z define a ◦ b = a + b − ab. Show that (Z, ◦, 0) is a monoid and that the map a → 1 − a is an isomorphism of the multiplicative monoid (Z, ·, 1) with (Z, ◦, 0).

Proof. Omitted. ¤

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