Test1. Due Friday, March 13, 2015.

Test1. Due Friday, March 13, 2015.

1. Euclidean algorithm and related.

(a) Suppose that a and b are two positive integers and that gcd(a, b) = d. Find all solutions m and n to am − bn = 0. Hint: Note that for each solution, am and bn are common multiples of a and b.

Solution: am − bn = 0 ↔ am = bn. Thus, am and bn are common multiples of a and b for every solution. Let

l = lcm(a, b) = ab d .

Then all common multiples of a and b are of the form kl, where k is any integer.

If am = bn, then am = bn = kl for some k ∈ Z. This is equivalent to m = kb/d and n = ka/d. On the other hand, for any k ∈ Z, m = kb/d and n = ka/d satisfy am = bn = kl. Thus all solutions are m = kb/d and n = ka/d for k ∈ Z.

(b) If aM − bN = d, find all solutions m and n to am − bn = d. Hint: Note that a(M − m) − b(N − n) = 0.

Solution: If aM − bN = d and am − bn = d, then by subtraction, a(M − m) − b(N − n) = d − d = 0, so M − m and N − n must be of the form kb/d and ka/d, respectively, where k ∈ Z. That is, m = M − kb/d and n = N − ka/d. In words, if aM − bN = d, then a(M + kb/d) − b(N + ka/d) = d, so (M + kb/d, N + kb/d) is a solution for every k ∈ Z and all solutions are of this form.

(c) Show that 1966 and 2017 are relatively prime and compute integers m and n such that 1966m + 2017n = 1.

Solution:

2017 = 1 ∗ 1966 + 51 1966 = 38 ∗ 51 + 28

51 = 1 ∗ 28 + 23 28 = 1 ∗ 23 + 5 23 = 4 ∗ 5 + 3

5 = 1 ∗ 3 + 2 3 = 1 ∗ 2 + 1 2 = 2 ∗ 1 + 0 so gcd(2017, 1966) = 1.

A solution for m and n may be computed by backtracking in the above compu- tation or by multiplying 2 × 2 matrices.

Backtracking:

1 = 3 − 1 ∗ 2

= 3 − 1 ∗ (5 − 1 ∗ 3)

= 2 ∗ 3 − 1 ∗ 5

= 2 ∗ (23 − 4 ∗ 5) − 1 ∗ 5

= 2 ∗ 23 − 9 ∗ 5

= 2 ∗ 23 − 9(28 − 1 ∗ 23)

= 11 ∗ 23 − 9 ∗ 28

= 11 ∗ (51 − 1 ∗ 28) − 9 ∗ 28

= 11 ∗ 51 − 20 ∗ 28

= 11 ∗ 51 − 20(1966 − 38 ∗ 51)

= 771 ∗ 51 − 20 ∗ 1966

= 771 ∗ (2017 − 1 ∗ 1966) − 20 ∗ 1966

= 771 ∗ 2017 − 21 ∗ 1966

= (−791) ∗ 1966 + 771 ∗ 2017 Thus m = −791 and n = 771 is a solution.

(d) Referring to (c) above, compute a solution where m > 0 and m is as small as possible and a solution where n > 0 and n is as small as possible. Hint: If you use the extended Euclidean algorithm, the solution you find will satisfy one of these conditions, so you need only compute a second solution.

Solution: Any solution is of the form m = −791 + 2017k and n = 771 − 1966k.

If k = 0, n is positive. The next smaller value of n is 771−1966 < 0, so (−791, 771) is the solution with the smallest positive value for n. The next larger value of m is −791 + 2017 = 1226, for which n = 771 − 1966 = −1195, so (1226, −1195) is the solution with the smallest positive value for m.

2. True or false.

(a) In the group Z _m , if 0 ≤ a ≤ b < m, then hai = hbi → a = b.

Solution: False. In fact, hai = hbi → gcd(a, m) = gcd(b, m).

(b) If the least common multiple of two positive integers a and b equals a or b, then either a|b or b|a.

Solution: True. ab/ gcd(a, b) = a → gcd(a, b) = b, so b|a. ab/ gcd(a, b) = b → gcd(a, b) = a, so a|b.

(c) Suppose that gcd(a, b) = 1 for positive integers a and b. Then for any positive integers m and n, a ^m and b ⁿ are relatively prime.

Solution: True. a = Q K

i=i p ^k _i

and b = Q L

i=i q ^l _i

, where each p _i and each q _i are

distinct primes. We know that no p _i equals some q _i . Thus a ^m is a product of K

distinct primes p i to higher powers and b ⁿ is a product of L distinct primes q i to

higher powers. The primes remain distinct so gcd(a ^m , b ⁿ ) = 1.

(d) For positive integers a, b and c, suppose that gcd(a, b) = 1 and that c|ab. Then c = d ₁ d ₂ where d ₁ |a, d ₂ |b and gcd(d ₁ , d ₂ ) = 1.

Solution: True. a = Q K

i=1 p ^k _i

and b = Q L

i=1 q _i ^l

where the p _i ’s and q _i ’s are distinct primes. If d|a, then d = d ₁ d ₂ , where d ₁ = Q K

i=1 p ^k _i

, 0 ≤ k ⁰ _i ≤ k _i and d ₂ = Q L

i=1 q ^l _i

, 0 ≤ l _i ⁰ ≤ l _i . d ₁ |a, d ₂ |b and gcd(d ₁ , d ₂ ) = 1.

(e) If σ ∈ S _n , then σ and σ ⁻¹ always have the same number of orbits.

Solution: True. The inverse of a cycle (i 1 , i ₂ , . . . i _k ) is (i _k , i _k−1 , . . . i ₂ , i ₁ ), which is a cycle. σ(i) = i ↔ σ ⁻¹ (i) = i. Thus, σ and σ ⁻¹ have the same orbits. This is stronger than “the same number of orbits”.

(f) If σ ∈ S n , then σ and σ ² always have the same number of orbits.

Solution: False. Counter-example. In S ₄ , σ = (1, 2, 3, 4) has one orbit, but σ ² = (1, 3)(2, 4) has two orbits.

(g) In D ₅ , let H = hρ ² , τ ρ ³ i. Then H = D ₅ .

Solution: True Since gcd(2, 5) = gcd(1, 5) = 1, hρi = hρ ² i, so H contains all powers of ρ. ρ ² ∈ H and τ ρ ³ ∈ H implies that τ ρ ³ ρ ² = τ ∈ H. Thus H contains ρ ⁱ and τ ρ ⁱ for 0 ≤ i < 5, so H = D ₅ .

(h) If H and K are subgroups of G, then HK is a subgroup of G.

Solution: False. If h ₁ , h ₂ ∈ H and k ₁ , k ₂ ∈ K, then there is no reason why (h ₂ k ₂ )(h ₁ k ₁ ) should equal h ₃ k ₃ for some h ₃ ∈ H and k ₃ ∈ K. Counter-example.

In S ₃ , let H = {ι, (1, 2)} and let K = {ι, (1, 3)}. Then HK = {ι, (1, 2), (1, 3), (1, 3, 2)}

However, (1, 2)(1, 3)(1, 2)(1, 3) = (1, 2, 3) / ∈ HK.

(i) If H and K are normal subgroups of G, then HK is a subgroup of G.

Solution: True. In fact, it’s true if just one of H and K is normal. H normal implies that gH = Hg for all g ∈ G. In particular, kH = Hk for any k ∈ K. If h ₁ , h ₂ ∈ H and k ₁ , k ₂ ∈ K, then

(h ₁ k ₁ )(h ₂ k ₂ ) = h ₁ (k ₁ h ₂ )k ₂

= h ₁ (h ₃ k ₁ )k ₂ because H is normal. h ₃ ∈ H

= (h ₁ h ₃ )(k ₁ k ₂ ) ∈ HK.

If h ∈ H and k ∈ K,

(hk) ⁻¹ = k ⁻¹ h ⁻¹

= h ₁ k ⁻¹ because H is normal. h ₁ ∈ H

(j) If H and K are normal subgroups of G, then HK is a normal subgroup of G.

Solution: True. HK is Normal if gHK = HKg for all g ∈ G. This is equivalent to gHKg ⁻¹ = HK for all g ∈ G. If ghkg ⁻¹ ∈ gHKg ⁻¹ , then

ghkg ⁻¹ = (gh)kg ⁻¹

= (h ₁ g)kg ⁻¹ since H is normal

= h ₁ (gk)g ⁻¹

= h ₁ (k ₁ g)g ⁻¹ since K is normal

= (h ₁ k ₁ )gg ⁻¹

= h 1 k 1 ∈ HK.

In the above, h ₁ ∈ H and k ₁ ∈ K. To prove that HK is normal, we need H and K to be normal.

3. For each set G and binary operation ∗, decide the following. Is G a group. If not, what properties fail? If so, is G Abelian? If it is Abelian, is it cyclic?

(a) G = Q ⁺ and ∗ is ordinary multiplication.

Solution: G is a non-cyclic Abelian group.

(b) G =  1 n 0 1

   n ∈ Z



and ∗ is matrix multiplication.

Solution: G is a cyclic group. In fact (G, ∗) ' (Z, +). If f : Z → G is defined by

f (n) =  1 n 0 1

 , then f is an isomorphism.

(c) G =  a b c d

 

 a, b, c, d ∈ Z and ad − bc = ±1



and ∗ is matrix multiplication.

Solution: G is a non-Abelian group (and cannot be cyclic, of course).

(d) G = {a + b √

3 ∈ R ^∗ | a, b ∈ Q} and ∗ is ordinary multiplication. Note that G excludes 0.

Solution: G is a non-cyclic Abelian group. The binary operation is well-defined, since a + b √

3)(c + d √

3) = (ac + 3bd) + (ad + bd) √ 3 ∈ G.

(a + b √

3)(a − b √

3) = a ² − 3b ² . If a ² − 3b ² = 0, then a = 0 → b = 0 and b = 0 → a = 0. If ab 6= 0, then √

3 = ^a _b ∈ Q, which is false. Thus, a + b √

3 = 0 ↔ a = b = 0. Then (a + b √

3) ⁻¹ = a ⁰ − b ⁰ √

3, where a ⁰ = _a

−3b ^a

and b ⁰ = _a

−3b ^b

. (e) G = {n ∈ Z 60 | gcd(n, 60) = 1}. ∗ is addition.

Solution: G is not a group. In fact, G is not even closed under addition, since

gcd(7, 60) = gcd(13, 60) = 1, but gcd(7 + 13, 60) = gcd(20, 60) = 20.

(f) G = {n ∈ Z 60 | gcd(n, 60) = 1}. ∗ is multiplication.

Solution: G is in fact an Abelian group. If m and n are both relatively prime to 60, then neither m nor n contain prime factors in common with 60. Thus, mn contains no prime factor in common with 60. ∗ is clearly commutative. If gcd(a, 60) = 1, then gcd(a ⁿ , 60) = 1 for all n ≥ 0 (see the appropriate true/false question above). If m is the smallest integer such that a ^m ≡ a ^h mod 60 for some h < m, then a ^m−h ≡ 1 mod 60 = a ⁰ , so h = 0. Thus a ^m−1 is the multiplicative inverse of a. |G| = ϕ(60) = ϕ(5)ϕ(3)ϕ(4) = 4 ∗ 2 ∗ 2 = 16. G is not cyclic. In fact, the maximum order of any a ∈ G is 4.

4. Let σ =

 i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

σ(i) 16 13 4 18 6 14 19 2 8 5 7 12 9 17 11 3 10 20 15 1



(a) Compute the decomposition of σ into disjoint cycles.

Solution: σ = µ 1 µ 2 µ 3 µ 4 , where µ 1 = (1, 16, 3, 4, 18, 20), µ 2 = (2, 13, 9, 8), µ ₃ = (5, 6, 14, 17, 10) and µ ₄ = (7, 19, 15, 11)

(b) How many orbits does σ have.

Solution: σ has 5 orbits. Each of the four disjoint cycles above comprise 19 of the 20 numbers between 1 and 12. σ(12) = 12, so {12} is the fifth orbit.

(c) Is σ an even or an odd permutation?

Solution: µ 1 is a 6-cycle (odd), mu ₂ is a 4-cycle (odd), mu ₃ is a 5-cycle (even) and mu ₄ is a 4-cycle (odd). Thus, the sign of σ is (−1)(−1)(1)(−1) = −1, so σ is odd.

(d) Compute |hσi|.

µ ₁ , µ ₂ , µ ₃ and µ ₄ have orders 6, 4, 5 and 4, respectively. The least common multiple of these integers is the least common multiple of 4 and the least common multiple of 6 and 5, which is the least common multiple of 4 and 30, which is 60.

Thus, |hσi| = 60.

(e) Compute the cycle decomposition of σ ² .

Solution: Since the cycles permute disjoint sets of numbers, they commute.

Thus, σ ² = (µ ₁ ) ² (µ ₂ ) ² (µ ₃ ) ² (µ ₄ ) ² .

• µ ² ₁ = (1, 3, 18)(4, 20, 16)

• µ ² ₂ = (2, 9)(8, 13)

• µ ² ₃ = (5, 14, 10, 6, 17)

• µ ² ₄ = (7, 15)(11, 19)

Thus σ ² = (1, 3, 18)(4, 20, 16)(2, 9)(8, 13)(5, 14, 10, 6, 17)(7, 15)(11, 19)

(f) Let µ = (1, 13, 7, 18, 3, 20, 5, 11)(2, 15, 5, 9, 7, 1, 10, 19) Is µ even or odd. You must give a reason.

Solution: µ is the product of two k-cycles, so it is even. The fact that k = 8,

making both cycles odd permutations, is not relevant.

5. In S _n for n > 2, let H = {ι, (1, 2)}, where ι is the identity in S _n and (1, 2) is a transposition. H is clearly a subgroup of S _n . Prove that H is not a normal subgroup of S _n . Hint: It suffices to find a single π ∈ S _n such that πH 6= Hπ.

Solution: (1, 3)H = {(1, 3), (1, 3)(1, 2)} = {(1, 3), (1, 2, 3)} and H(1, 3) = {(1, 3), (1, 2)(1, 3)} = {(1, 3), (1, 3, 2)}. Since (1, 3)H 6= H(1, 3), H is not normal.

6. A _n and 3-cycles. We know that 3-cycles are even permutations. The object is to show that A n is generated by 3-cycles. This problem is broken into parts to assist you.

(a) In S 4 , write the double transition (1, 2)(3, 4) as the product of two 3-cycles. Hint:

A 3-cycle will leave one of the four numbers fixed.

Solution: The hint is useful. Select the first 3-cycle to place 1 into position 2. Select the second 3-cycle to leave position 2 fixed and to place 2 into position 1. The first 3-cycle can be σ 1 = (1, 2, 3). The second 3-cycle σ 2 must satisfy (a) σ ₂ (2) = 2 to keep position 2 fixed, and (b) σ ₂ (3) = 1 to move 2 (which is in position 3 after σ ₁ is applied) into position 1. That is, σ ₂ = (3, 1, 4). There are choices for σ 1 , but given σ 1 , σ 2 is uniquely determined.

Check: (3, 1, 4)(1, 2, 3) = (1, 2)(3, 4). It works! Of course, other choices for σ ₁ are possible. They are: (1, 2, 4), (2, 1, 3), (2, 1, 4), (3, 4, 1), (3, 4, 2), (4, 3, 1), (4, 3, 2).

(b) In S _n for n > 3, suppose that i, j, k and l are distinct integers between 1 and n.

Write µ = (i, j)(k, l) as the product of two 3-cycles.

Solution: Replace 1, 2, 3 and 4 above by i, j, k and l. Then µ = (k, i, l)(i, j, k) = (i, j)(k, l).

(c) In S _n for n > 2, suppose that i, j and k are distinct integers between 1 and n.

Let µ = (i, k)(i, j). Compute the cycle decomposition of µ.

Solution: I did this one in class. µ = (i, j, k).

(d) Show that any even permutation is a product of 3-cycles. Hint: If µ ∈ A _n , then µ is a product of 2k transpositions for some k > 0 (unless µ = ι). Show that µ is a product of at most 2k 3-cycles.

Solution: An even permutation µ can be written as a product of an even number of swaps (transpositions), say 2k swaps. Then µ is a product of k pairs of swaps. Consider each pair:

Case 1: The two swaps involve just two distinct numbers, say i and j. This gives (i, j)(i, j) = ι, the identity permutation. That is, this pair may be deleted (or written as (i, j, k)(i, k, j) for some k not equal to i or j).

Case 2. The two swaps involve three distinct numbers, say i, j and k. As shown above (i, k)(i, j) is a 3-cycle. Note that the first swap may be written as (i, j) without loss of generality where i is the number that is repeated in the second swap.

Case 3 The two swaps involve four distinct numbers, say i, j, k and l. As shown

above, this is the product of 2 3-cycles.

Thus, µ may be written as the product of at most 2k 3-cycles. It may also be written as the product of exactly 2k 3-cycles. Why? Case 1, the trivial case, can be written as the product of 2 3-cycles. Case 2: A single 3-cycle (i, j, k) may also be written as (i, k, j)(i, k, j), which is the square of the inverse of (i, j, k). Case 3:

This is already the product of 2 3-cyles.