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2.3
2.3
J.A.Beachy
1
Permutations
from A Study Guide for Beginner’s by J.A.Beachy,
a supplement to Abstract Algebra by Beachy / Blair
1 2 3 4 5 6 7 8 9
17. For the permutation σ =
, write σ as a product of
7 5 6 9 2 4 8 1 3
disjoint cycles. What is the order of σ? Write σ as a product of transpositions. Is σ
an even permutation? Compute σ −1 .
Solution: Starting with 1, we have σ(1) = 7, σ 2 (1) = σ(7) = 8, and σ 3 (1) =
σ(8) = 1, so the first cycle is (1, 7, 8). The smallest number not in this cycle is
2, and starting with 2 we get the cycle (2, σ(2)) = (2, 5). Continuing, we get σ =
(1, 7, 8)(2, 5)(3, 6, 4, 9). By Proposition 2.3.8, σ has order 12, since lcm[3, 2, 4] = 12.
To write σ as a product of transpositions, we can simply write (1, 7, 8)(2, 5)(3, 6, 4, 9) =
(1, 7)(7, 8)(2, 5)(3, 6)(6, 4)(4, 9). Since σ can be expressed as the product of 6 transpositions, it is an even permutation.
Finally, we have σ −1 = (1, 8, 7)(2, 5)(3, 9, 4, 6). (Note that since the cycles are disjoint
they commute with each other, and so here the order of the cycles is not important.)
1 2 3 4 5 6 7 8 9
18. For the permutations σ =
and
2 5 1 8 3 6 4 7 9
1 2 3 4 5 6 7 8 9
τ =
, write each of these permutations as a product
1 5 4 7 2 6 8 9 3
of disjoint cycles: σ, τ , στ , στ σ −1 , σ −1 , τ −1 , τ σ, τ στ −1 .
Solution: σ = (1, 2, 5, 3)(4, 8, 7); τ = (2, 5)(3, 4, 7, 8, 9); στ = (1, 2, 3, 8, 9);
στ σ −1 = (1, 8, 4, 7, 9)(3, 5); σ −1 = (1, 3, 5, 2)(4, 7, 8); τ −1 = (2, 5)(3, 9, 8, 7, 4);
τ σ = (1, 5, 4, 9, 3); τ στ −1 = (1, 5, 2, 4)(7, 9, 8).
19. Let σ = (2, 4, 9, 7, )(6, 4, 2, 5, 9)(1, 6)(3, 8, 6) ∈ S9 . Write σ as a product of disjoint
cycles. What is the order of σ? Compute σ −1 .
Solution: We have σ = (1, 9, 6, 3, 8)(2, 5, 7), so it has order 15 = lcm[5, 3], and σ −1 =
(1, 8, 3, 6, 9)(2, 7, 5).
1 2 3 4 5 6 7 8 9 10 11
20. Compute the order of τ =
. For σ =
7 2 11 4 6 8 9 10 1 3 5
(3, 8, 7), compute the order of στ σ −1 .
Solution: Since τ = (1, 7, 9)(3, 11, 5, 6, 8, 10), it has order 6 = lcm[3, 6]. Then
στ σ −1 = (3, 8, 7)(1, 7, 9)(3, 11, 5, 6, 8, 10)(3, 7, 8) = (1, 3, 9)(8, 11, 5, 6, 7, 10), so the cycle structure of στ σ −1 is the same as that of τ , and thus στ σ −1 has order 6.
21. Prove that if τ ∈ Sn is a permutation with order m, then στ σ −1 has order m, for any
permutation σ ∈ Sn .
Solution: Assume that τ ∈ Sn has order m. It follows from the identity (στ σ −1 )k =
στ k σ −1 that (στ σ −1 )m = στ m σ −1 = σ(1)σ −1 = (1). On the other hand, the order of
2.3
J.A.Beachy
2
στ σ −1 cannot be less than n, since (στ σ −1 )k = (1) implies στ k σ −1 = (1), and then
τ k = σ −1 σ = (1).
22. Show that S10 has elements of order 10, 12, and 14, but not 11 or 13.
Solution: The element (1, 2)(3, 4, 5, 6, 7) has order 10, the element (1, 2, 3)(4, 5, 6, 7)
has order 12, and (1, 2)(3, 4, 5, 6, 7, 8, 9) has order 14. On the other hand, since 11 and
13 are prime, any element of order 11 or 13 would have to be a cycle, and there are
no cycles of that length in S10 .
23. Let S be a set, and let X ⊆ S. Let G = {σ ∈ Sym(S) | σ(X) = X}. Prove that G is
a group of permutations.
Solution: If σ, τ ∈ G, then στ (X) = σ(τ (X)) = σ(X) = X, which shows that
στ ∈ G. It is clear that 1S (X) = X, and thus 1S ∈ G. Finally, if σ ∈ G, then
σ −1 (X) = σ −1 (σ(X)) = X, which shows that σ −1 ∈ G.
24. Let G be a group of permutations, with G ⊆ Sym(S), for the set S. Let τ be a fixed
permutation in Sym(S). Prove that
τ Gτ −1 = {σ ∈ Sym(S) | σ = τ γτ −1 for some γ ∈ G}
is a group of permutations.
Solution: If σ1 , σ2 ∈ τ Gτ −1 , then we can write σ1 = τ γ1 τ −1 and σ2 = τ γ2 τ −1 for
some γ1 , γ2 ∈ G. Then σ1 σ2 = τ γ1 τ −1 τ γ2 τ −1 = τ (γ1 γ2 )τ −1 belongs to τ Gτ −1 since
γ1 γ2 ∈ G. We have 1S = τ τ −1 = τ 1S τ −1 , and so 1S ∈ τ Gτ −1 . If σ = τ γτ −1 , then
σ −1 = (τ γτ −1 )−1 = (τ −1 )−1 γ −1 τ −1 = τ γ −1 τ −1 , which shows that σ −1 ∈ τ Gτ −1 since
γ −1 ∈ G.
ANSWERS AND HINTS
25. Consider the following permutations in S7 .
1 2 3 4 5 6 7
1 2 3 4 5 6 7
σ=
and τ =
3 2 5 4 6 1 7
2 1 5 7 4 6 3
Compute the following
products.
1 2 3 4
Answer: (a) στ =
2 3 6 7
1 2 3
−1
Answer: (c) στ σ =
1 3 2
5
4
4
7
6
1
5
6
7
5
6 7
4 5
26. Using the permutations σ and τ from Problem 25, write each of the permutations στ ,
τ σ, τ 2 σ, σ −1 , στ σ −1 , τ στ −1 and τ −1 στ as a product of disjoint cycles. Write σ and τ
as products of transpositions.
Answers: σ = (1, 3, 5, 6) = (1, 3)(3, 5)(5, 6) τ = (1, 2)(3, 5, 4, 7) = (1, 2)(3, 5)(5, 4)(4, 7)
στ = (1, 2, 3, 6)(4, 7, 5)
στ σ −1 = (2, 3)(4, 7, 5, 6)
τ −1 στ = (2, 7, 3, 6)
28. Let σ = (3, 6, 8)(1, 9, 4, 3, 2, 7, 6, 8, 5)(2, 3, 9, 7) ∈ S9 .
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J.A.Beachy
(b) Is σ an even permutation or an odd permutation?
(c) What is the order of σ in S9 ?
Answer: 12
29. Let σ = (2, 3, 9, 6)(7, 3, 2, 5, 9)(1, 7)(4, 8, 7) ∈ S9 .
(b) Is σ an even permutation or an odd permutation?
(c) What is the order of σ in S9 ?
Answer: 15
3
Answer: Odd
Answer: Even