MATH 3005 Homework Solution
Han-Bom Moon
Homework 5 Solution
Chapter 5.
1. Let
"
α=
1 2 3 4 5 6
2 1 3 5 4 6
#
"
and β =
1 2 3 4 5 6
6 1 2 4 3 5
#
.
Compute each of the following.
(a) α−1
"
1 2 3 4 5 6
2 1 3 5 4 6
#
"
1 2 3 4 5 6
1 6 2 3 4 5
#
"
1 2 3 4 5 6
6 2 1 5 3 4
#
α−1 =
(b) βα
βα =
(c) αβ
αβ =
2. Let
"
α=
1 2 3 4 5 6 7 8
2 3 4 5 1 7 8 6
#
"
and β =
1 2 3 4 5 6 7 8
1 3 8 7 6 5 2 4
Write α, β, and αβ as
(a) products of disjoint cycles;
α = (12345)(678), β = (23847)(56), αβ = (12485736)
(b) products of 2-cycles.
α = (15)(14)(13)(12)(68)(67), β = (27)(24)(28)(23)(56),
αβ = (16)(13)(17)(15)(18)(14)(12)
3. Write each of the following permutations as a product of disjoint cycles.
(a) (1235)(413)
(15)(234)
(b) (13256)(23)(46512)
(124)(35) or (124)(35)(6)
1
#
.
MATH 3005 Homework Solution
Han-Bom Moon
(c) (12)(13)(23)(142)
(1423)
6. What is the order of each of the following permutations?
"
#
1 2 3 4 5 6
(a)
2 1 5 4 6 3
Let α be the permutation. Then α = (12)(356).
α2 = ((12)(356))2 = (12)2 (356)2 = (365),
α3 = ((12)(356))3 = (12)3 (356)3 = (12),
α4 = (356), α5 = (12)(365), α6 = e.
Thus |α| = 6.
Indeed, you may find a pattern. If α = α1 α2 · · · αk where αi are disjoint
cycles of length `i , then the order of α is lcm(`1 , `2 , · · · , `k ).
"
#
1 2 3 4 5 6 7
(b)
7 6 1 2 3 4 5
Let β be the permutation. Then β = (1753)(264). By the same idea, you may
find |β| = lcm(4, 3) = 12.
11. Determine whether the following permutations are even or odd.
(a) (135)
(135) = (15)(13)
So it is an even permutation.
(b) (1356)
(1356) = (16)(15)(13)
It is an odd permutation.
(c) (13567)
(13567) = (17)(16)(15)(13)
It is an even permutation.
(d) (12)(134)(152)
(12)(134)(152) = (12)(14)(13)(12)(15)
It is an odd permutation.
(e) (1243)(3521)
(1243)(3521) = (13)(14)(12)(31)(32)(35)
It is an even permutation.
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MATH 3005 Homework Solution
Han-Bom Moon
15. Let n be a positive integer. If n is odd, is an n-cycle an odd or an even permutation? If n is even, is an n-cycle an odd or an even permutation?
Note that an n-cycle α = (a1 a2 · · · an ) is a product
(a1 an )(a1 an−1 ) · · · (a1 a2 ).
So it is a product of (n − 1) 2-cycles. Therefore if n is odd, then it is an even
permutation and if n is even, then it is an odd permutation.
34. What cycle is (a1 a2 · · · an )−1 ?
The cycle α := (a1 a2 · · · an ) is a function where α(a1 ) = a2 , α(a2 ) = a3 , · · · ,
α(an−1 ) = an , and α(an ) = a1 . So α−1 (an ) = an−1 , · · · , α−1 (a2 ) = a1 , and
α−1 (a1 ) = an . Therefore α−1 = (an an−1 · · · a2 a1 ).
35. Let G be a group of permutations on a set X. Let a ∈ X and define stab(a) =
{α ∈ G | α(a) = a}. We call stab(a) the stabilizer of a in G (since it consists of all
members of G that leave a fixed). Prove that stab(a) is a subgroup of G.
First of all, stab(a) 6= ∅ because e ∈ stab(a). Let α, β ∈ stab(a). Then α(a) = a
and β(a) = a. αβ(a) = α(β(a)) = α(a) = a so αβ ∈ stab(a). Also for α ∈ stab(a),
α−1 (a) = a because a = α−1 (α(a)) = α−1 (a). Therefore α−1 ∈ stab(a) as well. So
stab(a) ≤ G.
57. Viewing the members of D4 as a group of permutations of a square labeled 1, 2,
3, 4 as described in Example 3, which geometric symmetries correspond to even
permutations?
If you denote all elements of D4 as a permutation on the vertex set {1, 2, 3, 4} as
in Example 3, we have
R0 = e, R90 = (1234) = (14)(13)(12), R180 = (13)(24),
R270 = (1432) = (12)(13)(14),
H = (12)(34), V = (14)(23), D = (24), D0 = (13).
Therefore R0 , R180 , H, and V are even permutations.
69. Prove that every element of Sn (n > 1) can be written as a product of elements of
the form (1k).
Because all permutations α ∈ Sn is a product of 2-cycles, if each 2-cycle (ij) is
a product of 2-cycles of the form (1k), we are able to obtain the conclusion. But
(ij) = (1i)(1j)(1i), so we have the result.
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