Solutions to Assignment 7
• Optional: Calculate the following products of permutations:
– (1, 2, 3)(2, 4, 5) = (1, 4, 5, 2, 3)
– (3, 4)(1, 2, 3, 4, 5, 6)(4, 3) = (1, 2, 4, 3, 5, 6)
– (3, 7)(1, 2, 3, 4, 5, 6)(7, 3) = (1, 2, 7, 4, 5, 6)
– (2, 5, 9)−1 (1, 2, 3, 4, 5, 6)(2, 5, 9) = (1, 5, 3, 4, 9, 6)
1. Show that a cycle of length k is odd if k is even and even if k is odd.
We can write (a1 , a2 , . . . , ak ) as a product of k − 1 permutations:
(a1 , a2 , . . . , ak ) = (a1 , a2 )(a1 , a3 ) . . . (a1 , ak ).
If k is even, then k − 1 is odd. If k is odd, then k − 1 is even.
2. Is the permutation
(1, 9)(2, 10, 11, 6, 3)(4, 5, 12, 7) ∈ S12
odd or even? Describe a general method to decide whether a product of cycles is odd or even.
We can classify each cycle in the product as even or odd using the previous exercise:
(1, 9) (2, 10, 11, 6, 3) (4, 5, 12, 7) ∈ S12 .
{z
} | {z }
| {z } |
odd
even
odd
Since signature is a homomorphism,
sgn((1, 9)(2, 10, 11, 6, 3)(4, 5, 12, 7)) = sgn((1, 9)) sgn((2, 10, 11, 6, 3)) sgn((4, 5, 12, 7))
= −1 · 1 · −1
=1
so the permutation is even.
By the same argument, a product of cycles is odd if and only if it contains an odd number of cycles of
even length.
3. Suppose that a ∈ Sn and that a = (a1 , . . . , ak ) is a cycle of length k. Show that for all b ∈ Sn , b−1 ab
is a cycle of length k. What cycle is it?
As suggested by the examples in the optional problem,
b−1 ab = (b1 , . . . , bk ),
where bi = b(ai ) is the result of applying the permutation b to ai . That is,
(b−1 ab)(bi ) = b(a(b−1 (bi ))) = b(a(ai )) = b(ai+1 )
for i < k and
(b−1 ab)(bk ) = b(a(b−1 (bk ))) = b(a(ak )) = b(a1 ).
If x 6∈ {b1 , . . . , bk }, then b−1 (x) 6∈ {a1 , . . . , ak }, so
(b−1 ab)(x) = b(a(b−1 (x))) = b(b−1 (x)) = x.
This is the definition of the cycle (b1 , . . . , bk ).
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4. We label the axes of reflection as follows:
V
D1
H
D2
Then the elements of D8 are
D8 = {R0 , R90 , R180 , R270 , D1 , D2 , H, V }
and each element corresponds to a map of the square as follows:
ρR0 (x, y) = (x, y)
ρH (x, y) = (x, −y)
ρR90 (x, y) = (−y, x)
ρV (x, y) = (−x, y)
ρR180 (x, y) = (−x, −y)
ρD1 (x, y) = (y, x)
ρR270 (x, y) = (y, −x)
ρD2 (x, y) = (−y, −x).
Then the orbits are:
ρR90 (x) = ρV (x)
ρR180 (x) = ρD2 (x)
x = ρR0 (x) = ρD1 (x)
ρR270 (x) = ρH (x)
R90 (y)
D1 (y)
V (y)
R0 (y)
R180 (y)
H(y)
D2 (y)
R270 (y)
R0 (z) = R90 (z) = R180 (z) = R270 (z) = H(z) = V (z) = D1 (z) = D2 (z)
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and the stabilizers are:
Stab(x) = {R0 , D1 }
Stab(y) = {R0 }
Stab(z) = D8 .
Therefore,
| Orb(x)| · | Stab(x)| = 4 · |{R0 , D1 }| = 4 · 2 = 8
| Orb(y)| · | Stab(y)| = 8 · |{R0 }| = 8 · 1 = 8
| Orb(z)| · | Stab(z)| = 1 · |D8 | = 1 · 8 = 8
5. Show that H = Stab(0).
By definition,
Stab(0) = {fm,b | fm,b (0) = 0}
= {fm,b | b = 0} = H
Show that the left cosets of H are all of the form
Suppose that a, b ∈ R and that a 6= 0. Let t = f −1 (0) =
−b
a .
Then
fa,b ◦ H = {fa,b ◦ fm,0 | m 6= 0}
= {fma,mb | m 6= 0}.
On one hand, if fma,mb ∈ fa,b ◦ H, then
fma,mb (t) = ma ·
−b
+ mb = 0,
a
so fa,b ◦ H ⊂ Lt . On the other hand, if fa0 ,b0 ∈ Lt , then
fa0 ,b0 (t) = 0
a0 t + b0 = 0
b0
a0
=
a
b
0
so if m = aa , then a0 = ma and b0 = mb. This implies that Lt ⊂ fa,b ◦ H, and every left coset is of the
form Lt for some t. Furthermore, Lt = f1,−t ◦ H, so every set Lt is a left coset.
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