Mathematics 1214: Introduction to Group Theory
Homework exercise sheet 3
Due 12:50pm, Friday 12th February 2010
1. Insert the word “odd” or “even” in the gaps and prove the resulting statements.
(a) The product of two even permutations is
(b) The product of three odd permutations is
.
.
(c) The product of an even and an odd permutation is
.
Solution (a) The product of two even permutations is even, since if α, β are even permutations then σ(α ◦ β) = σ(α) · σ(β) = 1 · 1 = 1.
(b) The product of three odd permutations is odd since if α, β, γ are odd permutations then
σ(α ◦ β ◦ γ) = σ(α) · σ(β ◦ γ) = σ(α) · σ(β) · σ(γ) = (−1) · (−1) · (−1) = −1.
(c) The product of an even and an odd permutation is odd since if α is even and β is odd then
σ(α ◦ β) = σ(α) · σ(β) = 1 · (−1) = −1 and σ(β ◦ α) = σ(β) · σ(α) = (−1) · 1 = −1. [We do
need both parts for a complete proof here].
1 2 3 4 5 6 7 8
2. Express the permutation
as a product of disjoint cycles. Determine
4 5 8 2 1 7 6 3
whether this permutation is odd or even using two different methods: by counting inversions,
and by examining your product of disjoint cycles.
1 2 3 4
Solution We have
4 5 8 2
in this permutation, so it is odd.
(6 7) has even length, so they are
5 6 7 8
= (1 4 2 5)(3 8)(6 7). There are 15 inversions
1 7 6 3
Alternatively, each of the three cycles (1 4 2 5), (3 8) and
all odd permutations. Hence their product is odd.
3. Consider the permutation in S10 given by
α = (4 2 1)(5 4 9 10)(2 3 4)(7 1)(3 6).
Without rewriting α, compute σ(α). Then write α in 2-row notation and as a product of
disjoint cycles.
Solution We have
σ(α) = σ(4 2 1)σ(5 4 9 10)σ(2 3 4)σ(7 1)σ(3 6) = 1 · (−1) · 1 · (−1) · (−1) = −1.
Using the usual rules for composing permutations (working from right to left) gives
1 2 3 4 5 6 7 8 9 10
α = (1 7 4)(2 3 6 9 10 5) =
.
7 3 6 1 2 9 4 8 10 5
4. (a) Let n, k ∈ N, let a1 , a2 , . . . , ak be distinct elements of {1, 2, . . . , n}, and consider the cycle
α = (a1 a2 . . . ak ) in Sn . What is α−1 ?
[As always, prove that your answer is correct].
(b) If (G, ∗) is a group and x1 , x2 , . . . , xn ∈ G, then the following identity holds:
−1
−1
−1
(x1 ∗ x2 ∗ · · · ∗ xn−1 ∗ xn )−1 = x−1
n ∗ xn−1 ∗ · · · ∗ x2 ∗ x1 .
Consider the following permutation in S16 :
α = (11 7)(1 3 15 2 4)(6 12 13 9)(5 8).
Use (a) and the identity above to compute α−1 , writing your answer as a product of
disjoint cycles.
Solution Recall that α(ai ) = ai+1 if 1 ≤ i < k, and α(ak ) = a1 , and α(x) = x if x 6∈
{a1 , . . . , an }.
Let β = (ak ak−1 . . . a2 a1 ). Then β(ai ) = ai−1 if 1 < i ≤ k, and β(a1 ) = ak , and β(x) = x if
x 6∈ {a1 , . . . , ak }.
If x ∈ {1, 2, . . . , n} with x 6∈ {a1 , . . . , an } then (β ◦α)(x) = β(α(x)) = β(x) = x and (α◦β)(x) =
α(β(x)) = α(x) = x.
If x = ai for some i with 1 < i < k, then (β ◦ α)(x) = β(α(ai )) = β(ai+1 ) = ai = x and
(α ◦ β)(x) = α(β(ai )) = α(ai−1 ) = ai = x.
If x = a1 then (β ◦ α)(x) = β(α(a1 )) = β(a2 ) = a1 = x and (α ◦ β)(x) = α(β(a1 )) = α(ak ) =
a1 = x.
If x = ak then (β ◦ α)(x) = β(α(ak )) = β(a1 ) = ak = x and (α ◦ β)(x) = α(β(ak )) = α(ak−1 ) =
ak = x.
Hence for every x ∈ {1, 2, . . . , n}, we have (β◦α)(x) = x = e(x) and (α◦β)(x) = x = e(x) where
e is the identity element of Sn . So β ◦ α = e and α ◦ β = e. So α−1 = β = (ak ak−1 . . . a2 a1 ).
(b) By the identity in the question, taking x1 = (11 7), x2 = (1 3 15 2 4), x3 = (6 12 13 9) and
x4 = (5 8), we have
−1
α−1 = (11 7)(1 3 15 2 4)(6 12 13 9)(5 8)
= (5 8)−1 (6 12 13 9)−1 (1 3 15 2 4)−1 (11 7)−1 .
Inverting each of these cycles using (a) gives α−1 = (8 5)(9 13 12 6)(4 2 15 3 1)(7 11), and this
is a product of disjoint cycles.
5. Let n ∈ N, and let α be the cycle in Sn given by α = (1 2 . . . n). For k ∈ N, let us write
αk = α ◦ α ◦ · · · ◦ α,
{z
}
|
k times
and let us also write α0 = e where e is the identity mapping in Sn .
(a) Show that αn = e and that αs ◦ αt = αs+t for any integers s, t ≥ 0.
(b) Show that if s, t are integers with 0 ≤ s, t < n and αs = αt , then s = t.
(c) Prove that composition is an operation on the set Cn = {e, α, α2 , . . . , αn−1 }, and deduce
that (Cn , ◦) is an abelian group of order n.
(d) Write down the Cayley tables of the groups (Cn , ◦) for n = 1, 2, 3, 4.
Solution (a) Since α is a cycle of length n, it is easy to see that αn (x) = x for all x ∈
{1, 2, . . . , n} [for example, by drawing a picture]. Hence αn = e.
By the associativity of composition, we have
αs ◦ αt = (α
· · ◦ α}) ◦ (α
· · ◦ α}) = α
· · ◦ α} = αs+t
| ◦ ·{z
| ◦ ·{z
| ◦ ·{z
s times
t times
s + t times
for s, t ≥ 1. Since α0 = e, this equation clearly also holds if either s or t is zero, hence it is
true for all integers s, t ≥ 0.
2
(b) Observe that αs (1) = s + 1 and αt (1) = t + 1 for 0 ≤ s, t < n. If αs = αt then this implies
that s + 1 = t + 1, so s = t.
(c) To see that ◦ is an operation on Cn = {α0 , α1 , . . . , αn−1 }, we must show that β, γ ∈ Cn =⇒
β ◦ γ ∈ Cn . So let s, t ∈ {0, 1, . . . , n − 1} and consider β = αs and γ = αt . If s + t < n then
β ◦ γ = αs ◦ αt = αs+t ∈ Cn . If s + t ≥ n then we have n ≤ s + t < 2n, so 0 ≤ s + t − n < n,
and
β ◦ γ = αs+t = αs+t−n ◦ αn = αs+t−n ◦ e = αs+t−n ∈ Cn .
Hence, in either case, β ◦ γ ∈ Cn , so ◦ is an operation on Cn .
To see that (Cn , ◦) is a group:
• ◦ is associative on Sn , so it is certainly associative on Cn
• e ∈ Cn and it is the identity element for all of Sn , so it is an identity element for Cn
• if β ∈ Cn then β = αs for some integer s with 0 ≤ k < n, and γ = αn−s satisfies
β ◦ γ = αn = γ ◦ β, so γ = β −1 and γ ∈ Cn (in the case s = 0, this is because
αn = e ∈ Cn ).
To see that (Cn , ◦) is an abelian group, observe that if β, γ ∈ Cn then β = αs and γ = αt for
some integers s, t ≥ 0, so
β ◦ γ = αs ◦ αt = αs+t = αt+s = αt ◦ αs = γ ◦ β.
By (b), no two of the mappings e, α, α2 , . . . , αn−1 are equal. Hence the order of Cn is |Cn | = n.
(d) We have
n = 1:
◦
n = 3:
e
◦
e
e
e
α
α2
2
e
e
α
α
α
α
α2
e
α2
α2
e
α
n = 2:
n = 4:
3
◦
e
α
e
e
α
α
α
e
◦
e
α
α2
α3
e
e
α
α2
α3
α
α
α2
α3
e
α2
α2
α3
e
α
α3
α3
e
α
α2