Mathematics 1214: Introduction to Group Theory
Tutorial exercise sheet 3
−1
1 2 3 4 5 6 7
1. (a) Compute
, writing your answer in 2-row notation
3 1 7 6 5 4 2
and as a product of disjoint cycles. Is this an even or an odd permutation?
(b) Let n ∈ N, and consider a permutation
1
2
...
α=
α(1) α(2) . . .
n
.
α(n)
Explain how to find α−1 in 2-row notation.
−1
1 2 3 4 5 6 7
3 1 7 6 5 4 2
Solution (a) We have
=
=
3 1 7 6 5 4 2
1 2 3 4 5 6 7
1 2 3 4 5 6 7
= (1 2 7 3)(4 6). The sign of a permutation of length k is
2 7 1 6 5 4 3
(−1)k−1 , so σ((1 2 7 3)(4 6)) = σ((1 2 7 3))σ((4 6)) = (−1)3 · (−1)1 = 1. Hence
this permutation is even. Alternatively, we could count the number of inversions
1 2 3 4 5 6 7
in
, which is 12, to deduce that this permutation is even.
2 7 1 6 5 4 3
α(1) α(2) . . . α(n)
(b) Swapping the rows of α gives us β =
and (by stack1
2
...
n
ing αand β) we have
α ◦β = e and β ◦ α = e where
e is the identity permutation
1 2 ... n
α(1) α(2) . . . α(n)
e =
=
. Permuting the columns of a
1 2 ... n
α(1) α(2) . . . α(n)
permutation in 2-row form doesn’t change the mapping, so we can then (optionally, I guess) permute the columns so that the first row of β becomes (1 2 . . . n).
This is the 2-row form of α−1 .
2. A deck of 52 playing cards is shuffled repeatedly using the following procedure:
Swap the top two cards, then cut the deck by moving the top 26 cards to the bottom.
How many of these shuffles must be performed before the cards are back in their
original order?
Solution We represent these shuffling operations using permutations in S52 , identifying the kth card from the top of the deck with the number k ∈ {1, 2, . . . , 52}.
Swapping the first two cards is then represented by the permutation (1 2).
Cutting the deck in the manner described is represented by
(1 27)(2 28)(3 29) . . . (26 52),
which is a product of disjoint transpositions. So the net effect of one shuffle, first
swapping and then cutting, is
α = (1 27)(2 28)(3 29) . . . (26 52)(1 2) = (1 27)(2 28)(1 2)(3 29)(4 30) . . . (26 52)
= (1 28 2 27)(3 29)(4 30) . . . (26 52).
The latter expression is a product of disjoint cycles, so we can compute powers of
α easily:
α2 = (1 28 2 27)2 (3 29)2 (4 30)2 . . . (26 52)2 = (1 2)(27 28),
α3 = (1 28 2 27)3 (3 29)3 (4 30)3 . . . (26 52)3 = (1 27 2 28)(3 29)(4 30) . . . (26 52),
α4 = (1 28 2 27)4 (3 29)4 (4 30)4 . . . (26 52)4 = e.
So it takes four of these shuffles to return the cards to their original order.
3. Let n, k ∈ N, let a1 , a2 , . . . , ak be distinct elements of {1, 2, . . . , n}, and consider
the cycle α = (a1 a2 . . . ak ) ∈ Sn . Prove that if β ∈ Sn then
β ◦ α ◦ β −1 = (b1 b2 . . . bk )
where bi = β(ai ) for 1 ≤ i ≤ k.
Solution Let γ = (b1 b2 . . . bk ). If 1 ≤ i < k then
(β ◦ α ◦ β −1 )(bi ) = (β ◦ α)(β −1 (β(ai ))) = β(α(ai )) = β(ai+1 ) = bi+1 = γ(bi ),
and
(β ◦ α ◦ β −1 )(bk ) = (β ◦ α)(β −1 (β(ak ))) = β(α(ak )) = β(a1 ) = b1 = γ(bk ),
and if x 6∈ {b1 , b2 , . . . , bk } then β −1 (x) 6∈ {a1 , a2 , . . . , ak } [why?], so α fixes β −1 (x),
meaning that α(β −1 (x)) = β −1 (x). Hence
(β ◦ α ◦ β −1 )(x) = (β ◦ α)(β −1 (x)) = β(α(β −1 (x))) = β(β −1 (x))x = γ(x).
So, for every x ∈ {1, 2, . . . , n}, we have (β◦α◦β −1 )(x) = γ(x). Hence β◦α◦β −1 = γ.
4. Let n ∈ N.
(a) Show that if α ∈ Sn , then α is an even permutation if and only if α−1 is an
even permutation.
(b) Show that if α, β ∈ Sn then σ(β ◦ α ◦ β −1 ) = σ(α).
Solution (a) We have σ(α)σ(α−1 ) = σ(α ◦ α−1 ) = α(e) = 1. Since σ(α) and
σ(α−1 ) are in {1, −1}, this implies that σ(α) = σ(α−1 ). So if α is even, then α−1
is even. And if α−1 is even then α is even.
(b) We have σ(β) = σ(β −1 ) by (a), and σ(β)2 = 1 since σ(β) ∈ {1, −1}. So
σ(β ◦ α ◦ β −1 ) = σ(β)σ(α)σ(β −1 ) = σ(β)2 σ(α) = σ(α).
2
5. Let n ∈ N,
(a) Let An = {α ∈ Sn : α is an even permutation}. Prove that (An , composition)
is a group. [Hint: use Exercise 4(a).]
(b) Let Bn = {β ∈ Sn : β is an odd permutation}. Is Bn a group under composition?
(c) Suppose that n ≥ 2. Show that the mapping
T : An → Bn ,
α 7→ α ◦ (1 2)
is a bijection. [Along the way, you should prove that T is well-defined by
checking that if α ∈ An then T (α) ∈ Bn .]
(d) Deduce that if n ≥ 2, then |An | = |Bn |, and hence compute the order of An .
What happens when n = 1?
Solution (a) If α, β ∈ An then α ◦ β ∈ An since σ(α ◦ β) = σ(α)σ(β) = 1 · 1 = 1.
So ◦ is an operation on An . Moreover, the identity permutation e is even, so
e ∈ An and e is clearly an identity element for (An , ◦). If α ∈ An then α−1 ∈ An
by Exercise 4(a). So (An , ◦) is a group.
(b) If n = 1 then Bn = ∅ so there is no identity element and Bn is not a group under
any operation on Bn . [Actually, there’s only one possible operation on ∅, namely
the “empty mapping” ∅ → ∅.] If n ≥ 2 then (1 2) ∈ Bn but (1 2)(1 2) = e 6∈ Bn ,
so composition is not an operation on Bn . So Bn is certainly not a group under
composition.
(c) If α ∈ An then σ(T (α)) = σ(α ◦ (1 2)) = σ(α) · σ((1 2)) = 1 · (−1) = −1, so
T (α) ∈ Bn . If T (α1 ) = T (α2 ) then since (1 2) ◦ (1 2) = e, we have
α1 = α1 ◦ (1 2) ◦ (1 2) = T (α1 ) ◦ (1 2) = T (α2 ) ◦ (1 2) = α2 ◦ (1 2) ◦ (1 2) = α2 .
So T is injective. If β ∈ Bn then σ(β ◦ (1 2)) = σ(β) · σ((1 2)) = (−1) · (−1) = 1,
so β ◦ (1 2) ∈ An and T (β ◦ (1 2)) = β. So T is surjective. Hence T is bijective.
(d) If n ≥ 2, then since there is a bijection from An → Bn , we have |An | = |Bn |.
Moreover, Sn = An ∪ Bn and An ∩ Bn = ∅. So n! = |Sn | = |An | + |Bn | = 2|An |, so
|An | = 21 n!.
If n = 1 then A1 = {e} = S1 so |A1 | = 1.
6. Let n, k ∈ N with k ≥ 2, let a1 , a2 , . . . , ak be distinct elements of {1, 2, . . . , n} and
let α be the cycle in Sn given by α = (a1 a2 . . . ak ). Recall that α2 is defined to
be α ◦ α. Show that if k is even then α2 is the product of two disjoint cycles.
What happens if k is odd? [Justify your answer].
3
Solution If k is even then
α2 : a1 7→ a3 7→ a5 7→ . . . 7→ ak−1 7→ a1
and
α2 : a2 7→ a4 7→ . . . 7→ ak 7→ a2 .
Hence α2 = (a1 a3 . . . ak−1 )(a2 a4 . . . ak ), and these two cycles are plainly disjoint.
On the other hand, if k is odd then
α2 : a1 7→ a3 7→ . . . 7→ ak 7→ a2 7→ a4 7→ . . . 7→ ak−1 7→ a1 .
So α2 = (a1 a3 . . . ak a2 a4 . . . ak−1 ) is a cycle.
4