MATH 3320, SPRING 2012:
EVEN AND ODD PERMUTATIONS
AND THE ALTERNATING GROUP An
PR HEWITT
In chapter 6 we define the symmetric group Sn to be the group of all permutations
of the set {1, 2, . . . , n}. Recall that a permutation of a set X is simply a bijection
from X to itself. Thus Sn is a group under composition.
If f ∈ Sn then we have already encountered its representation in permutation
notation:
1
2
···
n
f=
f (1) f (2) · · · f (n)
Since we can choose f (1) arbitrarily; then choose f (2) independently so long
as f (2) 6= f (1); and so on we find that there are exactly n! permutations of
{1, 2, . . . , n}. That is, |Sn | = n!.
We next encounter a more efficient notation for permutations. First of all a
k-cycle permutes a certain k points “cyclically”
f : x1 → x2 → x3 → · · · xk → x1
but fixes every other point. We abbreviate such a k-cycle f as (x1 x2 · · · xk ). Two
cycles (x1 x2 · · · xk ) and (y1 y2 · · · y` ) are said to be disjoint in case {x1 , x2 , . . . , xk }∩
{y1 , y2 , . . . , y` } = ∅. In this case one can prove that the cycles commute. In
chapter 6 we learned how to express any permutation as a product of disjoint
cycles. Note that fixed points of a permutation are 1-cycles. We don’t write them
in our cycle notation but they are there, nonetheless.
The 2-cycles are especially useful, and they get a special name: transpositions.
Chapter 6 contains a proof of the following important theorem.
Theorem 1. Sn is generated by its transpositions.
Proof. In other words we have to prove that if f ∈ Sn then there are transpositions
t1 , . . . , tk ∈ Sn such that f = t1 · · · tk . In the text this is proved by showing
explicitly how to express a k-cycle as a product of k − 1 transpositions, and then
using the fact that every permutation is a product of disjoint cycles. This is an
important proof and you should make sure that you understand it.
We take a different approach — we will take this opportunity to prove this by
induction on n. If n = 1 then Sn = {e} and the identity is the empty product (of
anything!). If this seems a bit funny then consider the case n = 2:
S2 = {e, (12)} = h(12)i.
Now suppose that n > 2. There are two cases:
Date: 20 February 2012.
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2
PR HEWITT
Case 1: f (n) = n. In this case we may regard f ∈ Sn−1 . By induction there are
transpositions t1 , . . . , tk ∈ Sn−1 such that f = t1 · · · tk . Since these transpositions
are also in Sn we have the required decomposition of f .
Case 2: f (n) = k < n. In this case consider g = (kn)f . We compute that
g(n) = (kn)(f (n)) = (kn)(k) = n.
This g ∈ Sn−1 and as above we have that
g = (kn)f = t1 · · · tk
for some transpositions tj . Thus
f = (kn)g = (kn)t1 · · · tk ,
as required.
A given permutation can be expressed as a product of transpositions in many
ways. In fact we have already seen an example of this when we were studying
the dihedral group D3 : (23)(12)(23) = (13). Exercise 2 below investigates another
example of this.
What is uniquely determined is the parity of the number of transpositions. In
other words a given permutation f is either a product of an even number of transpositions or a product of an odd number of transpositions, but never both. Somewhat
more precisely we have the following theorem.
Theorem 2. Suppose f = t1 · · · tk ∈ Sn , where each of the tj is a transposition. If
f is the product of m disjoint cycles (including fixed points) then
k ≡ n + m mod 2.
The proof we give is based on that given in Fraleigh’s A First Course in Abstract
Algebra, 7th edition. This is an excellent book and it is worth comparing the way
Fraleigh does things to the way Armstrong does.
Fraleigh uses the following lemma. It is instructive to look at the diagrams he
draws to illustrate the proof.
Lemma 3. If t is any transposition and if g is a product of p disjoint cycles
(including fixed points) then the number of disjoint cycles tg is p ± 1.
Proof. Say t = (ab). There are two cases.
Case 1: a and b are in the same cycle for g. Let’s say that
g = (ax1 · · · xk by1 · · · y` ) · · ·
(Possibly k or ` or both equal 0.) In this case we compute that
tg = (ax1 · · · xk )(by1 · · · y` ) · · ·
That is, tg has exactly p + 1 disjoint cycles.
Case 2: a and b are in different cycles for g. Let’s say that
g = (ax1 · · · xk )(by1 · · · y` ) · · ·
(Again, possibly k or ` or both equal 0.) In this case we compute that
tg = (ax1 · · · xk by1 · · · y` ) · · ·
That is, tg has exactly p − 1 disjoint cycles.
EVEN AND ODD PERMUTATIONS
3
Proof of theorem 2. We can apply lemma 3 k times, starting with g = e, to the
relation
f = t1 · · · tk = t1 · · · tk e.
Since the identity has exactly n disjoint cycles (all of them fixed points!) and since
m − 1 ≡ m + 1 mod 2 we find that
m ≡ n + k mod 2
n + m ≡ 2n + k ≡ k mod 2.
Corollary 4. The product of two even or two odd permutations is even. The
product of an even and an odd permutation, in either order, is odd.
We let An denote the set of even permutations. You will be asked to us the
corollary to prove that An is a subgroup.
Theorem 5. |An | = n!/2.
Proof. We have seen that An ∩ (Sn − An ) = ∅. We establish a bijection between An
and Sn − An . Define φ(f ) = (12)f . This is a bijection Sn → Sn , since φ ◦ φ is the
identity on Sn . Since (12) is odd we find that φ(An ) ⊂ Sn − An and φ(Sn − An ) ⊂
An . Hence φ establishes the required bijection.
Exercises. Turn these in Monday, 27 February as assignment 7.
(1) Prove that two cycles commute if and only if they are disjoint.
Update: The “only if” part of this statement is false. Find a counterexample!
(2) Work through theorem (1) with the example f = (x1 x2 · · · xk ). What
factorization of f into transpositions do you obtain? Is it the same one
given in the text?
(3) Prove by induction that An is generated by its 3-cycles.
(4) Prove the corollary.
(5) Use the corollary to prove that An is a subgroup of Sn . Hint: Apply
exercise 5 from chapter 5.