The Alternating Groups
The symmetric group Sn consists of all permutations of a set of n elements. Any set of n elements will do,
but we usually use the set
S = {1 , 2 , ..., n}.
The alternating group An is the group of even permutations in Sn . Our object is to prove
Theorem. If n ≥ 5, the alternating group An is a simple group.
This theorem supplies us with an infinite number of simple groups, of orders 12 n! = 60, 360, 2520, ... The
first two groups, A5 and A6 , appear also as P SL2 (F ). A4 is not a simple group.
We’ll use the customary convention for operating with permutations: A composition of functions is to be
read in the reverse of the usual order: f g means first apply f , then g. To make this work notationally, one
has to let the functions act on the right:
(i)f g = ((i)f )g.
The type t of a permutation p lists the lengths of the disjoint cycles making up p in increasing order, 1-cycles
being included. Thus the type of the permutation p = (56 )(923 )(71 ) in S9 is t = (1, 1, 2, 2, 3).
Lemma 1. The permutations of a given type t form one conjugacy class in the symmetric group Sn .
For example, p = (162 )(45 ) and p = (16 )(243 ) are conjugate elements of S6 , because they both have type
(1, 2, 3).
The proof of this lemma is not difficult, but some confusion among indices can be avoided by considering
permutations of two separate sets:
Lemma 2. Let p be a permutation of S of type t, and let α : S −→ S be a bijective map from S to another
set S .
(i) If p sends i → j , then α−1 pα sends (i)α → (j )α
(ii) q = α−1 pα is a permutation of S of type t.
(iii) For any permutation q of S of type t, there is a bijective map α : S −→ S such that q = α−1 pα.
Lemma 1 follows from Lemma 2 by setting S = S .
In this lemma, α−1 pα stands for composition of functions in the reverse order: first apply α−1 , then p, then
α. So if we denote (i)α by i , then (i) follows from the computation
(i )α−1 pα = (i)pα = (j )α = j .
Part (ii) of the lemma becomes clear when one thinks of α simply as an operation which renames the index
i as i = (i)α. To prove (iii), we write p and q as products of disjoint cycles, including 1-cycles, with the
lengths in increasing order. Then we define α to be the map which preserves this ordering of S and S . For
example, let S be the set {r , s, t , u, v , w }. Let p = (3 )(45 )(162 ), and q = (w )(u s)(r t v ). Then α sends
3 →
w, 4 → u, etc... �
Lemma 3. If n ≥ 5, the 3-cycles form a single conjugacy class in the alternating group An .
The 3-cycles form two conjugacy classes in A3 and in A4 .
Proof. Let p denote the cycle (123 ), and let q = (i j k ). Let τ denote the transposition (45 ). By Lemma
1, there is a permutation α such that q = α−1 pα. If α is odd, then τ α is even. We note that p = τ −1 pτ .
Therefore q = α−1 (τ −1 pτ )α = (τ α)−1 p(τ α). We replace α by τ α. Thus there always is an even permutation
α such that q = α−1 pα, which means that q is in the conjugacy class of p in the alternating group. �
1
The symmetric group Sn consists of all permutations of a set of n elements. Any set of n elements will do,
but we usually use the set
S = {1 , 2 , ..., n}.
The alternating group An is the group of even permutations in Sn . Our object is to prove
Theorem. If n ≥ 5, the alternating group An is a simple group.
This theorem supplies us with an infinite number of simple groups, of orders 12 n! = 60, 360, 2520, ... The
first two groups, A5 and A6 , appear also as P SL2 (F ). A4 is not a simple group.
We’ll use the customary convention for operating with permutations: A composition of functions is to be
read in the reverse of the usual order: f g means first apply f , then g. To make this work notationally, one
has to let the functions act on the right:
(i)f g = ((i)f )g.
The type t of a permutation p lists the lengths of the disjoint cycles making up p in increasing order, 1-cycles
being included. Thus the type of the permutation p = (56 )(923 )(71 ) in S9 is t = (1, 1, 2, 2, 3).
Lemma 1. The permutations of a given type t form one conjugacy class in the symmetric group Sn .
For example, p = (162 )(45 ) and p = (16 )(243 ) are conjugate elements of S6 , because they both have type
(1, 2, 3).
The proof of this lemma is not difficult, but some confusion among indices can be avoided by considering
permutations of two separate sets:
Lemma 2. Let p be a permutation of S of type t, and let α : S −→ S be a bijective map from S to another
set S .
(i) If p sends i → j , then α−1 pα sends (i)α → (j )α
(ii) q = α−1 pα is a permutation of S of type t.
(iii) For any permutation q of S of type t, there is a bijective map α : S −→ S such that q = α−1 pα.
Lemma 1 follows from Lemma 2 by setting S = S .
In this lemma, α−1 pα stands for composition of functions in the reverse order: first apply α−1 , then p, then
α. So if we denote (i)α by i , then (i) follows from the computation
(i )α−1 pα = (i)pα = (j )α = j .
Part (ii) of the lemma becomes clear when one thinks of α simply as an operation which renames the index
i as i = (i)α. To prove (iii), we write p and q as products of disjoint cycles, including 1-cycles, with the
lengths in increasing order. Then we define α to be the map which preserves this ordering of S and S . For
example, let S be the set {r , s, t , u, v , w }. Let p = (3 )(45 )(162 ), and q = (w )(u s)(r t v ). Then α sends
3 →
w, 4 → u, etc... �
Lemma 3. If n ≥ 5, the 3-cycles form a single conjugacy class in the alternating group An .
The 3-cycles form two conjugacy classes in A3 and in A4 .
Proof. Let p denote the cycle (123 ), and let q = (i j k ). Let τ denote the transposition (45 ). By Lemma
1, there is a permutation α such that q = α−1 pα. If α is odd, then τ α is even. We note that p = τ −1 pτ .
Therefore q = α−1 (τ −1 pτ )α = (τ α)−1 p(τ α). We replace α by τ α. Thus there always is an even permutation
α such that q = α−1 pα, which means that q is in the conjugacy class of p in the alternating group. �
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