Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-1 29
| n
Theorem 18. The set of all even permutations in S n is a normal subgroup. Further O(A n ) = 2
.
Proof. Let A n be the subset of S n consisting of all even permutations. Since
(i) the product of two even permutations is an even permutation.
(ii) the inverse of an even permutation is an even permutation,
it follows that A n is a subgroup of S n .
To prove that A n is a normal subgroup of S n , we proceed as follows :
Let W be the group of real numbers 1 and −1 under multiplication. Define
by
f : S n → W
f(α) = 1
if α is an even permutation
f(α) = −1 if α is odd permutation
Then it can be verified that f is homomorphism of S n of W. The kernel (null space) of f is given by
K = {α ∈ S n : f(α) = eW = 1 }
= {α ∈ S n : f(α) = 1}
= { α : α is even }
= A n .
Thus A n , being the kernel of a homomorphism is a normal subgroup of S n .
Moreover, by Isomorphism Theorem,
Therefore,
Sn
≅ W.
A
n
S
O(W) = O
A
=
But O(W) = 2, therefore,
or
2 =
n
n
O(Sn
)
O(A )
n
O(Sn
)
O(A )
O(S
)
O(A n ) = n
| n =
2 2
This completes the proof of the theorem.
n
Definition. The normal subgroup of A n formed by all even permutation in S n is called the Alternating
Group of degree n.
| n
We have shown above that order of A n is 2
.