Page 104 - DMTH403_ABSTRACT_ALGEBRA

Unit 8: Permutation Groups

Theorem 6: The set A,, of even permutations in S,, forms a normal subgroup of S , of order n! . Notes
2
n
Proof: Consider the signature function,
sign : S  (1, –1).
n
Note that (1, –1) is a group with respect to multiplication. Now, Im (sign) = (1, –1). Let us obtain
Ker (sign).
Ker (sign) = { f  S | sign f = 1 }
n
= (f  S | f is even)
n
= A.

 A  S .
n
Further, by the Fundamental Theorem of Homomorphism

S /A  (1, –1).
n
n
o(S )
n
 o(S /A ) = 2, that is, o(A )  2.
n
n
n
o(S ) n!
 o(An) = n  .
2 2
Note that this theorem says that the number of even permutations in S, equals the number of
odd permutations in S,.
Theorem 6 leads us to the following definition.

Definition: A,, the group of even permutations in S , is called the alternating group of
n
degree n.
Let us look at an example that you have already seen in previous units, A . Now, Theorem 6 says
3
3!
that o(A ) = 2  3. Since (1 2 3) = (1 3) (1 2), (1 2 3)  A . Similarly,
3
3
(1 3 2)  A . Of course, I  A . 3
3
 A = {I, (1 2 3), (1 3 2)).
3
A fact that we have used in the example above is that an r-cycle is odd if r is even, and even if r
is odd. This is because (i i .... i ,) = (i i,) (i i ) . . . . . . (i i ), a product of (r – 1) transpositions.
r
2
1 2
1
1
r-1
1
Now, for a moment, let us go back to Unit 4 and Lagrange’s theorem. This theorem says that the
order of the subgroup of a finite group divides the order of the group. We also said that if
n |o(G), then G need not have a subgroup of order n. Now that you know what A looks like, we
4
are in a position to illustrate this statement.
We will show that A has no subgroup of order 6, even though 6 | o (A ). Suppose such a
4
4
subgroup H exists. Then o(H) = 6, o (A ) = 12.  (A : H | = 2.  H  A4 (see Theorem 3, Unit 5).
4
4
Now, A /H is a group of order 2.
4
(Hg) = H  g  A . (Remember H is the identity of A /H.)
2
4
4
2
 g  H  g  A .
4
Now, (1 2 3) E A .  (1 2 3)* = (1 3 2)  H.
4
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