You have seen that a permutation in Sn can be written as a product of transpositions. From E 10 you can see that the factors in the product are not uniquely determined. But all such representations have one thing in common – if a permutation in Sn is the product of an odd number of transposition in one such representation, then it will be a product of a odd number of transpositions in any such representation, similarly, if ∈ Sn is a product of an even number of transpositions in one representation, then f is a product of an even number of transpositions in any such representation. To see this fact we need the concept of the signature or sign function.
Definition: The signature of f ∈ Sn (n ≥ 2) is define to be Sign f =
n 1 j i,Π
= j i
f(i) f(j)
−
− i<j for example, for f = (1 2 3) ∈ S3,
sign f =
1 2
f(1) f(2)
−
−
1 3
f(1) f(3)
−
−
2 3
) 2 f(
) 3 f(
−
−
=
− 1 2
3
− 2 2
1
− 1 3
1 = 1.
Similarly, if f = (1 2), ∈ S3, then Sign f =
1 2
f(1) f(2)
−
−
1 3
f(1) f(3)
−
−
2 3
f(2) f(3)
−
−
=
− 1 2
1
− 2 2
3
− 1 1
3 = -1.
Henceforth, whenever we talk of sign f, we shall assume that f ∈ Sn for some n ≥ 2.
Try this simple exercise now.
SELF ASSISMENT EXERCISE11
What is the signature of 1 ∈ Sn?
Have you noticed that the signature define a function
Sign: Sn → Z? We will now show that this function is a homomorphism.
Theorem 4: Let f,g ∈Sn. Then sign (f ∘g) = (sign g).
Proof: By definition, Sign fog =
n 1 j i,Π
= j i
f(g(i)) f(g(j))
−
−
i<j
j
Πi,
g(i) g(j)
f(g(i)) f(g(j))
−
−
j
Πi, i j
g(i) g(j)
−
−
Now, as i and j take all possible pairs of distinct values from 1 to n, so do g(i) and g(j) since g is a bijection
∴ i,Π<j
g(i) g(j)
f(g(i)) f(g(j))
−
− = sign f.
∴ sign (fog) = (sign f) (sign g).
Now we will show that Im (sign) = {1, - 1}.
Theorem 5: a) If t ∈ Sn is a transposition, then sign t = - 1.
b) sign f = 1 or –1 ∀ f ∈ Sn.
c) Im (sing) = {1, - 1}.
Proof: a) Let t = (p q), where p < q.
Now, only one factor of sign t involves both p and q, namely, p
q t(p) t(q)
−
− = p q
q p
−
− = -1.
Every factor of signt that doesn’t contain p or q equals 1, since j
i t(j) t(i)
−
−
j i
j i
−
− = 1, if i, j≠p, q.
The remaining factors contain either p or q. but not both. These can be paired together to form one of the following products.
p i
t(p) t(i)
−
− q i
t(q) t(i)
−
− = p i
q i
−
− q i
p i
−
− = 1, if i> q,
p i
t(p) t(i)
−
− i q
t(i) t(q)
−
− = p i
q i
−
− i q
i p
−
− = 1, if q> i> p,
i p
t(i) t(p)
−
− i q
t(i) t(q)
−
− = i p
i q
−
− i q
i p
−
− = 1, if i< p.
Taking the values of all the factors of sign t, we see that sign t = -1.
b) Let f ∈ Sn. By Theorem 3 we know that f = t1t2 …..tr for some transpositions
t1, ….. tr in Sn
∴ sign f = sign (t1 t2 …. tr)
= (sign t1) (sign t2) ….. sign (tr), by Theorem 4.
= ( - 1)r, by (a) above.
∴sign f = 1 or – 1.
c) We know that Im (sign) ⊆ {1, - 1}.
We also know that sign t = -1, for any transposition t; and sign 1
= 1.
∴ {1, -1} ⊆ Im (sign).
∴Im (sign) = {1, - 1}
Now, we are in a position to prove what we said at the beginning of this section.
Theorem 6: Let f ∈ Sn and let f = t1t2 …..tr =t1t2 …ts
X two factorization of f into a product of transpositions. Then either both r and s are even integers, or both are odd integers.
Proof: We apply the function Sign: Sn → {1, - 1} to f = t1t2 …..tr. By Theorem 5 we see that
Sign f = (sign t1) (sign t2) ….. (sign t1) = (-1)r.
∴sign (t1t2 …..ts) = ( - 1) r
, substituting(t1t2 …..ts for f.
that is, ( - 1)s = ( - 1)r.
This can only happen if both s and r are even, or both are odd.
So, we have shown that for f ∈ Sn, the number of factors occurring in any factorization of f into transpositions is always even or always odd.
Therefore, the following definition is meaningful.
Definition: A permutation f ∈ Sn is called even if it can be written as a product of an even number of transposition. f is called off if it can be represented as a product of an odd number of transpositions.
For example, (1 2) ∈ s3 is an odd permutation. In fact, any transposition is an odd permutation. On the other hand, any 3-cycle is an even permutation, since (i jk) (i k) (i j).
Now, see if you’ve understood what odd and even permutations are SELF ASSISMENT EXERCISE 12
Which of the permutation in E 8 and 9 are odd?
SELF ASSISMENT EXERCISE 13
If f,g ∈ Sn are odd, then is f∘g odd too?
SELF ASSISMENT EXERCISE 14
Is the identity permutation odd or even?
Now, we define an important subset of Sn namely, An = {f ∈ Sn| f is even}.
We’ll sow that An ∆Sn, and that o(An) = 2
n!, for n ≥ 2.
Theorem 7: The set An, of even permutations in Sn, forms a normal subgroup of Sn of order
2 n!
Proof: Consider the signature function, Sign: Sn → {1, - 1}.
Note that {1, - 1} is a group with respect to multiplication. Now Theorem 4 says that sign is a group homomorphism and Theorem 5 says that Im (sign) = {1, - 1}. Let us obtain Ker (sign).
Ker (sign) = {f ∈Sn | sign f = 1}
= {f ∈ Sn | f is even}
= An.
∴An ∆Sn.
Further, by the Fundamental Theorem of Homomorphism Sn/An ≃ {1, - 1}.
∴o(Sn/An) = 2, that is, ) o(A
) o(S
n
n = 2.
∴ o(An) = 2
) o(Sn
= 2 n!.
Note that this theorem says that the number of even permutations in Sn equals the number of odd permutations in Sn.
Theorem 7 leads us to the following definition.
Definition: An, the group of even permutations in Sn, is called the alternating group of degree n.
Let us look at an example that you already seen in previous units, A3. Now, Theorem 7 says that o(A3) =
2
! 3 = 3.
Since (1 2 3) = (1 3) (1 2), (1 2 3) ∈ A3. Similarly, (1 3 2) ∈ A3. Of course, 1 ∈ A3.
∴ A3 = {1, (1 2 3) = (1 3 2)}.
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 (i1i2….ir) = (i1ir) (i1ir-1)
…… (i1i2), a product of (r – 1) transpositions. Use this fact to do the following exercise.
SELF ASSISMENT EXERCISE 15
Write down all the elements of A4.
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 A4 looks like, we are in a position to illustrate this statement.
We will show that A4 has no subgroup of order 6, even though 6 | o (A4).
Suppose such a subgroup H exists. Then o(H) = 6, o(A4) = 12. ∴|A4 : H| = 2. ∴ H ∆A4 (see Theorem 3 unit 5). Now A4/H is a group of order 2. Therefore, by E 8 of unit 4,
(Hg)2 = H ∀ g ∈ A4. (Remember H is the identity of A4/H.)
∴ g2 ∈ H ∀ g ∈ A4.
Now, (1 2 3) ∈ A4 ∴ (1 2 3)2 = (1 3 2) ∈ H.
Similarly, (1 3 2)2 = (1 2 3) ∈H. By the same reasoning (1 4 2), (1 2 4) (1 4 3) (1 3 4), (2 3 4), (2 4 3) are also distinct element of H. of course, 1
∈H.
Thus, H contains at least 9 elements.
∴o(H) ≥ 9. This contradicts our assumption that o(H) = 6.
Therefore, A4 has no subgroup of order 6.
We use A4 to provide another example too. (See how useful A4 is!) In Unit 5 we’d said that if H ∆N and N∆G, then H need not be normal in G. well, here’s the example.
Consider the subset Y4 = {1(1 2) (3 4), (1 4) (2 3), (1 3) (2 4)) of A4. SELF ASSISMENT EXERCISE 16
Check that (V4, ∘) is a normal subgroup of A4.
Now, let H = {1,(1 2) (3 4)}. Then H is a subgroup of index 2 in V4. ∴ H ∆V4.
So, H ∆V4, V4 ∆A4. But H∆A4. why? Well, (1 2 3) ∈ A4 is such that (1 2 3)-1 (1 2) (3 4) (1 2 3) = (1 3 (2 4) ∉ H.
And now let us see why permutation groups are so important in group theory.