1. Symmetric group
1.1. Groups. Let G be a set with a binary operation, i.e., a function G × G −→ G,
denoted as (g, h) 7−→ g · h. Assume that
(G1) the operation is associative, i.e.,
g · (h · k) = (g · h) · k
for all g, h, k ∈ G;
(G2) there exists an element 1 ∈ G, called the identity such that
g·1=1·g =g
for any g ∈ G;
(G3) for any g ∈ G there exists an element g −1 called the inverse of g such that
g · g −1 = g −1 · g = 1.
Then G is called a group and the binary operation is the group multiplication in G.
A group G is finite if G is a finite set. Otherwise, we say that the group is
infinite. The number of elements in G is called the order of G.
1.1. Example. Let R be the real numbers. With addition (a, b) 7−→ a + b as the
group operation it is a group.
1.2. Example. Let R∗ be the nonzero real numbers. With multiplication (a, b) 7−→
a · b as the group operation it is a group.
1.3. Example. Let R∗+ be the nonzero positive real numbers. With multiplication
(a, b) 7−→ a · b as the group operation it is a group.
1.4. Example. Let S be a set. Consider the family S of all bijections of S −→ S.
Since the composition of two bijections is a bijection, the composition (φ, ψ) 7−→
φ ◦ ψ defines a binary operation on the set S. The identity map is the identity for
this operation and the inverse of a bijection φ is the inverse map φ−1 . In this way,
S becomes a group.
Clearly, the group S is finite if and only if the set S is finite.
Let G and H be two groups and φ : G −→ H a map such that
φ(g · h) = φ(g) · φ(h)
for any g, h ∈ G. Then φ is called a group homomorphism.
If φ : G −→ H is a group homomorphism, we have
φ(g) = φ(g · 1) = φ(g) · φ(1).
Multiplying this equality by φ(g)−1 from the left, we get
1 = φ(g)−1 φ(g) = φ(g)−1 φ(g)φ(1) = φ(1),
i.e., φ maps the identity in G into the identity in H. This also implies that for any
g ∈ G we have
φ(g) · φ(g −1 ) = φ(g · g −1 ) = φ(1) = 1
and by multiplication from the left by φ(g)−1 we see that
φ(g −1 ) = φ(g)−1 ,
i.e., φ maps the inverse of g into the inverse of φ(g).
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2
A subgroup H of G is a subset of G which is closed under multiplication, contains
the identity and for any g ∈ H, g −1 is also in H. Clearly, the subgroup H with the
induced multiplication is also a group.
Let φ : G −→ H be a group homomorphism. Let
ker φ = {g ∈ G | φ(g) = 1}.
Then ker φ is a subgroup of G. We call it the kernel of φ.
1.5. Example. The map a 7−→ a2 is a homomorphism of the multiplicative group
R∗ of nonzero real numbers into the multiplicative group R∗+ of nonzero positive
numbers. The kernel of this homomorphism is the subgroup of R∗ consisting of ±1.
1.2. Symmetric group. Let n be a positive integer and Fn = {1, 2, 3, . . . , n}. A
bijection σ : Fn −→ Fn is called a permutation. As we remarked before, the set
of all permutations with composition as a binary operation forms a group. This
group is denoted by Sn and called the symmetric group.
As we remarked before Sn is a finite group. More precisely, we have the following
result.
1.6. Lemma. The order of the group Sn is n!.
Proof. Clearly, if n = 1, the group S1 consists only of the identity. Hence, the
statement holds in this case.
Let H be the set of all permutations σ of Fn such that σ(n) = n. Then, by
restricting them to Fn−1 we get permutations of Fn−1 . Clearly, this map is a
bijection of H onto Sn−1 . Therefore, by the induction assumption H has the order
(n − 1)!.
Let si be a permutation of Fn such that si (i) = n, si (n) = i and si (k) = k for
all k different from i and n. Clearly, s2i = 1, for all 1 ≤ i ≤ n.
Consider the sets si · H in Sn . If si · H intersects sj · H, there would exists a
permutation σ equal to si · h and sj · h0 for some h, h0 ∈ H. Therefore, σ would
map n into i and into j. Hence, i would be equal to j and the sets would be equal.
Therefore, if i 6= j, the sets si · H and sj · H are disjoint.
If si · h = si · h0 , by multiplying with si from the left, we get h = h0 . Hence,
the map h 7−→ si · h is a bijection of H onto si · H. Therefore, si · H has (n − 1)!
elements. It follows that the union of all si · H has n · (n − 1)! = n! elements.
Let σ be an arbitrary permutation. Asssume that σ(n) = j. Then sj · σ is in H.
Therefore, σ is in sj · H.
Hence, Sn is s1 · H ∪ s2 · H ∪ · · · ∪ sn · H and its order is n!.
1.7. Example. Let n = 2. Then S2 has two elements. One of them is the identity,
the other is the map which maps 1 into 2, and 2 into 1. Clearly, this group is
commutative.
1.8. Example. Let n ≥ 3. Then Sn is not commutative. To see this, consider
the permutation σ defined by σ(1) = 2, σ(2) = 1 and σ(i) = i for i ≥ 3; and
permutation τ defined by τ (1) = 1, τ (2) = 3, τ (3) = 2 and τ (i) = i for i ≥ 4.
Then
(τ · σ)(1) = τ (σ(1)) = τ (2) = 3,
and
(σ · τ )(1) = σ(τ (1)) = σ(1) = 2.
Hence τ · σ 6= σ · τ , and the group Sn is not commutative.
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1.3. Parity. Let σ be a permutation. A pair (i, j) is called an inversion for σ if
i < j and σ(i) > σ(j).
If σ is the identity, the number of inversions is 0. On the other hand, if σ is a
permutation such that σ(i) = n − i, every pair (i, j) such that i < j is an inversion
and the number of inversions is 1 + 2 + · · · + n − 1 = n(n−1)
.
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We say that the permutation σ is even (resp. odd) if it contains an even (resp. odd)
number of inversions.
If σ is a permutation and I the number of its inversions, we define
(σ) = (−1)I .
Clearly is a function on Sn with values in ±1.
1.9. Theorem. Them map from the symmetric group Sn into the group {±1} of
two elements is a group homomorphism.
Consider the polynomial F (x1 , x2 , . . . , xn ) in variables x1 , x2 , . . . , xn given as the
product of factors xi − xj for all (i, j) such that i < j, i.e.,
Y
F (x1 , x2 , . . . , xn ) =
(xi − xj ).
i<j
Clearly, if we consider F (xσ(1) , xσ(2) , . . . , xσ(n) ) it is the product of xσ(i) − xσ(j)
for i < j. Therefore, this term appears in the product for F (x1 , x2 , . . . , xn ) if
σ(i) < σ(j). If not, its negative appears in this product. This immediately implies
that
F (xσ(1) , xσ(2) , . . . , xσ(n) ) = (σ)F (x1 , x2 , . . . , xn ).
Therefore,
F (xσ(1) , xσ(2) , . . . , xσ(n) )
.
(σ) =
F (x1 , x2 , . . . , xn )
Therefore, we have
F (xσ(τ (1)) , xσ(τ (2)) , . . . , xσ(τ (n)) )
F (x1 , x2 , . . . , xn )
F (xσ(τ (1)) , xσ(τ (2)) , . . . , xσ(τ (n)) ) F (xσ(1) , xσ(2) , . . . , xσ(n) )
=
F (xσ(1) , xσ(2) , . . . , xσ(n) )
F (x1 , x2 , . . . , xn )
(σ · τ ) =
=
F (xσ(τ (1)) , xσ(τ (2)) , . . . , xσ(τ (n)) )
(σ).
F (xσ(1) , xσ(2) , . . . , xσ(n) )
If we put yi = xσ(i) for 1 ≤ i ≤ n, we get
F (yτ (1) , yτ (2) , . . . , yτ (n) )
F (xσ(τ (1)) , xσ(τ (2)) , . . . , xσ(τ (n)) )
=
= (τ ).
F (xσ(1) , xσ(2) , . . . , xσ(n) )
F (y1 , y2 , . . . , yn )
It follows that
(σ · τ ) = (σ)(τ )
what proves the theorem.
This implies that the kernel An of is a subgroup of Sn . It consists of all even
permutations and it is called the alternating group.
If n = 1, we have S1 = A1 .
Assume that n > 1. Consider the permutation ti which switches i with i + 1 and
leaves all other elements of (1, 2, . . . , n) unchanged. Then the pair (i, i + 1) is the
only inversion in ti , i.e., (ti ) = −1.
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Clearly, the map φi : σ 7−→ σ · ti of Sn into itself satisfies φ2i = identity.
Therefore, φi is a bijection of Sn into itself. Since ti is odd, it maps an even
permutation into an odd permutation and vice versa. Therefore, the set of all odd
permutations is equal to φi (An ) = An · ti . It follows that
Sn = An ∪ An · ti
for any 1 ≤ i < n.
This implies the following result.
1.10. Lemma. Let n > 1. Then the order of An is
n!
2 .
A permutation τ is a transposition if there exist i < j such that τ (i) = j for
τ (j) = i and τ (k) = k for all other k.
1.11. Lemma. A transposition is an odd permutation.
Proof. Let τ be a transposition of i and j with i < j. If (k, l) are two indices such
that k < l and they are different from either i or j, τ (k) = k and τ (l) = l. So, (k, l)
is not an inversion for τ . If we consider the pair (i, l) for i < l, it is an inversion
for all l ≤ j. This gives j − i inversions. Moreover, the pair (k, j) for k < j is
an inversion for all k ≥ i. This gives also j − i inversions, but both sets contain
common pair (i, j). So the total number of inversions in τ is 2(j − i) − 1. Hence τ
is odd.
Next result follows from the proof of 1.6.
1.12. Lemma. Any permutation different from identity is a product of transpositions.
Proof. The proof is by induction in n. If n is 1 the statement is obvious. Assume
that n > 1. By the induction assumption, the statement holds for any permutation
in the subgroup H in Sn . since the permutations si are transpositions, the equality
of Sn with the union of all si · H, 1 ≤ i ≤ n, implies the result.
Clearly, a permutation σ can be written as a product of transpositions in a
number of different ways. If σ = t1 · t2 · · · tn is such product, we have
(σ) = (t1 )(t2 ) · (tn ) = (−1)n .
Hence, the parity of σ is equal to the parity of n.
This implies the following result.
1.13. Lemma. An odd permutation is a product of odd number of transpositions.
An even permutatioin is a product of even number of transpositions.