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Theorem. Every permutation is a product of transpositions. In other words, every permutation α ∈ S_{n}, n > 1, is a product of transpositions or 2-cycles.
Proof.
By a previous result, every permutation on a finite set can be written or expressed as a product of disjoint cycles, that is, in the form, (a_{1}a_{2}…a_{k})(b_{1}b_{2}…b_{t})…(c_{1}c_{2}…c_{s}).
However, every permutation is a product of cycles: (a_{1}a_{2}...a_{n}) = (a_{1}a_{n})(a_{1}a_{n-1})...(a_{1}a_{3})(a_{1}a_{2})
And therefore, (a_{1}a_{2}…a_{k})(b_{1}b_{2}…b_{t})…(c_{1}c_{2}…c_{s}) = (a_{1}a_{k})(a_{1}a_{k-1})…(a_{1}a_{2})(b_{1}b_{t})(b_{1}b_{t-1})…(b_{1}b_{2})(c_{1}c_{s})(c_{1}c_{s-1})…(c_{1}c_{2})∎
Examples:
Definition. A permutation that can be written or expressed as a product of an even number of transpositions or 2-cycles is called an even permutation. A permutation that can be written or expressed as a product of an odd number of transpositions or 2-cycles is called an odd permutation.
Lemma. The identity permutation is an even permutation. An even permutation can be obtained as the composition of an even and only an even number of transpositions. More formally, if β_{1}β_{2}...β_{n} = e where the β's are transpositions or 2-cycles, then n is even (n % 2 = 0).
Proof.
β^{n-1}β^{n} can be expressed in one of the following forms:
If we are in the first case, e = (ab)(ab) ⇒ β^{1}β^{2}…β^{n-2} = e. By induction hypothesis, n-2 < n, and therefore n-2 is even ⇒ n is even.
In the other cases, we replace the form of β^{n-1}β^{n} on the right by its counterpart on the left. We obtain a product of n transpositions where the rightmost occurrence of a is in the second place from the rightmost 2-cycle of the product. We repeat the process with β_{n-2}β_{n-1} and the rightmost occurrence of a is in the third 2-cycle from the right of the product. This means that each time you apply any of these identities, the rightmost occurrence of a is one transposition further to the left.
Eventually either we will get two identical adjacent transpositions (ax)(ax) = id, so they disappear and we can apply induction to the shorter string of transpositions; or we end up with an “a” only appearing in the leftmost transposition of the entire product of transpositions, but such a product does not fix a, so it is not the identity, ⊥.
Theorem. A permutation cannot be written or expressed as a product of both an odd and an even number of transpositions or 2-cycles..
Proof. Let α = β_{1}β_{2}…β_{r} = ζ_{1}ζ_{2}…ζ_{s}
e = αα^{-1} = ζ_{1}ζ_{2}…ζ_{s}β_{r}^{-1}…β_{2}^{-1}β_{1}^{-1} = [A transposition is its own inverse] ζ_{1}ζ_{2}…ζ_{s}β_{r}…β_{2}β_{1}. By the previous theorem, r+s is even ⇒ r and s are both even or both odd.
Theorem. The alternating group of degree n, A_{n}, is the subgroup of all even permutations in the symmetric group S_{n}, A_{n} = {σ ∈ S_{n} | σ is even} ≤ S_{n}
Proof.
Let τ = (1243)(67), then τ^{-1} = (76)(3421). Let σ = σ_{1}···σ_{l}, σ^{-1} = σ_{l}^{-1}···σ_{2}^{-1}σ_{1}^{-1} = σ_{l}···σ_{2}σ_{1}
Theorem. For n > 1, A_{n} has order n!/2, |A_{n}| = n/2.
Proof. Let A_{n} be the set of even permutations in S_{n}, and B_{n} be the set of odd permutations.
Let’s define a one-to-one function from A_{n} onto B_{n}, and therefore A_{n} and B_{n} has the same number of elements.
First, we define the function Φ: A_{n} → B_{n}, by Φ(α)=(12)α, α is even so it can be written or expressed as a product of an even number of transpositions, so (12)α ∈ B_{n}.
is Φ a one-to-one function? Φ(α) = Φ(β) ⇒ (12)α = (12)β ⇒ [S_{n} is a group, and by the Cancellation property] α = β.
is Φ onto? If α ∈ B_{n}, (12)^{-1}α = (12)α ∈ A_{n} ⇒ Φ((12)α) =(12)(12)α = α
Hence, S_{n} is split equally in both even and odd permutations, |S_{n}| = n! = 2|A_{n}| ⇒ |A_{n}| = n!/2.
Exercises.
σ: a_{1} → a_{2} → a_{3}··· a_{k} → a_{1}, then σ^{-1}: a_{k} → a_{k-1} → a_{k-2} ··· a_{1} → a_{k} which is nothing more than σ written backwards.