IDA: Proof

Proof

Suppose the permutation g can be written as the product c₁ ··· c_k of transpositions with k even, and suppose that g can also be written as the product of transpositions d₁ ··· d_m with m odd, then

e = c₁ ··· c_kd_m ··· d₁

expresses the identity as the product of an odd number of transpositions. We will show that this is impossible.

So assume the identity element e is a product of an odd number of transpositions t_i. We choose such a product

e = t₁ ··· t_2m+1

with m minimal. It is obvious that m > 0.

We may assume that t₁ = (1,2).

We may assume that there is some l > 0 with t₁ up to t_l all moving 1, i.e., all of the form t_i = (1,a_i), and that t_l+1 up to t_2m+1 all fix 1.

There is an index i with 2 < i l such that t_i = t₁.

Final contradiction.

If t₁ = (i,j), conjugate left-hand side and right-hand side by (1,i)(2,j).

Applying the formulas

(a,b)(1,c) = (1,c)(a,b) and (a,b)(1,b) = (1,a)(a,b),

where a, b and c are different numbers in {1, ..., n}, we can shift all transpositions which contain 1 to the front without violating the minimality of m.

Observe that we must have

t₁ ··· t_l(1) = 1.

Thus 2 = a₁ supp(t₂ ··· t_l), and at least one of the a_i (i > 1) is equal to a₁.

We have t_i = t₁ = t₁^-1, and, because of minimality of m, also t₂

t₁. Hence,

e = t₁ ··· t_{2m + 1} = t₁(t₂ ··· t_{i - 1})t₁^-1 t_{i + 1} ··· t_{2m + 1} = s₂ ··· s_{i - 1}t_{i + 1} ··· t_{2m + 1},

where s_j = t₁t_jt₁^-1, with 1 < j < i, is also a transposition. We have written e as a product of 2m - 1 transpositions. This contradicts the minimality of m.