The collection of elements of the symmetric group which are made by multiplying together an even number of permutations forms a subgroup of $S_{n}$ .

This proves that the alternating_group $A_{n}$ is well-defined, if it is given as "the subgroup of $S_{n}$ containing precisely that which is made by multiplying together an even number of transpositions".

Proof

Firstly we must check that "I can only be made by multiplying together an even number of transpositions" is a well-defined notion; this is in fact true.

We must check the group axioms.

Identity: the identity is simply the product of no transpositions, and $0$ is even.

Associativity is inherited from $S_{n}$ .

Closure: if we multiply together an even number of transpositions, and then a further even number of transpositions, we obtain an even number of transpositions.

Inverses: if $σ$ is made of an even number of transpositions, say $τ_{1} τ_{2} \dots τ_{m}$ , then its inverse is $τ_{m} τ_{m - 1} \dots τ_{1}$ , since a transposition is its own inverse.