Prove that the set An of all even permutations of degree n is a group of order for the product of permutations.
Answers
Answer:
It is known that the order of the alternating group An of order n is n!2. In Herstein's Abstract Algebra, it is proved by the First Homomorphism Theorem. I tried to find an alternative proof which needs not using the group homomorphism. I think the following rules may be helpful:
The product of two even permutations is even.
The product of two odd permutations is even.
The product of an even permutation by an odd one (or of an odd one by an even one) is odd.
Intuitively, since the product of an odd(resp. even) permutation and a 1-cycle is even(resp. odd), a half of all the permutations should be even. Then we get n!2.
What's more, the theorem mentioned in this question may be related. I don't know if one can turn the argument above into a proof.