Question

prove that in a group of even elements at least one element apart from the identity element must equal to its identity ​

Answer 1

Answer:Define an equivalence relation on the set G underlying your group by declaring x∼y iff either x=y or x=y−1 . This partitions G into equivalence classes of size 1 and 2, so the order of G is a sum of 1's and 2's. Since the equivalence class of the identity e has size 1 and the order of G is even, there must be another equivalence class of size 1. The element in that class is its own inverse, so its square is the identity.

(Cauchy's theorem also says that if a prime p divides the order of a group G then G has an element of order p , which is exactly what you're asking in the case p=2 . But Cauchy's theorem (and its stronger cousin the Sylow theorems) is overkill here.)

Explanation: