If G be a group of even order , prove that G contains an odd number of elements of order 2 .

 Solution : Since G is a group of even order , Clearly 2 divides o(G).

By Cauchy’s theorem we can easily conclude that G has an element of order 2 .

But our aim is to show that the number of elements of order 2 in G is odd .

Keeping this in mind , let us construct two sets S and T as

S = {a ϵ G : a ≠ a^-1}  and T = {a ϵ G : a = a^-1} 

Let x ϵ S . Then x ≠ x^-1 => x^-1  ≠ [ x^-1]^-1 in G and this shows that  x^-1 ϵ S . Thus x and x^-1 forms a pair in S such that each is the inverse of other . Therefore the number of elements in S is even .        

Since , o(G) = |S| + |T| and o(G) is even, it follows that the number of elements in T is even .                                                            …→(1)  

 a ϵ T => a = a^-1  => a² = e_G  implies either  a = e_G or o(a) = 2 .

This shows that T contains the identity element of G and all the elements of order 2 of G .                                                          ...→(2)

From (1) and (2) we can conclude that the number of elements of order 2 in G is odd .