A finite group of even order has an odd number of elements of order 2 [duplicate]

Solution 1:

Let $A$ be the set of all elements of order greater than $2$, and recall that $x$ and $x^{-1}$ have the same order. So convince yourself $$ A=\bigcup_{x\in A}\{x,x^{-1}\}. $$

Conclude that $|A|$ is even. Now why does that imply that there are an odd number of elements of order $2$?

Solution 2:

now in the group identity element have order one .thus we are left with odd number of elements.also for each element of the group has a unique inverse. so only even number of elements will covered by the elements which have different inverses from themselves.and we know odd minus even is odd thus we are left with odd number of elements which are the inverses of themselvesi.e which have order two.