Question

Prove that a group of even order must have an element of order 2

Answer 1

Pair up if possible each element of G with its inverse, and observe that

g2≠e⟺g≠g−1⟺there exists the pair(g,g−1)

Now, there is one element that has no pairing: the unit e (since indeed e=e−1⟺e2=e), so since the number of elements of G is even there must be at least one element more, say e≠a∈G , without a pairing, and thus a=a−1⟺a2=e