Proof:Finite group with even order has at least one element x with x = x^{-1}

Let be a finite group, and let be the set of all the elements of

The set can be divided up into pairs so that element is paired off with its own inverse. Prove that:

If the order of is even, there is at least one element in such that (neutral element) and (inverse element)

Is it possible to kindly tell me how I can prove this?

Re: Proof:Finite group with even order has at least one element x with x = x^{-1}

consider the set {g,g^{-1}}. this set has two elements if g ≠ g^{-1}, and one element if g = g^{-1}.

so the set of all elements of order > 2 (where g ≠ g^{-1}) has an even number of elements, for any finite group G, call it 2k.

the set of all elements of order 1 (which is one of the elements g for which g = g^{-1}, since e = e^{-1}) has just one element, e.

therefore, the number of elements of order 2 (that is elements g with g ≠ e, and g = g^{-1}) is:

|G| - 2k - 1.

if |G| is an even number, then |G| = 2m, and the number of elements of order 2 is:

2m - 2k - 1 = 2(m - k) - 1, which is an odd number.

in particular, since 0 is an even number, this number cannot be 0, so we have at least ONE element of order 2.

Re: Proof:Finite group with even order has at least one element x with x = x^{-1}

Quote:

Originally Posted by

**Deveno** consider the set {g,g^{-1}}. this set has two elements if g ≠ g^{-1}, and one element if g = g^{-1}.

so the set of all elements of order > 2 (where g ≠ g^{-1}) has an even number of elements, for any finite group G, call it 2k.

the set of all elements of order 1 (which is one of the elements g for which g = g^{-1}, since e = e^{-1}) has just one element, e.

therefore, the number of elements of order 2 (that is elements g with g ≠ e, and g = g^{-1}) is:

|G| - 2k - 1.

if |G| is an even number, then |G| = 2m, and the number of elements of order 2 is:

2m - 2k - 1 = 2(m - k) - 1, which is an odd number.

in particular, since 0 is an even number, this number cannot be 0, so we have at least ONE element of order 2.

Thank you Deveno. That's all I wanted to know.