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

Let $\displaystyle G$ be a finite group, and let $\displaystyle S$ be the set of all the elements of $\displaystyle G$

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

If the order of $\displaystyle G$ is even, there is at least one element $\displaystyle x$ in $\displaystyle G$ such that $\displaystyle x \neq e$ (neutral element) and $\displaystyle x = x^{-1}$(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.