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

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

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

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

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

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

2. ## 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.

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

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.