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

• Nov 8th 2012, 05:31 PM
x3bnm
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?
• Nov 8th 2012, 09:43 PM
Deveno
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.
• Nov 9th 2012, 09:12 AM
x3bnm
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.