Proof Even Order Groups Have Element of Order 2

[clkt]

How do I proof that groups of an even order must have an element of order 2? I have a vague idea, but I don't know how to put my idea together.
Aside from identity, there are an odd number of elements in my group. So one element will not have a partner and will have to be multiplied by itself to cancel out. That element must have an order of 2 such that its square = identity. But how can I create the scenario where all elements have to pair up and cancel out? Thanks in advance.

 

[AKG]

Aside from identity, there are an odd number of elements in my group. So one element will not have a partner and will have to be multiplied by itself to cancel out. That element must have an order of 2 such that its square = identity.

This is correct. Although the wording is informal, I'd consider this an adequate proof.

But how can I create the scenario where all elements have to pair up and cancel out?

Huh?

 

[clkt]

I guess my question is, what is the property of a group that dictates that every element must have a "partner" to cancel out with?

 

[StatusX]

Every element in a group has an inverse. Although not specified explicitly, it's easy to show this inverse is unique, and that the inverse of the inverse is the original element, which allows you to form pairs like you did above.

 

[clkt]

ahhhh, thank you!

 

[mathwonk]

can you prove a group whose order is divisible by three has an element of order 3?