Group Theory: Element of Order 2 in Groups of Even Order

[PsychonautQQ]

Homework Statement

If G is a group of even order, show that it has an element g not equal to the identity such that g^2 = 1.

Homework Equations

None

The Attempt at a Solution

What I wrote:

If |G| = n, then g^n = 1 for some g in G. Thus, (g^(n/2))(g^(n/2)) = 1, so g^(n/2) is the element of order 2.

Is this a flawed argument? there guaranteed to be an element such that g^n = 1?

 

[HallsofIvy]

The identity is, of course, its own inverse. Since the group is "of even order", i.e. it has an even number of elements, removing the identity leaves an odd number of elements. Pairing each number with its inverse, what happens?

 

[PsychonautQQ]

The identity is, of course, its own inverse. Since the group is "of even order", i.e. it has an even number of elements, removing the identity leaves an odd number of elements. Pairing each number with its inverse, what happens?

I see your arguement. Is this true though? Can't ab = 1, bc = 1, cd = 1 etc etc but ba not equal 1?

 

[Dick]

I see your arguement. Is this true though? Can't ab = 1, bc = 1, cd = 1 etc etc but ba not equal 1?

No, it's not possible. If ab=1 then ba=1. Prove it!