# Group Theory and Order of Elements

• Jul 2nd 2011, 11:26 AM
Kopeck
Group Theory and Order of Elements
Hello. I have a question about the order of elements in a group:

let G be a group of order 2n. Show that the number of elements in G with order 2 is odd.

Here is what I have done so far:

Let H be the subset of G consisting of the identity and all the elements of G with order 2. Then H is a subgroup of G and so, by Lagrange's Theorem, the order of H divides the order of G.

My strategy was to use this to arrive at the order of H is even, completing the proof. But I am stuck at this point.
• Jul 2nd 2011, 11:41 AM
Isomorphism
Re: Group Theory and Order of Elements
Hint: Prove that $\displaystyle |\{ a \in G | a \neq a^{-1} \}|$ is even. Check the parity of what remains, and notice that $\displaystyle e \notin \{ a \in G | a \neq a^{-1} \}$
• Jul 2nd 2011, 12:21 PM
Kopeck
Re: Group Theory and Order of Elements
Oh my goodness. This makes it very clear. Thank you for the hint Isomorphism.
• Jul 31st 2013, 07:27 PM
phys251
Re: Group Theory and Order of Elements
Hey, I'm trying to work this exact problem right now. I do not understand the conclusion of post #2. Why is the parity of the set of elements x such that x^2 is not e, necessarily even? This is where I'm stuck.
• Jul 31st 2013, 08:33 PM
Deveno
Re: Group Theory and Order of Elements
Don't inverse-pairs occur in pairs if an element is NOT its own inverse?
• Jul 31st 2013, 08:39 PM
phys251
Re: Group Theory and Order of Elements
Quote:

Originally Posted by Deveno
Don't inverse-pairs occur in pairs if an element is NOT its own inverse?

Huh?
• Jul 31st 2013, 09:17 PM
Deveno
Re: Group Theory and Order of Elements
Ok. Imagine this: you start with a finite group, G. First you take out all the elements that equal their own inverses (this will include the identity element).

Now, for any element g that is left, there is some OTHER element $\displaystyle g^{-1}$ (which is, by assumption, NOT g) also in the leftovers. So take them out as a pair. Keep doing this.

One of two things HAS to happen:

1) We wind up with a single element left over when we're done.

2) Everything comes out in pairs, so we wind up with NO elements left over.

Convince yourself that if #1 happens, that remaining element must be its own inverse, a contradiction, so #2 must occur, the number of elements who are NOT their own inverses must be EVEN.