Hint: Prove that is even. Check the parity of what remains, and notice that
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.
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 (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.