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I have to prove that a group of even order must contain a odd number of elements of order 2 and I know that if i take all the elements with order $>2$ that that subgroup has a even number of elements. But where do i go from here?

R.vW
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You know that there is an even number of elements with order $>2$; since your group is of even order, there must be an even number of elements with order $\leq 2$. But there is only one element with order $1$; thus there is an odd number of elements with order $2$.

Arnaud D.
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  • How do we now know that there must be an even number of order $\le2$? – R.vW Oct 01 '16 at 16:28
  • There is an even number of elements and an even number of elements with order $>2$. And every elements has order either $>2$ or $\leq 2$... – Arnaud D. Oct 01 '16 at 16:30
  • Does a group of even order always have an even number of elements? (Just to be sure) – R.vW Oct 01 '16 at 16:32
  • The order of a group is the number of elements... – Arnaud D. Oct 01 '16 at 16:36
  • In my book it says "... and the smallest positive integer m such that $x^m = e$ is called the order of x." Is this the same and if so what point might I be missing? – R.vW Oct 01 '16 at 16:42
  • That's the definition of order of an element in the group; the order of a (finite) group is defined as the number of elements. – Arnaud D. Oct 01 '16 at 16:46