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I am working on the following problem from group theory:

If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.

That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$.

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    @joriki: Indeed. Of course, even more is true: Frobenius's Theorem gives that the number of solutions to $x^n=1$ in$a$finite group $G$ is always$a$multiple of $\gcd(n,|G|)$. For $n=2$, the theorem implies the case of $|G|$ even, while Lagrange implies the case of $|G|$ odd.2011-05-30

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Hint. Define an equivalence relation on $G$ by $x\sim y$ if and only if $x=y$ or $x=y^{-1}$. Then remember that an equivalence relation partitions a set.

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    @DavidHolden As one doesn't need anti-isomorphism, it suffices to note this is an involution and count fixpoints2015-06-20